Advanced Business Technical Math - cloned copy

3-D Printed Prosthetic Hand

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8red

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jgray — Wed, 31 Jul 2019 22:51:51 +0000

drone

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Area of a Triangle

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Triangular Painting Example

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Finding Angles of a Triangle

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Pythagorean Theorem

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Finding the Third Side of a Triangle

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Pythagorean Theorem Exercise

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Pythagorean Theorem Exercise for Leg b

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Finding the Perimeter of a Rectangle

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Checking the Perimeter of a Rectangle

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Finding the Lengths of a Triangle's Three Sides

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Perimeter of a Triangle

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Soccer Field

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Soccer Field Diagram

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Diagram of a Gazebo

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Solving a Pythagorean Application

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House with Ladder

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Sailboat with String of Lights

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Coins

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Biking from Riverside Park to the Beach

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Where Carina and Her Brother Will Meet When Driving

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Drivers Going in Opposite Directions

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Comparing Two Speeds

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Cruz's Triathlon Training

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Graph of an Inequality

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Number Line Showing an Inclusive Inequality

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Interpreting Number Lines

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[-3, infinity)

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(Negative Infinity, 2.5)

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(Negative Infinity, -3/5]

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Internal Between 2 and 5, Exclusive

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-3 < x < 4

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-6 ≤ x < -1

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0 ≤ x ≤ 2.5

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(Negative Infinity, 9/8]

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(Negative Infinity, 6)

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(-25, Infinity)

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(Negative Infinity, -5]

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(Negative Infinity, -16]

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[-17/5, Infinity)

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(2, Infinity)

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All Real Numbers

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Number Line

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[75, Infinity)

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Intersection

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x > -3 and x ≤ 2

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x < 2 and x ≥ -3

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x ≤ 1/2 and x < 4

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x ≤ 1/2 and x < 4

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x ≥ 6 and x ≤ 1

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Empty Number Line

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[1, 5)

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x ≥ 2, x ≤ -3/2

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x ≥ 2 or x ≤ -3/2

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x≤ 21/2, x ≥ -12

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x ≤ 21/2 or x ≥ -12

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Absolute Value

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Absolute Value

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Absolute Value

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(-7, 7)

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(-7, 7)test

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[2/5, 2]

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Absolute Value

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Absolute Value of x ≥ 5

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Absolute Value of u ≥ a

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|x| > 4

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|x| > 4 Test

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|2x -3| ≥ 5

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Pictograph not Aligned and Different Size

kmoshenko — Fri, 02 Aug 2019 21:04:12 +0000

090821-M-0440G-043

kqzheng — Tue, 06 Aug 2019 15:27:52 +0000

090821-M-0440G-043 HELMAND PROVINCE, Afghanistan (Aug. 21, 2009) Voting ballots organized and arranged for counting by Afghan presidential election workers at a local school in the Nawa District. The school was used as a polling site for the Afghanistan presidential elections. (U.S. Marine Corps photo by Staff Sgt. William Greeson/Released)

fig-ch_02_03_01-1

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fig-ch_02_04_02-1

kqzheng — Tue, 06 Aug 2019 15:27:58 +0000

CNX_Stats_C02_M04_020-1

kqzheng — Tue, 06 Aug 2019 15:27:58 +0000

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CNX_Stats_C02_M05a_026annoN-1

kqzheng — Tue, 06 Aug 2019 15:27:59 +0000

fig-ch02_11_01-1

kqzheng — Tue, 06 Aug 2019 15:27:59 +0000

CNX_Stats_C02_M03_101-1

kqzheng — Tue, 06 Aug 2019 15:28:00 +0000

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kqzheng — Tue, 06 Aug 2019 15:28:00 +0000

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kqzheng — Tue, 06 Aug 2019 15:28:01 +0000

fig-ch02_13_11-1

kqzheng — Tue, 06 Aug 2019 15:28:01 +0000

fig-ch02_13_02-1

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fig-ch02_13_08-1

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fig-ch_02_05_01-1

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fig-ch_02_05_02-1

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fig-ch_02_05_03-1

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fig-ch02_sol_01-1

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CNX_Stats_C02_M05_015-1

kqzheng — Tue, 06 Aug 2019 15:28:05 +0000

fig-ch02_13_01_rev-1

kqzheng — Tue, 06 Aug 2019 15:28:05 +0000

fig-ch02_13_03-1

kqzheng — Tue, 06 Aug 2019 15:28:05 +0000

fig-ch02_13_06-1

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fig-ch02_13_07-1

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fig-ch02_13_04-1

kqzheng — Tue, 06 Aug 2019 15:28:06 +0000

fig-ch02_13_09-1

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fig-ch02_13_10-1

kqzheng — Tue, 06 Aug 2019 15:28:07 +0000

fig-ch02_08_01-1

kqzheng — Tue, 06 Aug 2019 15:28:08 +0000

fig-ch02_08_02-1

kqzheng — Tue, 06 Aug 2019 15:28:08 +0000

fig-ch02_08_03-1

kqzheng — Tue, 06 Aug 2019 15:28:09 +0000

CNX_Stats_C02_M08_030-1

kqzheng — Tue, 06 Aug 2019 15:28:09 +0000

CNX_Stats_C02_M08_031-1

kqzheng — Tue, 06 Aug 2019 15:28:09 +0000

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kqzheng — Tue, 06 Aug 2019 15:28:10 +0000

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kqzheng — Tue, 06 Aug 2019 15:28:11 +0000

CNX_Stats_C02_M08_013-1

kqzheng — Tue, 06 Aug 2019 15:28:11 +0000

fig-ch02_09_01-1

kqzheng — Tue, 06 Aug 2019 15:28:12 +0000

fig-ch02_09_02-1

kqzheng — Tue, 06 Aug 2019 15:28:12 +0000

CNX_Stats_C02_M09_016-1

kqzheng — Tue, 06 Aug 2019 15:28:12 +0000

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kqzheng — Tue, 06 Aug 2019 15:28:13 +0000

CNX_Stats_C02_M09_018-1

kqzheng — Tue, 06 Aug 2019 15:28:13 +0000

CNX_Stats_C02_M09_019-1

kqzheng — Tue, 06 Aug 2019 15:28:13 +0000

fig-ch02_sol_04-1

kqzheng — Tue, 06 Aug 2019 15:28:14 +0000

fig-ch02_13_05-1

kqzheng — Tue, 06 Aug 2019 15:28:14 +0000

fig-ch02_14_01-1

kqzheng — Tue, 06 Aug 2019 15:28:15 +0000

CNX_Stats_C01_COs-1

kqzheng — Tue, 06 Aug 2019 19:45:51 +0000

fig-ch01_02_01n-1

kqzheng — Tue, 06 Aug 2019 19:45:52 +0000

CNX_Stats_C01_M05_001-1

kqzheng — Tue, 06 Aug 2019 19:45:52 +0000

CNX_Stats_C01_M05_002-1

kqzheng — Tue, 06 Aug 2019 19:45:53 +0000

fig-ch01_patchfile_01-1

kqzheng — Tue, 06 Aug 2019 19:45:53 +0000

fig-ch01_patchfile_02-1

kqzheng — Tue, 06 Aug 2019 19:45:53 +0000

fig-ch01_patchfile_03-1

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fig-ch01_patchfile_04-1

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fig-ch01_patchfile_05-1

kqzheng — Tue, 06 Aug 2019 19:45:54 +0000

fig-ch01_patchfile_06-1

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fig-ch01_patchfile_07-1

kqzheng — Tue, 06 Aug 2019 19:45:54 +0000

fig-ch01_patchfile_08-1

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fig-ch01_patchfile_09-1

kqzheng — Tue, 06 Aug 2019 19:45:55 +0000

CNX_Stats_C01_Screenshots-1

kqzheng — Tue, 06 Aug 2019 19:45:55 +0000

Print

kqzheng — Tue, 06 Aug 2019 19:45:56 +0000

CNX_Stats_C01_M11_001-1

kqzheng — Tue, 06 Aug 2019 19:45:57 +0000

CNX_Stats_C01_M11_002-1

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CNX_Stats_C01_M11_003s-1

kqzheng — Tue, 06 Aug 2019 19:45:58 +0000

prac1

kmoshenko — Thu, 08 Aug 2019 20:13:04 +0000

prac1

kmoshenko — Thu, 08 Aug 2019 20:26:19 +0000

DA1 Try This 1-6

kmoshenko — Thu, 08 Aug 2019 23:50:45 +0000

DA1 Try This 1-6

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DA1 Graphs Top 6

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Hornets Line Graph

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vancouver real estate line graph

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Van real estate reduced

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kmoshenko — Thu, 05 Sep 2019 23:06:59 +0000

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kmoshenko — Thu, 05 Sep 2019 23:31:01 +0000

DA1 Try This Tables & Graphs 1-3

kmoshenko — Thu, 05 Sep 2019 23:42:25 +0000

DA1 Try This Tables & Graphs 4-6

kmoshenko — Thu, 05 Sep 2019 23:47:13 +0000

DA1 Try This Tables & Graphs 4-6

kmoshenko — Thu, 05 Sep 2019 23:49:40 +0000

Data Analysis SurveyGraph1

kmoshenko — Fri, 21 Feb 2020 02:30:25 +0000

Data Analysis SurveyGraph1

kmoshenko — Fri, 21 Feb 2020 02:34:59 +0000

Data Analysis SurveyGraph2

kmoshenko — Fri, 21 Feb 2020 02:47:42 +0000

Data Analysis SurveyGraph3

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Data Analysis SurveyGraph4

kmoshenko — Fri, 21 Feb 2020 02:47:54 +0000

Data Analysis SurveyGraph5

kmoshenko — Fri, 21 Feb 2020 02:47:59 +0000

Data Analysis SurveyGraph5

kmoshenko — Fri, 21 Feb 2020 03:17:12 +0000

Data Analysis SurveyGraph3

kmoshenko — Fri, 21 Feb 2020 03:22:10 +0000

Data Analysis SurveyGraph4

kmoshenko — Fri, 21 Feb 2020 03:29:33 +0000

CNX_ElemAlg_Figure_05_00_001_img_new

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jgray — Wed, 26 Feb 2020 00:41:52 +0000

CNX_IntAlg_Figure_10_02_012

jgray — Wed, 26 Feb 2020 00:41:53 +0000

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jgray — Wed, 26 Feb 2020 00:41:53 +0000

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jgray — Wed, 26 Feb 2020 00:41:58 +0000

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jgray — Wed, 26 Feb 2020 00:41:58 +0000

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jgray — Wed, 26 Feb 2020 00:41:59 +0000

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jgray — Wed, 26 Feb 2020 00:41:59 +0000

8.1 Quartiles

kmoshenko — Tue, 09 Jun 2020 21:35:48 +0000

8.2 Percentile Formula

kmoshenko — Tue, 09 Jun 2020 22:24:05 +0000

2020-06-10 14_11_02-Ch12 Statisticsv2.docx - Word

kmoshenko — Wed, 10 Jun 2020 21:12:29 +0000

2020-06-10 14_12_52-Ch12 Statisticsv2.docx - Word

kmoshenko — Wed, 10 Jun 2020 21:13:23 +0000

2020-06-10 14_13_33-Ch12 Statisticsv2.docx - Word

kmoshenko — Wed, 10 Jun 2020 21:13:52 +0000

2020-06-10 15_00_50-Ch12 Statisticsv2.docx - Word

kmoshenko — Wed, 10 Jun 2020 22:01:51 +0000

Simple Interest Triangle

zpobanzaou — Thu, 11 Jun 2020 00:00:45 +0000

Simple Interest - P

zpobanzaou — Thu, 11 Jun 2020 00:27:37 +0000

2020-06-11 14_18_38-8.1 DA2 Images.docx - Word

kmoshenko — Thu, 11 Jun 2020 21:19:56 +0000

2020-06-11 14_26_02-8.1 DA2 Images.docx - Word

kmoshenko — Thu, 11 Jun 2020 21:26:35 +0000

2020-06-11 14_53_49-8.1 DA2 Images.docx - Word

kmoshenko — Thu, 11 Jun 2020 21:54:11 +0000

2020-06-11 14_54_42-8.1 DA2 Images.docx - Word

kmoshenko — Thu, 11 Jun 2020 21:55:01 +0000

2020-06-11 14_55_16-8.1 DA2 Images.docx - Word

kmoshenko — Thu, 11 Jun 2020 21:55:34 +0000

2020-06-11 14_56_13-8.1 DA2 Images.docx - Word

kmoshenko — Thu, 11 Jun 2020 21:56:31 +0000

2020-06-11 15_39_01-8.1 DA2 Images.docx - Word

kmoshenko — Thu, 11 Jun 2020 22:39:26 +0000

2020-06-11 15_39_35-8.1 DA2 Images.docx - Word

kmoshenko — Thu, 11 Jun 2020 22:39:49 +0000

2020-06-11 15_39_56-8.1 DA2 Images.docx - Word

kmoshenko — Thu, 11 Jun 2020 22:40:15 +0000

2020-06-11 15_40_24-8.1 DA2 Images.docx - Word

kmoshenko — Thu, 11 Jun 2020 22:40:45 +0000

2020-06-11 17_17_54-8.1 DA2 Images.docx - Word

kmoshenko — Fri, 12 Jun 2020 00:18:11 +0000

2020-06-11 17_18_19-8.1 DA2 Images.docx - Word

kmoshenko — Fri, 12 Jun 2020 00:18:34 +0000

present-future value

zpobanzaou — Fri, 12 Jun 2020 20:38:09 +0000

2020-06-12 14_07_30-DataAnalysis1 7 (Recovered).xlsx - Excel

kmoshenko — Fri, 12 Jun 2020 21:08:33 +0000

2020-06-12 14_25_46-DataAnalysis1 7 (Recovered).xlsx - Excel

kmoshenko — Fri, 12 Jun 2020 21:26:50 +0000

compound interest - annuity

zpobanzaou — Fri, 12 Jun 2020 21:59:11 +0000

from ABE Advanced Level Mathematics

Simple Interest - r

zpobanzaou — Tue, 16 Jun 2020 19:21:14 +0000

from ABE Advanced Level Mathematics

Simple Interest - t

zpobanzaou — Tue, 16 Jun 2020 19:27:44 +0000

from ABE Advanced Level Mathematics

9.4 Annuity Due diagram

kmoshenko — Tue, 23 Jun 2020 16:56:18 +0000

9.4 Ordinary Annuity diagram

kmoshenko — Tue, 23 Jun 2020 16:58:07 +0000

9.5 Loan Payment formula

kmoshenko — Wed, 24 Jun 2020 18:02:52 +0000

from ABE Advanced Level Financial Mathematics

9.5 ex 1 amort sched

kmoshenko — Wed, 24 Jun 2020 21:27:50 +0000

9.5 ex. 1 soln

kmoshenko — Wed, 24 Jun 2020 21:28:02 +0000

9.5 ex 1 amort sched

kmoshenko — Wed, 24 Jun 2020 22:00:38 +0000

9.5 ex. 2 soln

kmoshenko — Wed, 24 Jun 2020 22:01:01 +0000

9.5 ex 7 amort sched

kmoshenko — Wed, 24 Jun 2020 23:44:38 +0000

9.2 Present value diagram

kmoshenko — Fri, 26 Jun 2020 19:30:14 +0000

8.1 Quartiles 2

kmoshenko — Mon, 29 Jun 2020 16:57:49 +0000

8.2 Quartiles3

kmoshenko — Mon, 29 Jun 2020 17:27:32 +0000

8.2 Ex1 Range

kmoshenko — Mon, 29 Jun 2020 20:37:20 +0000

8.2 Try It 1 Range

kmoshenko — Mon, 29 Jun 2020 20:38:09 +0000

8.2 pop std dev formula

kmoshenko — Tue, 30 Jun 2020 16:20:41 +0000

8.2 sample std dev formula

kmoshenko — Tue, 30 Jun 2020 16:21:42 +0000

8.2 Sets A&B image

kmoshenko — Tue, 30 Jun 2020 21:06:10 +0000

8.2 Example 3

kmoshenko — Tue, 30 Jun 2020 21:06:34 +0000

8.2 Example 4

kmoshenko — Tue, 30 Jun 2020 21:06:45 +0000

7.2 histogram example

kmoshenko — Tue, 30 Jun 2020 22:26:44 +0000

8.3 Team Harmony Histogram

kmoshenko — Wed, 01 Jul 2020 00:03:05 +0000

8.3 Team Unity Histogram

kmoshenko — Wed, 01 Jul 2020 00:03:05 +0000

8.3 Team Extend Histogram

kmoshenko — Wed, 01 Jul 2020 00:03:06 +0000

8.3 Team Mix Histogram

kmoshenko — Wed, 01 Jul 2020 00:03:06 +0000

8.2 Example5

kmoshenko — Wed, 01 Jul 2020 01:07:53 +0000

8.2 Try It 5 Histogram

kmoshenko — Wed, 01 Jul 2020 01:07:54 +0000

8.2 Example5

kmoshenko — Wed, 01 Jul 2020 01:13:39 +0000

8.2 Example5

kmoshenko — Wed, 01 Jul 2020 01:19:55 +0000

8.2 Example5

kmoshenko — Wed, 01 Jul 2020 01:47:43 +0000

8.2 ex 5 solution

kmoshenko — Wed, 01 Jul 2020 01:47:44 +0000

8.2 Example5

kmoshenko — Wed, 01 Jul 2020 01:48:41 +0000

8.2 Try It 5 Histogram

kmoshenko — Wed, 01 Jul 2020 01:56:00 +0000

8.3 Normal Dist histogram

kmoshenko — Thu, 02 Jul 2020 18:07:59 +0000

8.3 Norm Distn4

kmoshenko — Thu, 02 Jul 2020 19:20:43 +0000

8.3 Norm Distn5

kmoshenko — Thu, 02 Jul 2020 19:20:44 +0000

Normal_Distribution_NIST (1)

kmoshenko — Thu, 02 Jul 2020 19:29:45 +0000

8.3 fig. 7

kmoshenko — Thu, 02 Jul 2020 20:36:31 +0000

8.3-Normal-Distn-std-deviation-wikimedia

kmoshenko — Thu, 02 Jul 2020 20:41:00 +0000

8.3 Standard_deviation_diagram.svg

kmoshenko — Thu, 02 Jul 2020 21:13:21 +0000

8.3 Example 1 Solutioniii

kmoshenko — Thu, 02 Jul 2020 22:15:31 +0000

8.3 Example 1 Solutionii

kmoshenko — Thu, 02 Jul 2020 22:15:32 +0000

8.4 Fig.1 Normal curve

kmoshenko — Fri, 03 Jul 2020 22:06:27 +0000

8.4 Fig.2 Normal curves

kmoshenko — Fri, 03 Jul 2020 22:10:31 +0000

8.4 Fig.4 Normal curves

kmoshenko — Fri, 03 Jul 2020 22:30:12 +0000

8.4 Fig.2 Normal curves

kmoshenko — Fri, 03 Jul 2020 22:30:13 +0000

8.4 Fig.3 Normal curves

kmoshenko — Fri, 03 Jul 2020 22:30:13 +0000

8.4 Fig.1 Normal curve

kmoshenko — Fri, 03 Jul 2020 22:30:15 +0000

8.4 Fig.5 Normal curves

kmoshenko — Fri, 03 Jul 2020 22:30:16 +0000

8.4 Fig.6 Normal curves

kmoshenko — Fri, 03 Jul 2020 22:30:17 +0000

8.3 Example 3 figure

kmoshenko — Wed, 08 Jul 2020 17:50:59 +0000

8.3 Try It 3 figure

kmoshenko — Wed, 08 Jul 2020 17:51:00 +0000

7.2 Exercise Set BC median age

kmoshenko — Thu, 09 Jul 2020 23:44:42 +0000

7.2 Exercise Set BC median agev2

kmoshenko — Thu, 09 Jul 2020 23:44:43 +0000

Average Commuting Time

kmoshenko — Thu, 23 Jul 2020 17:44:30 +0000

% of Population 65 and over

kmoshenko — Thu, 23 Jul 2020 17:44:40 +0000

Median Household Income 2105

kmoshenko — Thu, 23 Jul 2020 17:49:27 +0000

No. of Employeed Women who are Family Physicians

kmoshenko — Thu, 23 Jul 2020 17:51:57 +0000

2016 census

7.2 Ex Set Women Physicians Pie

kmoshenko — Thu, 23 Jul 2020 18:36:16 +0000

7.2 Ex Set StemnLeaf Seniors

kmoshenko — Thu, 23 Jul 2020 19:40:05 +0000

7.2 Ex Set Bball Bar graph

kmoshenko — Thu, 23 Jul 2020 21:26:42 +0000

7.2 Ex Set Bball Freq Dist

kmoshenko — Thu, 23 Jul 2020 21:26:44 +0000

7.2 Ex Set Bball tally

kmoshenko — Thu, 23 Jul 2020 21:26:44 +0000

7.2 Ex Set Bike Line

kmoshenko — Thu, 23 Jul 2020 22:41:06 +0000

7.2 Ex Set Bike Bar

kmoshenko — Thu, 23 Jul 2020 22:41:06 +0000

Fig. 1

kmoshenko — Mon, 27 Jul 2020 21:21:10 +0000

9.2 Compount Interest diagram

kmoshenko — Mon, 27 Jul 2020 21:27:41 +0000

9.3 Effective rates diagram

kmoshenko — Mon, 27 Jul 2020 23:38:43 +0000

7.2 histogram she size

kmoshenko — Wed, 29 Jul 2020 22:40:54 +0000

DA1 Try This #5

kmoshenko — Wed, 29 Jul 2020 23:25:28 +0000

DA1 Try This #6

kmoshenko — Wed, 29 Jul 2020 23:25:29 +0000

DA1 Try This #4

kmoshenko — Wed, 29 Jul 2020 23:25:30 +0000

8.1Ex Set#5 Table

kmoshenko — Thu, 30 Jul 2020 22:23:39 +0000

8.1Ex Set#2 Table

kmoshenko — Thu, 30 Jul 2020 22:23:40 +0000

8.1Ex Set#4 Table

kmoshenko — Thu, 30 Jul 2020 22:23:40 +0000

8.2 Ex Set#1 Table

kmoshenko — Thu, 30 Jul 2020 22:41:00 +0000

8.2 Ex Set#6 Histograms

kmoshenko — Thu, 30 Jul 2020 22:41:01 +0000

8.3Normal curve

kmoshenko — Fri, 31 Jul 2020 00:05:01 +0000

8.3 Ex Set#2 figure

kmoshenko — Fri, 31 Jul 2020 00:05:02 +0000

8.4 Ex 4b soln

kmoshenko — Fri, 31 Jul 2020 03:12:50 +0000

8.4 Ex 4soln

kmoshenko — Fri, 31 Jul 2020 03:12:53 +0000

8.4 zscore gen

kmoshenko — Fri, 31 Jul 2020 03:12:54 +0000

8.4 zscore pop

kmoshenko — Fri, 31 Jul 2020 03:12:54 +0000

8.4 zscore sample

kmoshenko — Fri, 31 Jul 2020 03:12:54 +0000

8.4 Ex 1 soln

kmoshenko — Fri, 31 Jul 2020 03:12:55 +0000

8.4 Ex 1 solnfig

kmoshenko — Fri, 31 Jul 2020 03:12:55 +0000

8.4 Ex 2soln

kmoshenko — Fri, 31 Jul 2020 03:12:56 +0000

8.4 Ex 3soln

kmoshenko — Fri, 31 Jul 2020 03:12:57 +0000

8.4 ztable

kmoshenko — Fri, 31 Jul 2020 16:10:53 +0000

8.4 zcore 142

kmoshenko — Fri, 31 Jul 2020 16:22:41 +0000

8.4 Ex 2soln

kmoshenko — Fri, 31 Jul 2020 18:06:02 +0000

8.4 Ex 3soln

kmoshenko — Fri, 31 Jul 2020 20:31:32 +0000

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imazur — Wed, 19 Aug 2020 07:27:06 +0000

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imazur — Wed, 19 Aug 2020 07:27:07 +0000

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imazur — Wed, 19 Aug 2020 07:27:07 +0000

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imazur — Wed, 19 Aug 2020 07:27:07 +0000

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imazur — Wed, 19 Aug 2020 07:27:07 +0000

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imazur — Wed, 19 Aug 2020 07:27:08 +0000

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imazur — Wed, 19 Aug 2020 07:27:08 +0000

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imazur — Wed, 19 Aug 2020 07:27:08 +0000

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imazur — Wed, 19 Aug 2020 07:27:08 +0000

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imazur — Wed, 19 Aug 2020 07:27:09 +0000

7.4 Qn4

hanil — Fri, 23 Oct 2020 03:52:59 +0000

7.2 figure5

hanil — Fri, 23 Oct 2020 03:57:45 +0000

7.2Figure6

hanil — Fri, 23 Oct 2020 04:00:10 +0000

8.1 Quartiles 3

kmoshenko — Tue, 20 Apr 2021 23:54:54 +0000

8.1 Quartiles 3

kmoshenko — Tue, 20 Apr 2021 23:55:10 +0000

8.1 Quartiles3i

kmoshenko — Wed, 21 Apr 2021 21:09:02 +0000

8.2 pop std dev formula

kmoshenko — Wed, 28 Apr 2021 19:02:44 +0000

8.2 sample std dev formula

kmoshenko — Wed, 28 Apr 2021 19:02:54 +0000

8.2 Ex Set#6 Histograms

kmoshenko — Thu, 29 Apr 2021 21:01:57 +0000

8.3 Norm Distn1

kmoshenko — Thu, 29 Apr 2021 21:19:06 +0000

8.3 PMP#3 diagram

kmoshenko — Fri, 30 Apr 2021 21:28:21 +0000

8.4 Fig.2

kmoshenko — Mon, 03 May 2021 19:10:58 +0000

8.4 neg zscore table

kmoshenko — Mon, 03 May 2021 22:02:33 +0000

8.4 normal distrn and rule

kmoshenko — Tue, 04 May 2021 21:06:22 +0000

8.4 Ex5partd

kmoshenko — Wed, 05 May 2021 21:44:02 +0000

8.4 Ex5partabc

kmoshenko — Wed, 05 May 2021 21:44:03 +0000

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imazur — Wed, 05 May 2021 21:46:16 +0000

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imazur — Wed, 05 May 2021 21:46:16 +0000

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8.4 Ex5partabc

kmoshenko — Wed, 05 May 2021 21:48:12 +0000

8.4 Ex5partd

kmoshenko — Wed, 05 May 2021 21:48:14 +0000

8.4 Ex5parta

kmoshenko — Thu, 06 May 2021 02:09:23 +0000

8.4 Ex5partb

kmoshenko — Thu, 06 May 2021 02:09:25 +0000

8.4 Ex5partc

kmoshenko — Thu, 06 May 2021 02:11:26 +0000

7.1 intro mountains

kmoshenko — Thu, 06 May 2021 21:27:25 +0000

7.2 intro graphs

kmoshenko — Thu, 06 May 2021 22:35:07 +0000

7.3 intro survey

kmoshenko — Thu, 06 May 2021 23:58:27 +0000

7.4 SurveyGraphs

kmoshenko — Fri, 07 May 2021 00:09:42 +0000

7.4 intro dice

kmoshenko — Fri, 07 May 2021 00:26:31 +0000

8.1 Quartiles ex4

kmoshenko — Fri, 07 May 2021 00:57:47 +0000

8.1 medianoddno

kmoshenko — Fri, 07 May 2021 17:59:31 +0000

8.1 Quartiles oddno

kmoshenko — Fri, 07 May 2021 17:59:32 +0000

8.1 Q2method2oddno

kmoshenko — Fri, 07 May 2021 21:36:17 +0000

8.1 medianmethod2oddno

kmoshenko — Fri, 07 May 2021 21:36:17 +0000

8.1 Q3method2oddno

kmoshenko — Fri, 07 May 2021 21:36:17 +0000

8.1 medianoddno1

kmoshenko — Fri, 07 May 2021 22:10:49 +0000

8.1 quartilesmethod1oddno

kmoshenko — Fri, 07 May 2021 22:10:49 +0000

8.1 DA2 Images

kmoshenko — Fri, 07 May 2021 22:10:50 +0000

8.1 Q3method1oddno

kmoshenko — Fri, 07 May 2021 22:10:50 +0000

8.1 Q1method1oddno

kmoshenko — Fri, 07 May 2021 22:10:52 +0000

8.1 medianoddno

kmoshenko — Fri, 07 May 2021 22:13:40 +0000

8.1 intro image exam

kmoshenko — Mon, 10 May 2021 16:07:52 +0000

8.1Q3 method2

kmoshenko — Mon, 10 May 2021 16:21:40 +0000

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8.2 intro image flower

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illustration of a police lineup

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8.3 bean machinepolice line intro

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8.3 ex set #7 image

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8.4 intro image exam

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9.1 intro piggy bank

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9.2 watercan image

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9.3 Intro image interest

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9.4 retire intro

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9.5 imagehousesale

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7.1 himalaya

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blank unit circle Test

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3620770052_ac67133361_o

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System of equations

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Graphs of equations

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48277390351_0b57c7188b_b

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4.5 try it 2

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4.5 1

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4.5 2

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7.1 Measures of Central Tendency

kmoshenko — Wed, 07 Aug 2019 18:04:26 +0000

Learning Objectives

By the end of this section it is expected that you will be able to:

Calculate three measures of central tendency: the mean, median and mode
Describe what the mean, median and mode tell us about a data set

When we discuss the mountains of the world we often hear mention of the world's highest peak, Mount Everest. Aside from the highest, there are other criteria which can be used to compare mountains. Perhaps we might want to determine the average height of the mountains within a specified mountain range. We may wish to determine who has made the most number of summit attempts of a particular mountain and what that number of attempts was. These kinds of questions can be answered with statistics.

Measures of Central Tendency

In our daily lives we encounter data and statistics on a regular basis. The data can be analysed to ascertain patterns and trends or to determine measures of central tendency. Three statistics that measure the center point of a set of data are mean, median and mode. Each of these measures of central tendency serves a different purpose and provides a different perspective. Consider a college level biology class of 44 students that receives feedback on their midterm exam. The instructor shares the following results with the class: the average score was 68%, the most common score was 48%, and the score that ranked right in the middle of the class was 75%. The three values each represent a different measure of central tendency for the exam scores. What can we conclude about how the class performed? An examination of each of the three measures will help answer this question.

Mean

In the college biology class the average score on the midterm exam was 68%. The average score is also referred to as the mean. The average score of 68% on the biology exam is the sum of all exam scores divided by the number of scores (44 students).

Mean

The mean is the sum of all data items divided by the number of data items. This can be expressed as :

[latex] mean = \bar{x} = \frac{\Sigma x}{n}[/latex]

The symbol ∑ is called the summation symbol and indicates that all data items are to be summed. The mean or average of a set of data is calculated by adding all of the values and dividing by the number of values.

Consider a runner who is training for a 10 km race. Her coach requires that she run an average distance of 15 km a day. Over a two week period the runner logged the following distances (in km) on her daily runs.

Su	M	T	W	Th	F	Sa	Su	M	T	W	Th	F	Sa
15	16	14	22	15	10	30	0	15	20	20	24	5	32

Did the runner meet the coach's requirement? To determine the answer we need to calculate the average or mean. To determine the average daily distance the calculation would be:

[latex] \bar{x} = \frac{15\, +\, 16 \,+ \,14 \,+\,22\, + \,15 \,+ \,10 \,+\, 30\, + \,0\, + \,15 \,+\, 20\, + \,20 \,+ \,24 \,+\, 5 \,+ \,32}{14} = \frac{238}{14} = 17 \,km [/latex]

On average the runner covered 17km/day which exceeds the coach's required 15km/day.

EXAMPLE 1

Fifteen students wrote a math test and received the following grades: 89, 45, 78, 76, 73, 98, 73, 92, 88, 73, 100, 51, 64, 80, 95. Solution To determine the mean: Add the grades and divide by the number of grades (15) (89+45+78+76+73+98+73+92+88+73+100+51+64+80+95)/15 = 1175/15 = 78.33... The average grade, rounded to the nearest whole number, was 78.

TRY IT 1

Find the mean of the data set: 5.7, 3.4, 7.8, 9.2, 3.8, 1.6

Show answer

7.3 Collecting Data

kmoshenko — Wed, 07 Aug 2019 18:06:20 +0000

Learning Objectives

By the end of this section it is expected that you will be able to:

State whether data is quantitative or qualitative
Describe the random sampling methods: simple random sampling, systematic sampling, cluster sampling and convenience sampling
Discuss potential problems that might arise when sampling from a population

Populations and Samples

In statistics, we generally want to study a population. You can think of a population as a collection of persons, things, or objects under study. It is often not feasible or possible to study the entire population. Instead we can select a sample. The idea of sampling is to select a portion (or subset) of the larger population and study that portion (the sample) to gain information about the population. Data are the result of sampling from a population. Because it takes a lot of time and money to examine an entire population, sampling is a very practical technique. If you wished to compute the overall grade point average at your school, it would make sense to select a sample of students who attend the school. The data collected from the sample would be the students' grade point averages. In elections, opinion poll samples of 1,000–2,000 people are taken. The opinion poll is supposed to represent the views of the people in the entire country.

Types of Data

Most data can be categorized as qualitative or quantitative. Qualitative data are the result of categorizing or describing attributes of a population using our senses such as sight or touch. Hair color, blood type, ethnic group, the car model that a person drives, and the street a person lives on are examples of qualitative data. Qualitative data are generally described by words or letters. For instance, hair color might be black, dark brown, light brown, blonde, gray, or red. Blood type might be AB+, O-, or B+. Quantitative data are always numbers. Quantitative data are the result of counting or measuring attributes of a population. Amount of money, pulse rate, weight, number of people living in your town, and number of students who take statistics are examples of quantitative data. Researchers often prefer to use quantitative data over qualitative data because it lends itself more easily to mathematical analysis. For example, it does not make sense to find an average hair color or median blood type.

EXAMPLE 1

Consider a high school math class and a sample of five student's backpacks. Determine whether the data is quantitative or qualitative. 1. One data set is the number of books students carry in their backpacks.Two students carry three books, one student carries four books, one student carries two books, and one student carries one book. 2. For the sample of five backpacks you weigh the backpacks and contents. The weights (in kilograms) of their backpacks are 3.2, 5, 4.8, 5.1, 2.3. 3. For the sample of five students you record the colour of the backpacks. The books are red, blue or black. Solution:

This is quantitative data.
This is quantitative data.
This is qualitative data.

TRY IT 1

Determine the correct data type (quantitative or qualitative).

the number of pairs of shoes you own
the colour of vehicle you drive
the distance it is from your home to the nearest grocery store
the number of classes you take per school year.
the model of calculator you use
weights of sumo wrestlers
total number of correct answers on a quiz
IQ scores

Show answer

7.2 Graphs and Tables

kmoshenko — Wed, 07 Aug 2019 18:07:00 +0000

Learning Objectives

By the end of this section it is expected that you will be able to:

Extract information from a table, a bar graph, a line graph or a pie graph
Create a stem and leaf graph from a set of data
Create a frequency distribution table from a set of data
Create a line graph, a bar graph and a pie graph (with or without technology)
Compare a bar graph to a histogram

We have seen how data can be represented numerically with measures such as the mean, median and mode. Data can be organized and displayed in visual formats that allow the user to more easily extract information. When we represent data graphically we can determine data clusters, make comparisons, or determine trends.

Displaying Data with Tables or Graphs

We will consider some graphical alternatives for displaying the information presented in the following paragraph:

According to Venture Kamloops, the six largest employers in Kamloops, British Columbia, along with the number of employees in parentheses are: Interior Health Authority (3398), School Distric#73 (1924), Thompson Rivers University (1092), Highland Valley Copper Mine (1351), the City of Kamloops (761) and Bc Lottery Corporation (440) http://venturekamloops.com/pdf/DBIK-Community-Facts-Residential-April-2017.pdf

Tables

Transferring the data to a table, as in Table 1, provides greater clarity. The reader can quickly determine the names of the employers and their corresponding number of employees. It is easier to determine the employer with the greatest and least number of employees.

Table 1

Employer	Number of Employees
Interior Health Authority	3398
School District #73	1924
Thompson Rivers University	1092
Highland Valley Copper Mine	1351
City of Kamloops	761
BC Lottery Corporation	440

Graphs

When the data is represented visually the reader can quickly retrieve information and make comparisons. Technology can be used to easily create a wide variety of graphs. The data from the table was entered into a spreadsheet and three graphs were generated. The results are displayed below as a bar graph (Fig. 1a), a pie graph (Fig. 1b), and a waterfall graph (Fig. 1c).

Bar Graph

[caption id="attachment_967" align="aligncenter" width="468"] Fig. 1a[/caption]

Circle or Pie Graph

[caption id="attachment_970" align="aligncenter" width="544"] Fig. 1b[/caption]

Waterfall

[caption id="attachment_972" align="aligncenter" width="469"] Fig. 1c[/caption] Consider each of the three graphs in Figures 1a, 1b and 1c (bar, pie and waterfall) to answer the following questions. Beside each answer indicate which of the three graph(s) provides the answer.

Which of the six employers has the most number of employees?
How many employees work for the largest employer?
Which of the six employers has the least number of employees?
How many employees work for the smallest employer?
Where does TRU place in the ranking of number of employees?
Which graph is the easiest to extract data from?

The answers to the six questions would be:

Interior Health has the most number of employees. This information is found in all three graphs.
Interior Health employs 3398 people. This information can only be determined using the pie graph
BC Lottery has the least number of employees. This information is found in all three graphs.
BC Lottery employs 440 people. This information can only be determined using the pie graph
TRU ranks fourth in the number of employees. This can be stated with certainty by using the bar graph or pie graph. The reader may not be so certain with the waterfall graph.
This depend on which information is required.

Note that there is not necessarily one form of graph that is better than the others. It is important to recognize that the way in which the information is presented will impact its use. By making one change, such as including the numerical values for the bar or waterfall graphs, the user would be able to obtain more exact information. When choosing and creating a table or graph it is important to know what kind of information is required. A decision can then be made as to how best to depict this. Since technology provides easily accessible tools for creating tables and charts, this section will focus on the features of different tables and graphs rather than on the manual construction of the graphs. We will now take a closer look at line graphs, bar graphs, and circle graphs as well as stem-and-leaf plots and frequency tables.

Stem-and-Leaf Graph

One simple graph, the stem-and-leaf graph or stemplot, is a good choice when the data sets are small. This graph indicates data clusters and can be used to determine the measures of central tendency. A stem-and-leaf graph divides each observation of data into a stem and a leaf. The leaf consists of one final significant digit. For example, 23 has a stem of 2 and a leaf 3. The number 432 has a stem of 43 and a leaf of 2. Likewise, the number 5,432 has a stem 543 and a leaf of two. The decimal 9.3 has a stem of nine and a leaf of three. To create the plot, write the stems in a vertical line from smallest to largest. Draw a vertical line to the right of the stems. Then write the leaves in increasing order next to their corresponding stem.

EXAMPLE 1

For Susan's spring pre-calculus class, scores for the first exam were as follows (ranked from lowest to highest): 33; 42; 49; 49; 53; 55; 55; 61; 63; 67; 68; 68; 69; 69; 72; 73; 74; 78; 80; 83; 88; 88; 88; 90; 92; 94; 94; 94; 94; 96; 100 a) Create a stem-and-leaf graph for the data. b) Describe where the data clusters. c) What percentage of the students obtained a score of 90 or better? d) What is the mean, median and mode? Solution a) To create the graph, rank the data from lowest to highest. Create the column for the stems. This will be the first digit in a two digit number and the first two digits in a three digit number. The stems will start at 3 and end at 10. For each data value, add each leaf to its corresponding stem. For the value 33. the stem is 3 and the leaf is 3. For the value 68 the stem is 6 and the leaf is 8. Since 68 occurs twice in the data set, for the stem of 6 there will be two leaves of 8.

Stem	Leaf
3	3
4	2 9 9
5	3 5 5
6	1 3 7 8 8 9 9
7	2 3 4 8
8	0 3 8 8 8
9	0 2 4 4 4 4 6
10	0

b) There appears to be two clusters of data. The stemplot shows that most scores fell in either the 60s or the 90's. c) Eight out of the 31 scores or approximately 26% were in the 90s or 100. d) The mean is 73.5. Since there are 31 students, the median is the 16th score, which is 73. The mode is 94 as it occurs 4 times.

TRY IT 1

For the Park City basketball team, scores for the last 30 games were as follows (from lowest to highest): 32; 32; 33; 34; 38; 40; 42; 42; 43; 44; 46; 47; 47; 48; 48; 48; 49; 50; 50; 51; 52; 52; 52; 53; 54; 56; 57; 57; 60; 61 a) Construct a stem-and-leaf graph for the data. b) In what percent of the games did the team score less than 40 points? c) Use the graph to determine the mean, median and mode.

Show answer a)

7.4 Survey Creation

kmoshenko — Wed, 07 Aug 2019 18:08:01 +0000

Learning Objectives

By the end of this section it is expected that you will be able to:

Design a statistical experiment, collect the data, and analyze and communicate the results

This section provides an overview of the steps involved in designing and conducting a survey.

Survey Steps

Determine the objective of the survey and the population to be surveyed.
Determine the survey questions.
Determine the sampling technique: simple random sampling, systematic sampling, cluster or convenience. Select the sample to be surveyed.
Collect the data using an appropriate instrument (interview, paper questionnaire, internet,...). Record the responses in a frequency table.
Analyze the data and communicate the results using descriptive statistics such as measures of central tendency and/or graphs. The graphs can be created using technology.

Math Support Survey

Determine the objective of the survey and the population to be surveyed.
- The goal of the survey is to determine the types of math support and frequency of use by the students enrolled in upgrading math at College ABC
Determine three survey questions.
- Three questions will be used:i. Which of the following support(s) do you access while you are enrolled in upgrading math at College ABC? Check all that apply: math learning lab, my math instructor, private tutor, internet, otherii. Which method of support do you most prefer? math learning lab, my math instructor, private tutor, internet, otheriii. How many hours a day (on average) do you work on math outside of classtime? 0; more than 0 but less than 1; 1-2; more than 2
Determine the sampling technique: simple random sampling, systematic sampling, cluster or convenience. Select the sample to be surveyed.
- Cluster sampling will be used. There are 12 sections of upgrading math during the winter semester at College ABC. Each of these represents a cluster. The clusters will be numbered 3-14 and three clusters will be selected by rolling three die. The total of the three die will be matched with the corresponding cluster.Upon completion of this step the three selected clusters are cluster numbers 5, 9 and 14.

Collect the data using an appropriate instrument (interview, paper questionnaire, internet,...). Record the responses in a frequency table.

Permission to visit these three classes will be obtained by contacting the instructors via email. Each cluster will be visited on a test day when attendance tends to be most reliable. All students in each cluster will be asked to complete a three question paper survey. Three frequency tables will be created to organize the results. In total 86 students will be surveyed.a) Which of the following support(s) do you access while you are enrolled in upgrading math at College ABC? Check all that apply: math learning lab, my math instructor, private tutor, internet, other

Type of Support	Tally	Frequency
Math Learning Lab	~~////~~ ~~////~~ ~~////~~ ~~////~~ ~~////~~ ~~////~~ ~~////~~ ~~////~~ ///	43
My Math Instructor	~~////~~ ~~////~~ ~~////~~ ~~////~~ ~~////~~ ///	28
Tutor	~~////~~ ////	9
Internet	~~////~~ ~~////~~ ~~////~~ ~~////~~ ~~////~~ ~~////~~ ~~////~~ ~~////~~ ~~////~~ ~~////~~ //	52
Other	~~////~~ ~~////~~ ~~////~~ ~~////~~ ~~//// ////~~ /	31

b) Which support do you prefer most?

Type of Support	Tally	Frequency
Math Learning Lab	~~////~~ ~~////~~ ~~////~~ ~~////~~ ~~////~~ //	27
My math instructor	~~////~~ ~~////~~ ~~////~~ /	16
Tutor	~~////~~ ///	8
Internet	~~////~~ ~~////~~ ~~////~~ ~~////~~ ////	24
Other	~~////~~ ~~////~~ /	11

c) How many hours per day, on average, do you work on math?

# of Hours	Frequency
0	31
between 0 and 1	18
1 to 2	27
More than 2	10

Analyze the data and communicate the results. Graphs can be created using technology.
- Pie and bar graphs will be used to organize and summarize the results of the survey.
  Bar Graphs
  The following three graphs were created using the Chart feature in Excel. Bar Graph#1 depicts the number of students on the horizontal axis and the type of support on the vertical axis. It is clear from the graph that the support that has been used by the most number of students is the internet and by the least number of students is a tutor. [caption id="attachment_1119" align="aligncenter" width="482"] Bar Graph#1[/caption] Bar Graph#2 depicts the total number of students using the support as compared against the number of students preferring the support. Forty-three students in total have used the math lab whereas twenty-seven of these would choose the math lab first for support. Although fifty-two students have accessed the internet for support, only twenty-four would choose this type of support first. The support that is most preferred is the math lab; least preferred is the tutor. [caption id="attachment_1121" align="aligncenter" width="480"] Bar Graph#2[/caption] Bar Graph#3 depicts the amount of hours per day, on average, that students spend studying math. [caption id="attachment_1131" align="aligncenter" width="476"] Bar Graph#3[/caption]
  Pie Graphs
  Graph#4 depicts which type of math support is most preferred by the students. The math lab is the largest sector followed closely by the Internet. [caption id="attachment_5500" align="aligncenter" width="470"] Pie Graph #4[/caption] Graph#5 depicts the number of hours per day, on average, spent working on math. [caption id="attachment_1137" align="aligncenter" width="472"] Pie Graph #5[/caption]

4.1 Solve Systems of Equations by Graphing

jgray — Tue, 30 Jul 2019 03:24:31 +0000

Learning Objectives

By the end of this section it is expected that you will be able to:

Determine whether an ordered pair is a solution of a system of equations
Solve a system of linear equations by graphing
Determine the number of solutions of linear system
Solve applications of systems of equations by graphing

Determine Whether an Ordered Pair is a Solution of a System of Equations

We learned before how to solve linear equations with one variable. Now we will work with systems of linear equations, two or more linear equations grouped together, witch is known as a system of linear equations.

System of Linear Equations

When two or more linear equations are grouped together, they form a system of linear equations.

We will focus our work here on systems of two linear equations in two unknowns. Later, you may solve larger systems of equations.

An example of a system of two linear equations is shown below. We use a brace to show the two equations are grouped together to form a system of equations.

$\left\{\begin{array}{c}2x+y=7\hfill \\ x-2y=6\hfill \end{array}$

A linear equation in two variables, like 2x + y = 7, has an infinite number of solutions. Its graph is a line. Remember, every point on the line is a solution to the equation and every solution to the equation is a point on the line.

To solve a system of two linear equations, we want to find the values of the variables that are solutions to both equations. In other words, we are looking for the ordered pairs (x, y) that make both equations true. These are called the solutions to a system of equations.

Solutions of a System of Equations

Solutions of a system of equations are the values of the variables that make all the equations true. A solution of a system of two linear equations is represented by an ordered pair (x, y).

To determine if an ordered pair is a solution to a system of two equations, we substitute the values of the variables into each equation. If the ordered pair makes both equations true, it is a solution to the system.

Let’s consider the system below:

$\left\{\begin{array}{c}3x-y=7\hfill \\ x-2y=4\hfill \end{array}$

Is the ordered pair $\left(2,-1\right)$ a solution?

The ordered pair (2, −1) made both equations true. Therefore (2, −1) is a solution to this system.

Let’s try another ordered pair. Is the ordered pair (3, 2) a solution?

The ordered pair (3, 2) made one equation true, but it made the other equation false. Since it is not a solution to both equations, it is not a solution to this system.

EXAMPLE 1

Determine whether the ordered pair is a solution to the system: $\left\{\begin{array}{c}x-y=-1\hfill \\ 2x-y=-5\hfill \end{array}$

a) $\left(-2,-1\right)$ b) $\left(-4,-3\right)$

Solution

a) (−2, −1) does not make both equations true. (−2, −1) is not a solution. b) (−4, −3) does not make both equations true. (−4, −3) is a solution.

TRY IT 1

Determine whether the ordered pair is a solution to the system: $\left\{\begin{array}{c}3x+y=0\hfill \\ x+2y=-5\hfill \end{array}.$

a) $\left(1,-3\right)$ b) $\left(0,0\right)$

Show answer

4.2 Solve Systems of Equations by Substitution

jgray — Tue, 30 Jul 2019 03:24:45 +0000

Learning Objectives

By the end of this section it is expected that you will be able to:

Solve a system of equations by substitution
Solve applications of systems of equations by substitution

Solving systems of linear equations by graphing is a good way to visualize the types of solutions that may result. However, there are many cases where solving a system by graphing is inconvenient or imprecise. If the graphs extend beyond the small grid with x and y both between −10 and 10, graphing the lines may be cumbersome. And if the solutions to the system are not integers, it can be hard to read their values precisely from a graph.

In this section, we will solve systems of linear equations by the substitution method.

Solve a System of Equations by Substitution

We will use the same system we used first for graphing.

$\left\{\begin{array}{c}2x+y=7\hfill \\ x-2y=6\hfill \end{array}$

We will first solve one of the equations for either x or y. We can choose either equation and solve for either variable—but we’ll try to make a choice that will keep the work easy.

Then we substitute that expression into the other equation. The result is an equation with just one variable—and we know how to solve those!

After we find the value of one variable, we will substitute that value into one of the original equations and solve for the other variable. Finally, we check our solution and make sure it makes both equations true.

EXAMPLE 1

How to Solve a System of Equations by Substitution

Solve the system by substitution. $\left\{\begin{array}{c}2x+y=7\hfill \\ x-2y=6\hfill \end{array}$

Solution

TRY IT 1

Solve the system by substitution. $\left\{\begin{array}{c}-2x+y=-11\hfill \\ x+3y=9\hfill \end{array}$

Show answer

4.3 Solve Systems of Equations by Elimination

jgray — Tue, 30 Jul 2019 03:24:57 +0000

Learning Objectives

By the end of this section it is expected that you will be able to:

Solve a system of equations by elimination
Solve applications of systems of equations by elimination
Choose the most convenient method to solve a system of linear equations

We have solved systems of linear equations by graphing and by substitution. Graphing works well when the variable coefficients are small and the solution has integer values. Substitution works well when we can easily solve one equation for one of the variables and not have too many fractions in the resulting expression.

The third method of solving systems of linear equations is called the Elimination Method. When we solved a system by substitution, we started with two equations and two variables and reduced it to one equation with one variable. This is what we’ll do with the elimination method, too, but we’ll have a different way to get there.

Solve a System of Equations by Elimination

The Elimination Method is based on the Addition Property of Equality. The Addition Property of Equality says that when you add the same quantity to both sides of an equation, you still have equality. We will extend the Addition Property of Equality to say that when you add equal quantities to both sides of an equation, the results are equal.

For any expressions a, b, c, and d,

$\begin{array}{cccc}\text{if}\hfill & \hfill a& =\hfill & b\hfill \\ \text{and}\hfill & \hfill c& =\hfill & d\hfill \\ \text{then}\hfill & \hfill a+c& =\hfill & b+d\hfill \end{array}$

To solve a system of equations by elimination, we start with both equations in standard form. Then we decide which variable will be easiest to eliminate. How do we decide? We want to have the coefficients of one variable be opposites, so that we can add the equations together and eliminate that variable.

Notice how that works when we add these two equations together:

$\begin{array}{ccccc}3x+y=5\hfill \\ \text{}{2x-y=0}\hfill \\ 5x\phantom{\rule{1.7em}{0ex}}=5\hfill \end{array}$

The y’s add to zero and we have one equation with one variable.

Let’s try another one:

$\left\{\begin{array}{c}x+4y=2\hfill \\ 2x+5y=-2\hfill \end{array}$

This time we don’t see a variable that can be immediately eliminated if we add the equations.

But if we multiply the first equation by −2, we will make the coefficients of x opposites. We must multiply every term on both sides of the equation by −2.

Now we see that the coefficients of the x terms are opposites, so x will be eliminated when we add these two equations.

Add the equations yourself—the result should be −3y = −6. And that looks easy to solve, doesn’t it? Here is what it would look like.

We’ll do one more:

$\left\{\begin{array}{c}4x-3y=10\hfill \\ 3x+5y=-7\hfill \end{array}$

It doesn’t appear that we can get the coefficients of one variable to be opposites by multiplying one of the equations by a constant, unless we use fractions. So instead, we’ll have to multiply both equations by a constant.

We can make the coefficients of x be opposites if we multiply the first equation by 3 and the second by −4, so we get 12x and −12x.

This gives us these two new equations:

$\left\{\begin{array}{c}\phantom{\rule{1.1em}{0ex}}12x-9y=30\hfill \\ -12x-20y=28\hfill \end{array}$

When we add these equations,

$\begin{array}{ccc}\left\{\begin{array}{c}\phantom{\rule{1.1em}{0ex}}12x-9y=30\hfill \\ \text{}{-12x-20y=28}\hfill \end{array}\\ \hfill -29y=58\end{array}$

the x’s are eliminated and we just have −29y = 58.

Once we get an equation with just one variable, we solve it. Then we substitute that value into one of the original equations to solve for the remaining variable. And, as always, we check our answer to make sure it is a solution to both of the original equations. Now we’ll see how to use elimination to solve the same system of equations we solved by graphing and by substitution.

EXAMPLE 1

How to Solve a System of Equations by Elimination

Solve the system by elimination. $\left\{\begin{array}{c}2x+y=7\hfill \\ x-2y=6\hfill \end{array}$

Solution

TRY IT 1

Solve the system by elimination. $\left\{\begin{array}{c}3x+y=5\hfill \\ 2x-3y=7\hfill \end{array}$

Show answer

4.4. Solve Applications with Systems of Equations

jgray — Tue, 30 Jul 2019 03:25:07 +0000

Learning Objectives

By the end of this section it is expected that you will be able to:

Translate to a system of equations
Solve direct translation applications
Solve geometry applications
Solve uniform motion applications

Previously in this chapter we solved several applications with systems of linear equations. In this section, we’ll look at some specific types of applications that relate two quantities. We’ll translate the words into linear equations, decide which is the most convenient method to use, and then solve them.

We will use our Problem Solving Strategy for Systems of Linear Equations.

Use a problem solving strategy for systems of linear equations.

Read the problem. Make sure all the words and ideas are understood.
Identify what we are looking for.
Name what we are looking for. Choose variables to represent those quantities.
Translate into a system of equations.
Solve the system of equations using good algebra techniques.
Check the answer in the problem and make sure it makes sense.
Answer the question with a complete sentence.

Translate to a System of Equations

Many of the problems we solved in earlier applications related two quantities.

Let’s see how we can translate these problems into a system of equations with two variables. We’ll focus on Steps 1 through 4 of our Problem Solving Strategy.

EXAMPLE 1

How to Translate to a System of Equations

Translate to a system of equations:

The sum of two numbers is negative fourteen. One number is four less than the other. Find the numbers.

Solution

TRY IT 1

Translate to a system of equations:

The sum of two numbers is negative twenty-three. One number is 7 less than the other. Find the numbers.

Show answer

4.5 Graphing Systems of Linear Inequalities

jgray — Tue, 30 Jul 2019 03:25:31 +0000

Learning Objectives

By the end of this section it is expected that you will be able to:

Determine whether an ordered pair is a solution of a system of linear inequalities
Solve a system of linear inequalities by graphing
Solve applications of systems of inequalities

Determine Whether an Ordered Pair is a Solution of a System of Linear Inequalities

The definition of a system of linear inequalities is very similar to the definition of a system of linear equations.

System of Linear Inequalities

Two or more linear inequalities grouped together form a system of linear inequalities.

A system of linear inequalities looks like a system of linear equations, but it has inequalities instead of equations. A system of two linear inequalities is shown below.

$\left\{\begin{array}{c}x+4y\ge 10\hfill \\ 3x-2y<12\hfill \end{array}$

To solve a system of linear inequalities, we will find values of the variables that are solutions to both inequalities. We solve the system by using the graphs of each inequality and show the solution as a graph. We will find the region on the plane that contains all ordered pairs $\left(x,y\right)$ that make both inequalities true.

Solutions of a System of Linear Inequalities

Solutions of a system of linear inequalities are the values of the variables that make all the inequalities true.

The solution of a system of linear inequalities is shown as a shaded region in the x-y coordinate system that includes all the points whose ordered pairs make the inequalities true.

To determine if an ordered pair is a solution to a system of two inequalities, we substitute the values of the variables into each inequality. If the ordered pair makes both inequalities true, it is a solution to the system.

EXAMPLE 1

Determine whether the ordered pair is a solution to the system. $\left\{\begin{array}{c}x+4y\ge 10\hfill \\ 3x-2y<12\hfill \end{array}$

a) (−2, 4) b) (3,1)

Solution

a) Is the ordered pair (−2, 4) a solution?

The ordered pair (−2, 4) made both inequalities true. Therefore (−2, 4) is a solution to this system.

b) Is the ordered pair (3,1) a solution?

The ordered pair (3,1) made one inequality true, but the other one false. Therefore (3,1) is not a solution to this system.

TRY IT 1

Determine whether the ordered pair is a solution to the system.

a) $\left(3,-1\right)$ b) $\left(6,-3\right)$

Show answer

8.1 Percentiles and Quartiles

kmoshenko — Thu, 04 Jun 2020 18:25:12 +0000

Learning Objectives

By the end of this section it is expected that you will be able to:

Describe the measures of location: percentile and quartile
Find the percentile represented by a given data value
Determine the first, second and third quartiles for a set of data

Measures of Central Tendency

The mean, median and mode, as measures of central tendency, provide us with a point of comparison. As an example, consider Company ABC where the average (mean) salary is $55,000/year. An employee earning $38,000/year might feel unjustly treated or at the very least the employee might explore the reasons for the substantial difference. If in the process the employee learns that the [pb_glossary id="4755"]median[/pb_glossary] salary at his workplace is $26,000/year the employee would learn that relative to everyone else this employee's salary is in the upper half of the employee group. To provide additional comparison the employee could consider other measures of position or location. Two such measures are percentiles and quartiles.

Percentiles

Percentiles are useful for comparing values. If a data item is in the 75th percentile then three-quarters of the values are less than this value. This is not to be confused with a score of 75%, which is something very different. A student could score 35% on an exam but be in the 75th percentile. This means that relative to the rest of the class the student had a score that was higher than 75% of the students.

Percentiles

Percentiles divide ordered data into hundredths. A data item is said to be in the k^th percentile of a data set if k% of the data items are less than the item.

The notation P_k can be used to represent the k^th percentile. A data set can be divided into one hundred equal parts by ninety-nine percentiles P₁ , P₂ , P₃ , ... P₉₉ . The 60^th percentile would be denoted P₆₀ . If an item is in the 60th percentile, then 60 percent of the data items are less than this item. Consider a set of math exam scores. A student scoring in the 60th percentile achieved a score equal to or higher than 60 percent of the other students. This does not mean that the student scored 60% on the exam. Perhaps the student's score was 78%, which would mean that 60 percent of the other students in the class had exam scores less than (or equal to) 78%. It is important to note that since percentiles divide a data set into one hundred equal parts, percentiles are best used with large data sets. Percentiles are mostly used with very large populations. For a specified percentile P_k if you were to say that k percent of the data values are less (and not the same or less) than a specified data value, it would be acceptable because removing one particular data value is not significant. Refer again to the employee earning $38,000/year at Company ABC. If the employee learns that their salary is in the 90th percentile then 90 percent of the other employees at Company ABC have a salary less than (or possibly equal to) this salary. In relation to the other employees this salary ranks among the upper portion of the employee group. Percentiles are useful for comparing values. For this reason, universities and colleges use percentiles on entrance exams. Rather than set one value as an acceptance score, a university may set a percentile target. Perhaps all students scoring in the the 80th percentile or above will receive an acceptance letter. Every year there is likely to be a different acceptance score. Students will be accepted based on their score relative to all other applicants.

Determining Percentiles

To determine the k^th percentile that is represented by a particular data item x, the following formula can be used. Step 1: If necessary order the data values from smallest to largest. Step 2: Determine the total number of data values, n. This will be the denominator in the formula. Step 3: Count the number of data values that are less than the value x. This will be the value in the numerator of the formula. Step 4: Calculate the percentile, k, that is associated with a score of x using the formula.

EXAMPLE 1

A class set of exam scores for 48 students are ranked from lowest to highest. Determine the percentiles associated with the scores of a) 39% b) 60% c) 94%.

39	54	59	65	75	79	84	92
42	54	60	67	76	80	86	92
43	55	60	69	76	80	88	94
48	57	60	69	77	82	88	95
51	57	63	72	77	83	89	96
51	59	65	72	78	83	91	97

Solution a) For a score of 39%: Step 1: The data values are already ordered from smallest to largest. Step 2: Determine the number of data values. Since there are 48 students n = 48. Step 3: We count 0 data values that are less than 39 Step 4: Calculate the percentile, k, that is associated with a score of x using the formula

k = (0/48)*100% = 0%.

This means that the student who scored 39% is in the 0 percentile. A score of 39% is not higher than any other score. b) For a score of 60%: There are 13 scores lower than 60% so k = (13/48)*100% = 27%. A score of 60% is in the 27th percentile which means that 27% (or just over one-fourth) of the test scores are less than 60%. c) For a score of 94%: There are 44 scores less than 94% so k = (44/48)*100% = 92%. A score of 94% is in the 92nd percentile which means that 92% of the test scores are less than 94%.

TRY IT 1

A set of assignment scores for a class of 32 students are provided in the table below. Determine the percentiles associated with the scores of a) 61% b) 79% c) 98%.

72	65	85	52	61	49	65	82
55	99	58	79	98	79	58	93
88	48	97	74	65	85	71	75
99	39	60	96	80	70	54	77

Show answer

9.1 Simple Interest

kmoshenko — Thu, 04 Jun 2020 20:56:14 +0000

Learning Objectives

By the end of this section it is expected that you will be able to:

Determine the simple interest earned on an investment or charged on a loan
Determine the principal amount, the interest rate, or the time for applications involving simple interest
Determine the maturity value of a loan that involves simple interest

Interest

Some people keep money at home in an easily accessible location, perhaps a piggy bank, a safe or locked box, or perhaps even a mattress. Although this provides instant access to funds it does not provide any return or earnings on this money. For that reason, most people hold their money in accounts or investments that provide some form of return or earning power. Interest is the price paid for the use of money. If you borrow money from another person or a lending institution, eventually you must pay back this amount plus the interest owing. When you deposit money in a bank, you are lending them money and after some time they will pay you interest on the money you lent them. The amount of interest you will owe or receive is determined by the principal, the interest rate, and the time (the length of the loan). The amount of money that you lend or borrow is called the principal. The length of the loan can range between a few days to several years. The interest rate is stated as an annual percentage. It may be simple interest or compound interest. With simple interest the interest is calculated only once during the entire time period of the loan or deposit. Simple interest is calculated solely on the principal investment or loan. With compound interest the interest is calculated more than once during the time period of the loan. It will be calculated on the principal as well as the accumulated interest This section will focus on simple interest and in the next section we will consider compound interest.

Simple Interest

Simple interest is calculated by finding the product of the principal (P), the rate (r), and the time (t).

Simple Interest

The simple interest formula is I = Prt where I = interest earned r = annual interest rate ( stated as a decimal) P = principal t = time (in years)

Interest rates are quoted for periods of one year and when used in a formula must be converted to a decimal fraction. The time must be expressed in the same unit of time as the interest rate so time must be stated in years or portions of a year. If you deposit money in a savings account earning 3% interest then the annual interest rate is 3% per year.

EXAMPLE 1

Jo borrows $2000 at an interest rate of 5% per year. How much interest will Jo owe after one year? Solution Identify the P, r, and t. P = $2000 r = 5% = 0.05 t = 1 year

Here,	[latex]P = \$2000[/latex]	[latex]I = Prt[/latex]
	[latex]r = 5\%\, or\, 0.05[/latex]	[latex]I = 2000 \left(5\%\right)\left(1\right)[/latex]	Replace P, r and t with their values
	[latex]t = 1\, year[/latex]	[latex]I = 2000 \left(0.05\right)\left(1\right)[/latex]	Change 5% to its decimal equivalent, 0.05
		[latex]I = 100[/latex]	[latex][/latex]

Start with the formula I = Prt

I = 2000 (5%) (1) Replace P, r, and t with their values I = 2000 (0.05) (1) Change 5% to its decimal equivalent, 0.05 I = 100 Jo will pay $100 in interest.

TRY IT 1

Terri borrowed $3200 at an interest rate of 4.75%. How much interest will Terri owe on the loan at the end of one year?

Show answer

2.5 Solve Inequalities

imazur — Tue, 30 Jul 2019 03:14:59 +0000

Learning Objectives

By the end of this section, you will be able to:

Graph inequalities on the number line
Solve inequalities using the Subtraction and Addition Properties of inequality
Solve inequalities using the Division and Multiplication Properties of inequality
Solve inequalities that require simplification
Translate to an inequality and solve

Graph Inequalities on the Number Line

Do you remember what it means for a number to be a solution to an equation? A solution of an equation is a value of a variable that makes a true statement when substituted into the equation.

What about the solution of an inequality? What number would make the inequality $x$ > $3$ true? Are you thinking, ‘x could be 4’? That’s correct, but x could be 5 too, or 20, or even 3.001. Any number greater than 3 is a solution to the inequality $x$ > $3$.

We show the solutions to the inequality $x$ > $3$ on the number line by shading in all the numbers to the right of 3, to show that all numbers greater than 3 are solutions. Because the number 3 itself is not a solution, we put an open parenthesis at 3. The graph of $x$ > $3$ is shown in (Figure). Please note that the following convention is used: light blue arrows point in the positive direction and dark blue arrows point in the negative direction.

The inequality $x$ > $3$ is graphed on this number line.

The graph of the inequality $x\ge 3$ is very much like the graph of $x$ > $3$, but now we need to show that 3 is a solution, too. We do that by putting a bracket at $x=3$, as shown in (Figure).

The inequality $x\ge 3$ is graphed on this number line.

Notice that the open parentheses symbol, (, shows that the endpoint of the inequality is not included. The open bracket symbol, [, shows that the endpoint is included.

EXAMPLE 1

Graph on the number line:

a) $x\le 1$ b) $x$ < $5$ c) $x$ > $-1$

Solution

a) $x\le 1$ This means all numbers less than or equal to 1. We shade in all the numbers on the number line to the left of 1 and put a bracket at $x=1$ to show that it is included.

b) $x$ < $5$ This means all numbers less than 5, but not including 5. We shade in all the numbers on the number line to the left of 5 and put a parenthesis at $x=5$ to show it is not included.

c) $x$ > $-1$ This means all numbers greater than $-1$, but not including $-1$. We shade in all the numbers on the number line to the right of $-1$, then put a parenthesis at $x=-1$ to show it is not included.

TRY IT 1

Graph on the number line: a) $x\le -1$ b) $x$ > $2$ c) $x$ < $3$

Show answer

4.5 3

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4.5 5

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9.2 Compound Interest

kmoshenko — Thu, 04 Jun 2020 20:57:26 +0000

Learning Objectives

By the end of this section it is expected that you will be able to:

Determine the compound amount (future value) of an investment or loan
Determine the interest component of an investment or loan that involves compound interest
Determine the present value of a compound amount

Compound Interest

We have seen that with simple interest an investment will earn interest on the original amount. For an investment of $100 earning 10% simple interest, the interest earned after one year will be $10 since 10% of $100 = $10. An investment will grow more quickly if the interest is calculated more often than once a year. Interest will not only be calculated on the principal amount but also on the previously earned interest. This process is referred to compounding. Figure 1 illustrates the process of compounding or earning interest on interest. Consider an investment of $100 that earns 10%/year with interest being compounded semiannually. With semiannual compounding the interest on the investment will be calculated twice during the year. [caption id="attachment_3260" align="aligncenter" width="1024"] Fig. 1[/caption] Using the simple interest formula I = Prt, at the end of six months (half a year) interest will be calculated as follows: I = $100 x 10% x 1/2 year = $5. Adding this $5 to the principal of $100 you will have $105 at the end of the first six months. At the end of the year interest will be calculated again on the $105: I = $105 x 10% x 1/2 year = $5.25. Adding this $5.25 to $105 you will have $110.25 at the end of the year. In this case you would be earning interest not only on the original principal of $100, but also on the previously earned interest of $5. When interest is earned on interest, we say the interest is compounded. The total amount of principal and accumulated interest at the end of a loan or investment is called the compound amount. Consider a $100 investment that earns 10%/year compounded annually. The table in Figure 2 shows how the value of the $100 investment will grow over a 6-year period.

Year	Amount at the beginning of the year	Earned Interest	Year End Total
1	$100	$10	$110
2	$110	$11	$121
3	$121	$12.10	$133.10
4	$133.10	$13.31	$146.41
5	$146.41	$14.64	$161.05
6	$161.05	$16.11	$177.16

Fig. 2

At the beginning of Year 1, $100 is invested, so the interest earned in the first year will be: I = Prt = $100 × 0.10 × 1 = $10. This is added to the original $100 to result in $110 at the end of Year 1. At the beginning of Year 2 the process will repeat but the principal P is now $110. I = Prt = $110 × 0.10 ×1 = $11 in interest so at the end of Year 2 there will be: $110 + $11 = $121 in the account. Notice that the compound amount at the end of the six year period is $177.16. The investment has earned an accumulated $77.16 in interest. If the investment had earned simple interest as opposed to compound interest it would have only earned:

I = Prt = 100 × 0.10 × 6 = $60 in interest.

The above method of calculating the compound amount is very time consuming. Fortunately, there is a mathematical formula that we can use when working with compound interest.

Compound Interest Formula

The compound interest formula is:

[latex] A = P \left ( 1 + \frac{r}{n}\right )^{nt} [/latex]	where,	A = total compound amount(includes principal and interest)
		P = principal
		r = annual interest rate
		n = number of times in one year that interest is calculated
		t = time (in years)

Since A includes both the principal and interest, to find the interest amount I calculate: [latex]I = A - P [/latex]

EXAMPLE 1

Find the compound amount and the interest earned on $100 compounded annually at 10% for 6 years. Solution

P = $100 r = 10% = 0.1 n = 1 (since the interest is calculated once a year) t = 6 years	[latex] A = P \left ( 1 + \frac{r}{n}\right )^{nt} [/latex] [latex] A = 100 \left ( 1 + \frac{0.1}{1}\right )^{1 \times 6} [/latex]	Replace the variables with their values
	[latex] A = 100 \left ( 1 + 0.1\right )^{6} [/latex]	[latex]\frac{0.1}{1} = 0.1 [/latex] and [latex] 1 \times 6 = 6 [/latex]
	[latex] A = 100 \left ( 1.1 \right )^{6} [/latex]	Raise [latex]\left( 1.1 \right)^{6} = 177.1561 [/latex]
	[latex] A = 100 \left ( 1.771 561 \right ) = 177.1561 [/latex]

The interest earned is [latex] A - P = \$ 177.16 - \$100 =[/latex] $77.16 The compound amount is $177.16

TRY IT 1

Kyle won $10,000 in a lottery and deposited the full amount in a 3 year investment at 3.8% compounded annually. Find the compound amount and the interest earned over the three years.

Show answer

9.3 Interest Rates

kmoshenko — Thu, 04 Jun 2020 20:58:22 +0000

Learning Objectives

By the end of this section it is expected that you will be able to:

State the difference between nominal and effective interest rates
Determine the effective annual interest rate
Determine the best option when comparing nominal interest rates under different situations

Nominal and Effective Rates of Interest

As consumers and investors we are inundated with all kinds of offers for both investments and financing. When we require a loan we seek the lowest possible interest rate but when we invest we want the highest rate of return. As we search for the best offer it is important to recognize that the advertised interest rate is not necessarily the true interest rate. An advertised rate of 10% with annual compounding works out to be equivalent to a rate of 10% annually. We refer to this stated rate of 10% as the nominal interest rate. An advertised rate of 10% with daily compounding works out to be equivalent to a rate of 10.52% since interest is calculated on the interest 365 times in one year. We refer to the 10.52% as the effective interest rate. The stated interest rate is referred to as the nominal interest rate, as it describes the named or numerical value. It is the rate that is usually stated in advertisements. The actual interest rate, or effective interest rate, reflects the real rate of return as it takes the compounding periods into account. With simple interest calculations (where there is no compounding), the stated annual interest rate indicates the true rate of return. If $1000 is borrowed at 6% for one year, then the interest owined will be I = Prt = $1000 × 0.06/yr × 1yr = $60. With compound interest calculations, the stated annual interest rate does not indicate the true interest cost. If $1000 is invested at 6% for one year compounded semiannually, then the interest owed will be $60.90 rather than $60. One way to determine the effective interest rate is to divide the total compound interest for the first year by the principal amount. If in the first year $60.90 is the interest charged on a principal of $1000 then the effective interest rate is $60.90/$1000 = 0.0609 = 6.09%. Although the nominal interest rate is 6%, the effective rate is 6.09%. It is also possible to use a formula to calculate the effective interest rate.

Effective Interest Rate Formula

The effective interest rate formula is:

[latex]f = \left( 1 + \frac{r}{n} \right)^{n} - 1[/latex]	where	f = effective interest rate
		r = nominal interest rate (annual interest rate)
		n = number of times in one year that interest is calculated

EXAMPLE 1

Use the formula to determine the effective interest rate for 6% compounded annually. Solution

f = ? r = 6% = 0.06 n = 1	[latex]f = \left( 1 + \frac{r}{n} \right)^{n} - 1[/latex]
	[latex]f = \left( 1 + \frac{0.06}{1} \right)^{1} - 1[/latex]	Replace the variables with their values
	[latex]f = \left( 1.06 \right)^{1} - 1[/latex]	Add [latex]1 + 0.06[/latex]
	[latex]f = 1.06 - 1[/latex]
	[latex]f = 0.06 = 6%[/latex]

The effective interest rate is 6%. Note that the nominal and effective rate are the same since the number of compounding period is one (n = 1).

TRY IT 1

What is the effective rate of 4% compounded yearly?

Show answer

9.4 Annuities

kmoshenko — Thu, 04 Jun 2020 21:01:14 +0000

Learning Objectives

By the end of this section it is expected that you will be able to:

Distinguish between an ordinary annuity and an annuity due
Determine the future value of an ordinary annuity
Determine the payment, given the future value for an ordinary annuity

Annuities

A common financial goal is to be comfortable in retirement. During our working lives we contribute to a retirement fund so that upon retirement we receive a financial payment at regular intervals. Financial transactions that involve a series of equal payments at equal intervals are called annuities. Other examples of annuities include payments on a loan, rental payments, and insurance premiums. The term of the annuity is the time from the beginning of the first payment interval to the end of the last payment interval. A payment interval is the time between successive payments. If, for example, a vehicle is purchased with monthly payments on a four-year loan then the term of the loan is 4 years and the payment interval is monthly. In some cases, as with salaries or a senior's pension, the payments are made at the end of a payment interval. This is referred to as an ordinary annuity. When payments are required at the beginning of a payment interval, as with many loans and mortgages, this is referred to as an annuity due. It is important to note that the term of the annuity does not necessarily coincide with the first and last payment. Consider a one year loan where 12 equal payments are made on the first of each month. This is an example of an annuity due. The term of the annuity is one year and the payment interval is one month. Refer to Figure 1. There are twelve payments, each occurring on the first day of the month. The first payment is made on Jan. 1 and the last payment is made on Dec. 1. Note that the final payment on Dec. 1 does not occur on the last day of the term of the annuity which is Dec. 31. [caption id="attachment_2949" align="aligncenter" width="1024"] Fig. 1[/caption] Consider a one year loan where 12 equal payments are made on the last day of each month. This is an example of an ordinary annuity. The term of the annuity is one year and the payment interval is one month. Refer to Figure 2. There are twelve payments, each occurring on the last day of the month. The first payment is made on Jan. 31 and the last payment is made on Dec. 31. Note that the first payment on Jan. 31 does not occur on the first day of the term of the annuity which is Jan. 1. [caption id="attachment_2950" align="aligncenter" width="1024"] Fig. 2[/caption] In this section we will only be concerned with ordinary simple annuities. As with all ordinary annuities the payments are made at the end of each payment interval. It is also the case that the compounding interval equals the payment interval. This means that if the payment interval is monthly then interest will also be compounded monthly.

Future Value of an Ordinary Annuity

Consider an investment that is in the form of an ordinary simple annuity. This means that a deposit is made at the end of regular intervals and interest is compounded at each of these intervals. The value of the annuity can grow substantially. The final amount of the annuity is called the future value of the annuity. It is the total of all annuity payments and the accumulated compound interest as illustrated in Figure 3.

[caption id="attachment_2847" align="aligncenter" width="477"] Fig. 3[/caption] To see how the annuity process works, consider the table in Figure 4 below. This table depicts an ordinary 6-month annuity commencing on Jan. 1 and ending on June 30. The monthly payments are $1000 and the annual interest rate is 6% compounded monthly.

Month	Balance on the first day of the month	Interest earned during the interval	Deposit at end of the month	Balance at end of the month
Jan	$0	$0	$1000	$1000
February	$1000	$5.00	$1000	$2005
March	$2005	$10.025 = $10.03	$1000	$3015.03
April	$3015.03	$15.075 = $15.08	$1000	$4030.11
May	$4030.11	$20.15	$1000	$5050.26
June	$5050.26	$25.25	$1000	$6075.51

Fig. 4 Since this is an ordinary annuity the payments are made at the end of the month. Interest is calculated as simple interest I = Prt where r = 0.06 and t = 1/12 year and P = the balance at the beginning of the month. The first payment of $1000 is deposited at the end of January and therefore no interest is earned for the month of January. At the beginning of February there is $1000 in the account. Interest for the month of February is I = $1000(0.06)(1/12) = $5. At the end of February there will be a deposit of $1000 so the balance at the end of February is $1000 + $5 + $1000 = $2005. At the beginning of March there is $2005 in the account. Interest for the month of March will be I = $2005(0.06)(1/12) = $10.025 or $10.03 (rounded off). At the end of March $1000 is deposited so the balance at the end of March is $2005 + $$10.03 + $1000 = $3015.03. At the beginning of April there is $3015.03 in the account. Interest for the month of April will be I = $3015.03(0.06)(1/12) = $15.075 or $15.08 (rounded off). At the end of April $1000 is deposited so the balance at the end of April is $3015.03 + $$15.08 + $1000 = $4030.11. The process is continued for the remaining two months. At the end of June, the balance will be $6075.51. This is the future value of the annuity, which is the total of all annuity payments and the accumulated compound interest. Notice that with an ordinary annuity the interest calculation is based on the balance at the beginning of the interval. Since the first payment does not occur until the end of the first payment interval there will not be any interest in the first payment interval. Although the term of the annuity is six months there will only be five intervals where interest is calculated. With an annuity due the payment would be made at the beginning of each payment interval so for a six month term there would be six interest calculations. Since many loans are set up as an annuity due it is advantageous to the lending institution (but not to the loan recipient). The calculation of the future value of an annuity can be very time consuming. Fortunately there is a formula for this.

Ordinary Annuity Formula

The ordinary annuity formula is:

[latex]A = \frac{nP \left[ \left( 1 + \frac{r}{n} \right)^{nt}-1\right]}{r}[/latex]

where

A = amount of annuity (Future Value) P = periodic payment amount r = annual interest rate n = number of compounding periods per year t = time (in years)

It is important to note that there are variations on how the ordinary annuity formula is written. This is due to the way in which the variables are defined. The formula that is provided in this section defines r as the annual interest rate, n as the number of compounding periods per year, and t as the time in years (term of the annuity in years).

EXAMPLE 1

Assume that the formula will be used to calculate the future value of a two year ordinary annuity that offers an annual interest rate of 6%, monthly payments of $1000, and monthly compounding. Define each of the variables but do not calculate the future value. Solution

[latex]A = \frac{nP \left[ \left( 1 + \frac{r}{n} \right)^{nt}-1\right]}{r}[/latex]

where

A = amount of annuity (Future Value) = unknown P = periodic payment amount = $1000 r = annual interest rate = 6% = 0.06 n = number of compounding periods per year = 12 t = time (in years) = 2 years

Note that the term of the annuity is 2 years. The interest calculation involves monthly compounding so n = 12 since there are 12 compounding periods in a year.

TRY IT 1

Assume that the formula will be used to calculate the future value of a 1 year ordinary annuity that offers an annual interest rate of 3%, semiannual payments of $500, and semiannual compounding. Define each of the variables but do not calculate the future value.

Show answer

9.5 Loans and Mortgages

kmoshenko — Thu, 04 Jun 2020 21:02:52 +0000

Learning Objectives

By the end of this section it is expected that you will be able to:

Determine the periodic payments on an installment loan
Determine the amount financed and the finance charge on an installment loan
Determine the payments and finance charge on a mortgage

Installment Loans

A loan is something that is borrowed. In the case where this is a sum of money the amount that will be paid by the borrower will include the original amount plus interest. Some loans require full payment on the maturity date of the loan. The maturity date is when all principal and/or interest must be repaid to the the lender. Consider a one year loan of $1000 at a simple interest rate of 5%. At the end of one year (the maturity date) the borrower will pay back the original $1000 plus the interest of $50 for a total of $1005. For major purchases such as vehicles or furniture there is a different type of loan, called the installment loan. The average consumer cannot afford to pay $25000 or more for a new vehicle and they may not want to wait three or four years until they have saved enough money to do so. The qualifying consumer has the option of paying for the item with an installment loan. Installment loans do not require full repayment of the loan on a specific date. With an installment loan the borrower is required to make regular (installment) payments until the loan is paid off. Each installment payment will include an interest charge. An installment loan can vary in length from a few years to perhaps twenty years or more (in the case of real estate). Consider an installment loan for a $4000 television. The purchaser takes out a $4000 loan with a four-year term at an interest rate of 4.5%. The monthly installment payments will be $91.21. Although the television has a purchase price of $4000, the total cost to the purchaser will be more than $4000. The total of the installment payments will be: Total Installment Payments = Number of Installment Payments x Payment Amount =

4 years x 12 payments/year x $91.21/mth = $4378.08

The $4000 television ends up costing $4378.08 because the consumer is charged interest. Each payment includes an interest component that adds to the overall cost of the item. The total of the interest charges is referred to as the finance charge on the loan.

Finance Charge

The finance charge is the sum of the interest charges on a loan. These interest charges are embedded in the installment payments. To calculate the finance charge:

Finance Charge = Total Installment Payments - Loan Amount

= (Number of Installment Payments x Payment Amount) - Loan Amount

For the $4000 television the finance charge will be calculated as follows:

Finance charge = Total Installment Payments - Loan Amount =

(4 years x 12 payments/year x $91.21/payment) - $4000 = $4378.08 - $4000 = $378.08

Over the 4-year term of the loan the purchaser will have paid the $4000 loan amount plus an additional $378.08 in interest (the finance charge). Sometimes the borrower will make an initial payment at the time of purchase. This is called a down payment. When a down payment is made the remaining amount is the amount financed or the loan amount.

Amount Financed

The amount financed or loan amount is the purchase price of the item less any down payment:

Amount Financed = Purchase Price - Down Payment

Consider the $4000 television. Assume the purchaser makes a down payment of $1500.

The amount financed is: Purchase Price - Down Payment = $4000 - $1500 = $2500.

In this case the purchaser borrows $2500 rather than $4000. The amount financed is therefore $2500. Assuming the same 4-year term and an interest rate of 4.5%, the installment payments on the $2500 will be reduced to $57.01 per month. In this case the finance charge will be calculated as follows: Finance charge = Total Installment Payments - Loan Amount =

(4 years x 12 payments/year x $57.01/payment) - $2500 = $2736.48 - $2500 = $236.48

With the down payment of $2500 the total finance charges will be reduced to $236.48 from $378.08. The total cost of the television to the purchaser will be:

Purchase Price + Finance Charge

= $4000 + $236.48 = $4236.48

Alternatively we can calculate:

Total Installment Payment + Down Payment

= $2736.48 + $1500 = $4236.48

As one can see, the finance charges are a hidden but added cost. This cost will become more pronounced with more expensive purchases such as with real estate.

Installment Loan Terminology

Total Installment Payments = Number of payments x Payment Amount Finance Charge = Total Installment Payments - Loan Amount Amount Financed or Loan Amount = Purchase Price of Item - Down payment

EXAMPLE 1

Paul purchased a home entertainment system at a total cost of $6000. He obtained a 3 year loan at an interest rate of 7.5%. His monthly payments will be $186.64 over three years. a) State the amount financed. b) Determine the total installment payments. c) Determine the finance charge. Solution a) Since there was no down payment the amount financed (or loan amount) will be $6000. b) The total installment payments will be: Number of payments x Payment Amount = 3 years x 12 payments/year x $186.64 = $6719.04 c) Finance Charge = Total installment payments - Loan Amount = $6719.04 - $6000 = $719.04

TRY IT 1

Cassie purchased a new washer and dryer at a total cost of $3800. She obtained a 4 year loan at an interest rate of 6.2%. Her monthly payments will be $89.59 over four years. a) State the amount financed. b) Determine the the total installment payments. c) Determine the finance charge.

Show answer

8.2 Measures of Dispersion

kmoshenko — Fri, 05 Jun 2020 22:44:26 +0000

Learning Objectives

By the end of this section the student should be able to:

Determine the range for a data set.
Determine the standard deviation for a data set.
Determine the standard deviation from a histogram

Measures of Dispersion or Spread

We have seen that measures of central tendency, including the [pb_glossary id="4798"]mean[/pb_glossary] and [pb_glossary id="4755"]median[/pb_glossary], are used to identify a central position within a data set. They indicate where the data clusters. Consider student A's scores on five tests: 32% 95% 89% 74% 55% The mean, or average, is (32 + 95 + 89 + 74 + 55)/5 = 69% and the median is 74%. Consider student B's scores on the same five tests: 68% 69% 72% 74% 62% The mean, or average, is (68 + 69 + 72 + 74 + 62)/5 = 69% and the median is 69%. Both student's have the same test average of 69% but there is a substantial difference in the spread or dispersion of their scores. Student A's test scores range from a low score of 32% to a high score of 95% so the spread in marks is 63 percentage points. Student B's test scores range from a low score of 62% to a high score of 74% so the spread in marks is 12 percentage points. When we analyze data it is important to consider how dispersed or spread out the data values are. In this section we will consider two measures of dispersion.

Range

Range is one measure of dispersion. A measure of dispersion is used to describe the spread of data.

Range

The range indicates the total spread in data values. It is the difference between the highest and lowest data values. Range = highest data value - lowest data value

EXAMPLE 1

The table shows the daily high temperature (°C) over a three week period. Determine the highest temperature, lowest temperature, and the range in daily high temperatures over the three weeks. Solution The highest temperature was 29º, the lowest temperature was 16°, and the range in temperatures was 29 - 16 = 13º.

TRY IT 1

The table shows the test score for a group of fourteen students. Determine the highest test score, the lowest test score, and the range in test scores for the group of 14 students.

Show answer

8.3 The Normal Curve

kmoshenko — Fri, 05 Jun 2020 22:48:02 +0000

Learning Objectives

After completing this section the student should be able to:

Recognize the characteristics of a normal distribution
Find scores at a designated standard deviation from the mean
Interpret and use the 68-95-99.7 Rule

The Normal Distribution

The Galton Board, invented by Sir Francis Galton, consists of a vertical board with interleaved rows of pegs. Beads are dropped from the top and, when the device is level, bounce either left or right as they hit the pegs. Eventually they are collected into bins at the bottom, where the height of bead columns accumulated in the bins approximate a normal distribution. (https://en.wikipedia.org/wiki/Bean_machine#/) In this section we will explore the normal distribution and the dispersion of data values around the mean. We have seen that the standard deviation provides a measure of the dispersion of the data values around the mean. If the standard deviation is zero then all data vales will equal the mean. The general idea seems to be that as the standard deviation increases the data will be more widely dispersed around the mean. We have also seen that data can be distributed in a variety of ways. Consider the histograms in Figures 1, 2 & 3. These histograms represent the evaluation scores (on a scale of 1 to 5) for three instructors. In all three cases a group of 10 students provided feedback for each of the instructors. [caption id="attachment_3114" align="aligncenter" width="563"] Fig. 1 Instructor A[/caption] [caption id="attachment_3110" align="aligncenter" width="552"] Fig. 2 Instructor B[/caption] [caption id="attachment_3124" align="aligncenter" width="537"] Fig. 3 Instructor C[/caption] Referring to Figure 1, Instructor A received each possible score two times. Figure 1 represents a uniform distribution since every data value occurs with the same [pb_glossary id="4846"]frequency[/pb_glossary]. Referring to Figure 2, Instructor B received a mix of scores. Figure 2 represents a skewed distribution where one tail of the distribution is stretched out more than the other. Referring to Figure 3, Instructor C also received a mix of scores. Figure 3 represents a symmetrical distribution. Data values occur most often in the centre of the distribution and spread out equally on either side. The histogram in Figure 3 is symmetric but because it represents a very small sample size it appears to be a series of rectangles stacked side by side. As the sample size increases a symmetrical distribution will become less boxy as illustrated in Figures 4 & 5. [caption id="attachment_3128" align="aligncenter" width="300"] Fig. 4[/caption] [caption id="attachment_3129" align="aligncenter" width="300"] Fig. 5[/caption] Eventually if we consider the entire population the distribution will approach what is called the normal distribution as in Figure 6. This distribution is also called the bell curve. [caption id="attachment_4843" align="aligncenter" width="296"] Fig. 6[/caption] The normal distribution models many aspects of real life, including height, blood pressure and IQ scores.

Normal Distribution

The normal distribution is also called the bell curve. In a normal distribution the data values are [pb_glossary id="5980"]symmetrical[/pb_glossary] around a vertical line drawn through its centre which is also where the mean is located. Half of the data values lie on either side of the mean. In a normal distribution the mean, median and mode will all be equal.

Normal Distribution and Standard Deviation

The normal distribution will have symmetry in relation to the mean but it could be flat or high, depending on the standard deviation. In Figure 7 the means are identical but the distribution in A has a smaller standard deviation. [caption id="attachment_3135" align="aligncenter" width="376"] Fig. 7[/caption] The standard deviation plays an important role in the normal distribution, as described by the 68-95-99.7 Rule.

68-95-99.7 Rule

According to the 68-95-99.7 Rule:

Approximately 68% (68.26%) of the data items lie within one standard deviation of the mean.
Approximately 95% (95.44%) of the data items lie within two standard deviations of the mean.
Approximately 99.7% of the data items lie within three standard deviations of the mean.

Figure 8 depicts the 68-95-99.7 Rule. [caption id="attachment_3140" align="aligncenter" width="519"] Fig. 8 68-95-99.7 Rule https://commons.wikimedia.org/wiki[/caption] In simple language what does this rule tell us? Refer to Figure 8 and consider the percentage of data items that lie within one standard deviation (σ) of the mean. This is the region that lies between -1σ and 1σ. If 34.1% lie on either side of the mean, then 34.1% + 34.1% = 68.2% or approximately 68% . For a population with a normal distribution, just over two-thirds of the data (68%) will lie within one standard deviation of the mean. Consider the percentage of data items that lie within two standard deviations of the mean. This is the region that lies between -2σ and 2σ. Between -2σ and 0, we find 13.6% + 34.1% or 47.7% of the data values. So between -2σ and 2σ we will have 47.7% + 47.7% = 95.4% or approximately 95% . For a population with a normal distribution, approximately 95% of the data values will lie within two standard deviations of the mean. Consider the percentage of data items that lie within three standard deviations of the mean. This is the region that lies between -3σ and 3σ. Between -3σ and 0, we find 2.1% + 13.6% + 34.1% or 49.8% of the data values. So between -3σ and 3σ we will have 49.8% + 49.8% = 99.6%. (Note: The values in Figure 8 are all rounded to the nearest tenth. The number is actually closer to 49.86% x 2 = 99.72%). For a population with a normal distribution, approximately 99.7% of the data values will lie within three standard deviations of the mean. Another way of stating this, since 99.7% of the data will lie within three standard deviations of the mean, then only 0.3 % of the data will not lie within three standard deviations.

Using the 68-95-99.7 Rule

When working with a population that has a normal distribution the 68-95-99.7 Rule can be used to determine the percentage of the population that will be within one, two or three standard deviations of the mean.

EXAMPLE 1

A certain segment of the economy has a normally distributed salary, with a mean salary of $45,000 and a standard deviation of $4000. a) Determine the salary that is one standard deviation above the mean. b) Determine the salary that is three standard deviations below the mean. c) Determine the salary range for the employees that lie within one standard deviation of the mean. What percent of the employees lie in this salary range? d) Determine the salary range for the employees that lie within two standard deviations of the mean. What percent of the employees lie in this salary range? e) What percent of the employees earn a salary less than $33,000? Solution a) $45,000 + $4000 = $49,000 b) $45,000 - (3 x $4000) = $33,000 c) $45,000 ± $4000 = $41,000 to $49,000. According to the 68-95-99.7 Rule, sixty-eight percent of the employees for this segment of the economy lie within this salary range. d) $45,000 ± (2 x $4000) = $37,000 to $53,000. According to the 68-95-99.7 Rule, ninety-five percent of the employees for this segment of the economy lie within this salary range. e) A salary of $33,000 is 3 standard deviations below the mean. According to the 68-95-99.7 Rule, 100% - 99.7% or 0.3% of the employees lie above or below three standard deviations from the mean. Dividing 0.3 in half, we determine that 0.15% of the employees earn a salary less than $33,000.

TRY IT 1

Birth weights for newborns follow a normal distribution with a mean birth weight of 3.4 kg and a standard deviation of 0.55 kg. (Source O'Cathain et al) a) Determine the birth weight that is two standard deviations above the mean. b) Determine the birth weight that is one standard deviation below the mean. c) Determine the weight range for newborns that lie within two standard deviations of the mean. What percent of the newborns lie in this weight range? d) Determine the weight range for newborns that lie within three standard deviations of the mean. What percent of the newborns lie in this weight range? e) What percent of newborns have a mean birth weight greater than 4.5 kg? O'Cathain A., Walters S.J., Nicholl J.P., Thomas K.J., & Kirkham M. Use of evidence based leaflets to promote informed choice in maternity care: randomised controlled trial in everyday practice. British Medical Journal 2002; 324: 643-646

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8.4 Z-Scores and the Normal Curve

kmoshenko — Fri, 03 Jul 2020 21:59:44 +0000

Learning Objectives

By the end of this section it is expected that you will be able to:

Convert a data item to a z-score
Solve applications using z-score tables

The Normal Curve

When a set of data values is normally distributed, the 68-95-99.7 Rule can be used to determine the percentage of values that lie one, two or three standard deviations from the mean. We will shift gears and explore how to determine where a specific data value lies in relation to all other values. As an example, a student who has written a college entrance exam may want to know where they placed in comparison to all other students. This section will explore how to determine this. Consider the normal curve which is an idealized representation of a normally distributed population. The normal curve, also called a bell-shaped curve, is represented in Figure 1. The area under the curve represents 100% (or 1.00) of the data (or population) and the mean score is 0. [caption id="attachment_3163" align="aligncenter" width="409"] Fig. 1[/caption] We have seen that the standard deviation plays an important role in the normal distribution. Refer to Figure 2 for the visual representation of the 68 – 95 - 99.7 Rule. For a normally distributed set of data:

Approximately 68% (68.26%) of the data items fall within one standard deviation of the mean.
Approximately 95% (95.44%) of the data items fall within two standard deviations of the mean.
Approximately 99.7% of the data items fall within three standard deviations of the mean.

[caption id="attachment_3166" align="aligncenter" width="1024"] Fig. 2[/caption]

Z–Scores

When a data set is normally distributed we can use a standardized score, called the z-score, to determine the number of standard deviations that a data value is from the mean. Reconsider an example from the previous section. Acertain segment of the economy has a normally distributed salary, with a mean salary of $45,000 and a standard deviation of $4000. Refer to Figure 3. [caption id="attachment_4896" align="aligncenter" width="378"] Fig. 3[/caption] With this information we are able to determine that a salary of $49,000 lies exactly one standard deviation above the mean since $45,000 + $4000 = $49, 000. In turn, using the 68-95-99.7 Rule we can determine that a salary of $49,000 is higher than 84% of the other salaries for this segment of the economy. The calculation would be 50% + (68%/2) = 84%. This calculation was possible since $49,000 was exactly one standard deviation away from the mean. Consider a salary which does not lie exactly one, two or three standard deviations from the mean, such as $38,500. The calculation does not appear so straightforward but as it turns out we can use a z-score for situations such as this. A z-score converts a data value and standardizes it so that we are able to determine how many standard deviations a specific data value will lie above or below the mean. Z-scores can be used in situations with a normal distribution. Consider a chemistry class with a set of test scores that is normally distributed. The average score is 76% and one student receives a score of 55%. Converting the 55% to a z-score will provide the student with a sense of where their score lies with respect to the rest of the class. We can also use z-scores to determine the percent of the data values that will lie between any two data values. Perhaps we wish to determine the percentage of students whose test scores lie between 70% to 85%. This can also be done using z-scores. We have seen that when calculating standard deviation we must consider whether we are working with the entire population or a sample of the population. We must do the same when calculating a z-score.

Formulas for Finding Z-Scores

The z-score represents the number of standard deviations a data value is from the mean value. The formula for z is: For a population we calculate the z-score using the population mean μ and standard deviation σ. The data value is represented by x. For a representative sample of the population we calculate the z-score using the sample mean and standard deviation.:

A z-score is similar to a percentile in that it is a measure of position. As a rule, z-scores above 2.0 (or below –2.0) are considered “unusual” values. According to the 68-95-99.7 Rule, in a normal population such scores would occur less than 5% of the time. Z-scores between -2.0 and 2.0 are considered “ordinary” values and these represent 95% of the values.

EXAMPLE 1

IQ scores are normally distributed. The mean IQ is 100 and the standard deviation is 15. a) If Frank has an IQ of 127, find his z-score. b) Intrepret the meaning of this z-score. c) Using the 68-95-99.7 Rule, how does Frank's IQ compare to the rest of the population? Solution a) We will use the z-score for a population: Frank's z-score is 1.8. b) For this z-score, the mean of 100 has been “standardized” to a value of 0 and the score of 127 has been standardized to a value of 1.8. This means that Frank’s IQ score is 1.8 (almost 2 standard deviations) higher than the average. c) Considering the 68-95-99.7 Rule, Frank's score lies within 2 standard deviations of the mean. His score is certainly better than at least 84% of the population but does not rank in the top 2.5% of the population.

TRY IT 1

Consider a chemistry class and a set of test scores with an average of 76% and a standard deviation of 7%. A student receives a test score of 55%. a) Determine the student's z-score. b) Intrepret the meaning of this z-score. c) Using the 68-95-99.7 Rule, how does the student's test score compare to the rest of the class?

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3.1 Use the Rectangular Coordinate System

jgray — Fri, 07 Aug 2020 23:03:13 +0000

Learning Objectives

By the end of this section it is expected that you will be able to:

Plot points in a rectangular coordinate system
Verify solutions to an equation in two variables
Complete a table of solutions to a linear equation
Find solutions to a linear equation in two variables

Plot Points on a Rectangular Coordinate System

Just like maps use a grid system to identify locations, a grid system is used in algebra to show a relationship between two variables in a rectangular coordinate system. The rectangular coordinate system is also called the xy-plane or the ‘coordinate plane.’ The horizontal number line is called the x-axis. The vertical number line is called the y-axis. The x-axis and the y-axis together form the rectangular coordinate system. These axes divide a plane into four regions, called quadrants. The quadrants are identified by Roman numerals, beginning on the upper right and proceeding counterclockwise. See (Figure 1). ‘Quadrant’ has the root ‘quad,’ which means ‘four.’ [caption id="" align="aligncenter" width="342"] Figure .1[/caption] In the rectangular coordinate system, every point is represented by an ordered pair. The first number in the ordered pair is the x-coordinate of the point, and the second number is the y-coordinate of the point.

Ordered pair

An ordered pair, $\left(x,y\right)$, gives the coordinates of a point in a rectangular coordinate system.

The first number is the x-coordinate.

The second number is the y-coordinate.

The phrase ‘ordered pair’ means the order is important. What is the ordered pair of the point where the axes cross? At that point both coordinates are zero, so its ordered pair is $\left(0,0\right)$. The point $\left(0,0\right)$ has a special name. It is called the origin.

The origin

The point $\left(0,0\right)$ is called the origin. It is the point where the x-axis and y-axis intersect.

We use the coordinates to locate a point on the xy-plane. Let’s plot the point $\left(1,3\right)$ as an example. First, locate 1 on the x-axis and lightly sketch a vertical line through $x=1$. Then, locate 3 on the y-axis and sketch a horizontal line through $y=3$. Now, find the point where these two lines meet—that is the point with coordinates $\left(1,3\right)$. [caption id="" align="aligncenter" width="301"] Figure .2[/caption] Notice that the vertical line through $x=1$ and the horizontal line through $y=3$ are not part of the graph. We just used them to help us locate the point $\left(1,3\right)$.

EXAMPLE 1

Plot each point in the rectangular coordinate system and identify the quadrant in which the point is located: A $\left(-5,4\right)$ B $\left(-3,-4\right)$ C $\left(2,-3\right)$ D $\left(-2,3\right)$ E $\left(3,\frac{5}{2}\right)$. Solution The first number of the coordinate pair is the x-coordinate, and the second number is the y-coordinate.

Since $x=-5$, the point is to the left of the y-axis. Also, since $y=4$, the point is above the x-axis. The point $\left(-5,4\right)$ is in Quadrant II.
Since $x=-3$, the point is to the left of the y-axis. Also, since $y=-4$, the point is below the x-axis. The point $\left(-3,-4\right)$ is in Quadrant III.
Since $x=2$, the point is to the right of the y-axis. Since $y=-3$, the point is below the x-axis. The point $\left(2,-3\right)$ is in Quadrant lV.
Since $x=-2$, the point is to the left of the y-axis. Since $y=3$, the point is above the x-axis. The point $\left(-2,3\right)$ is in Quadrant II.
Since $x=3$, the point is to the right of the y-axis. Since $y=\frac{5}{2}$, the point is above the x-axis. (It may be helpful to write $\frac{5}{2}$ as a mixed number or decimal.) The point $\left(3,\frac{5}{2}\right)$ is in Quadrant I.

TRY IT 1

Plot each point in a rectangular coordinate system and identify the quadrant in which the point is located: A $\left(-2,1\right)$ B $\left(-3,-1\right)$ C $\left(4,-4\right)$ D $\left(-4,4\right)$ E $\left(-4,\frac{3}{2}\right)$.

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3.2 Graph Linear Equations in Two Variables

jgray — Fri, 07 Aug 2020 23:03:31 +0000

Learning Objectives

By the end of this section it is expected that you will be able to:

Recognize the relationship between the solutions of an equation and its graph.
Graph a linear equation by plotting points.
Graph vertical and horizontal lines.

Recognize the Relationship Between the Solutions of an Equation and its Graph

In the previous section, we found several solutions to the equation $3x+2y=6$. They are listed in the table below. So, the ordered pairs $\left(0,3\right)$, $\left(2,0\right)$, and $\left(1,\dfrac{3}{2}\right)$ are some solutions to the equation $3x+2y=6$. We can plot these solutions in the rectangular coordinate system as shown in (Figure 1).

$3x+2y=6$
$x$	$y$	$\left(x,y\right)$
0	3	$\left(0,3\right)$
2	0	$\left(2,0\right)$
1	$\dfrac{3}{2}$	$\left(1,\dfrac{3}{2}\right)$

[caption id="" align="aligncenter" width="301"] Figure .1[/caption]

Notice how the points line up perfectly? We connect the points with a line to get the graph of the equation $3x+2y=6$. See (Figure 2). Notice the arrows on the ends of each side of the line. These arrows indicate the line continues.

[caption id="" align="aligncenter" width="301"] Figure .2[/caption]

Every point on the line is a solution of the equation. Also, every solution of this equation is a point on this line. Points not on the line are not solutions.

Notice that the point whose coordinates are $\left(-2,6\right)$ is on the line shown in (Figure 3). If you substitute $x=-2$ and $y=6$ into the equation, you find that it is a solution to the equation.

[caption id="" align="aligncenter" width="301"] Figure .3[/caption]

So the point $\left(-2,6\right)$ is a solution to the equation $3x+2y=6$. (The phrase “the point whose coordinates are $\left(-2,6\right)$” is often shortened to “the point $\left(-2,6\right)$.”)

So $\left(4,1\right)$ is not a solution to the equation $3x+2y=6$. Therefore, the point $\left(4,1\right)$ is not on the line. See (Figure 2). This is an example of the saying, “A picture is worth a thousand words.” The line shows you all the solutions to the equation. Every point on the line is a solution of the equation. And, every solution of this equation is on this line. This line is called the graph of the equation $3x+2y=6$.

Graph of a linear equation

The graph of a linear equation $Ax+By=C$ is a line.

Every point on the line is a solution of the equation.
Every solution of this equation is a point on this line.

EXAMPLE 1

The graph of $y=2x-3$ is shown.

For each ordered pair, decide:

a) Is the ordered pair a solution to the equation? b) Is the point on the line?

A $\left(0,-3\right)$ B $\left(3,3\right)$ C $\left(2,-3\right)$ D $\left(-1,-5\right)$

Solution

Substitute the x- and y- values into the equation to check if the ordered pair is a solution to the equation.

a) b) Plot the points A $\left(0,3\right)$, B $\left(3,3\right)$, C $\left(2,-3\right)$, and D $\left(-1,-5\right)$.

The points $\left(0,3\right)$, $\left(3,3\right)$, and $\left(-1,-5\right)$ are on the line $y=2x-3$, and the point $\left(2,-3\right)$ is not on the line.

The points that are solutions to $y=2x-3$ are on the line, but the point that is not a solution is not on the line.

TRY IT 1

Use the graph of $y=3x-1$ to decide whether each ordered pair is:

a solution to the equation.
on the line.

a) $\left(0,-1\right)$ b) $\left(2,5\right)$

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3.3 Graph with Intercepts - optional

jgray — Fri, 07 Aug 2020 23:03:44 +0000

Learning Objectives

By the end of this section it is expected that you will be able to:

Identify the $x$- and $y$- intercepts on a graph
Find the $x$- and $y$- intercepts from an equation of a line
Graph a line using the intercepts

Identify the x- and y- Intercepts on a Graph

Every linear equation can be represented by a unique line that shows all the solutions of the equation. We have seen that when graphing a line by plotting points, you can use any three solutions to graph. This means that two people graphing the line might use different sets of three points.

At first glance, their two lines might not appear to be the same, since they would have different points labeled. But if all the work was done correctly, the lines should be exactly the same. One way to recognize that they are indeed the same line is to look at where the line crosses the x- axis and the y- axis. These points are called the intercepts of the line.

Intercepts of a line

The points where a line crosses the x- axis and the y- axis are called the intercepts of a line.

Let’s look at the graphs of the lines in (Figure 1).

Examples of graphs crossing the x-negative axis.

[caption id="" align="aligncenter" width="648"] Figure .1[/caption]

First, notice where each of these lines crosses the $x$ negative axis. See (Figure 1).

Figure	The line crosses the x- axis at:	Ordered pair of this point
Figure (a)	3	\(\left(3,0\right)\)
Figure (b)	4	\(\left(4,0\right)\)
Figure (c)	5	\(\left(5,0\right)\)
Figure (d)	0	\(\left(0,0\right)\)

Figure

The line crosses the x- axis at:

Ordered pair of this point

Figure (a)

$\left(3,0\right)$

Figure (b)

$\left(4,0\right)$

Figure (c)

$\left(5,0\right)$

Figure (d)

$\left(0,0\right)$

Do you see a pattern?

For each row, the y- coordinate of the point where the line crosses the x- axis is zero. The point where the line crosses the x- axis has the form $\left(a,0\right)$ and is called the x- intercept of a line. The x- intercept occurs when $y$ is zero.

Now, let’s look at the points where these lines cross the y- axis. See the table below.

Figure	The line crosses the y-axis at:	Ordered pair for this point
Figure (a)	6	\(\left(0,6\right)\)
Figure (b)	\(-3\)	\(\left(0,-3\right)\)
Figure (c)	\(-5\)	\(\left(0,5\right)\)
Figure (d)	0	\(\left(0,0\right)\)

Figure

The line crosses the y-axis at:

Ordered pair for this point

Figure (a)

$\left(0,6\right)$

Figure (b)

$-3$

$\left(0,-3\right)$

Figure (c)

$-5$

$\left(0,5\right)$

Figure (d)

$\left(0,0\right)$

What is the pattern here?

In each row, the x- coordinate of the point where the line crosses the y- axis is zero. The point where the line crosses the y- axis has the form $\left(0,b\right)$ and is called the y- intercept of the line. The y- intercept occurs when $x$ is zero.

x- intercept and y- intercept of a line

The x- intercept is the point $\left(a,0\right)$ where the line crosses the x- axis.

The y- intercept is the point $\left(0,b\right)$ where the line crosses the y- axis.

EXAMPLE 1

Find the x- and y- intercepts on each graph.

Solution

a) The graph crosses the x- axis at the point $\left(4,0\right)$. The x- intercept is $\left(4,0\right)$. The graph crosses the y- axis at the point $\left(0,2\right)$. The y- intercept is $\left(0,2\right)$. b) The graph crosses the x- axis at the point $\left(2,0\right)$. The x- intercept is $\left(2,0\right)$ The graph crosses the y- axis at the point $\left(0,-6\right)$. The y- intercept is $\left(0,-6\right)$. c) The graph crosses the x- axis at the point $\left(-5,0\right)$. The x- intercept is $\left(-5,0\right)$. The graph crosses the y- axis at the point $\left(0,-5\right)$. The y- intercept is $\left(0,-5\right)$.

TRY IT 1

Find the x- and y- intercepts on the graph.

Show answer

3.4 Understand Slope of a Line - optional

jgray — Fri, 07 Aug 2020 23:04:04 +0000

Learning Objectives

By the end of this section it is expected that you will be able to:

Use $m=\dfrac{rise}{run}$ to find the slope of a line from its graph
Find the slope of horizontal and vertical lines
Use the slope formula to find the slope of a line between two points
Graph a line given a point and the slope
Solve slope applications

When you graph linear equations, you may notice that some lines tilt up as they go from left to right and some lines tilt down. Some lines are very steep and some lines are flatter. What determines whether a line tilts up or down or if it is steep or flat?

In mathematics, the ‘tilt’ of a line is called the slope of the line. The concept of slope has many applications in the real world. The pitch of a roof, grade of a highway, and a ramp for a wheelchair are some examples where you literally see slopes. And when you ride a bicycle, you feel the slope as you pump uphill or coast downhill.

In this section, we will explore the concept of slope.

The slope of a line is the ratio of the rise to the run. In mathematics, it is always referred to with the letter $m$.

Slope of a line

The slope of a line of a line is $m=\dfrac{\text{rise}}{\text{run}}$.

The rise measures the vertical change and the run measures the horizontal change between two points on the line.

Positive and negative slopes

We ‘read’ a line from left to right just like we read words in English. As you read from left to right, the line is going up; it has positive slope. The line is going down; it has negative slope.

Use $m=\dfrac{\text{rise}}{\text{run}}}$ to Find the Slope of a Line from its Graph

We’ll look at some graphs on the $xy$-coordinate plane and see how to find their slopes.

To find the slope, we must count out the rise and the run. But where do we start?

We locate two points on the line whose coordinates are integers. We then start with the point on the left and sketch a right triangle, so we can count the rise and run.

EXAMPLE 1

How to Use $m=\dfrac{\text{rise}}{\text{run}}$ to Find the Slope of a Line from its Graph

Find the slope of the line shown.

Solution

TRY IT 1

Find the slope of the line shown.

Show answer

3.5 Use the Slope–Intercept Form of an Equation of a Line -optional

jgray — Fri, 07 Aug 2020 23:04:20 +0000

Learning Objectives

By the end of this section it is expected that you will be able to:

Recognize the relation between the graph and the slope–intercept form of an equation of a line
Identify the slope and y-intercept form of an equation of a line
Graph a line using its slope and intercept
Choose the most convenient method to graph a line
Graph and interpret applications of slope–intercept
Use slopes to identify parallel lines
Use slopes to identify perpendicular lines

Recognize the Relation Between the Graph and the Slope–Intercept Form of an Equation of a Line

We have graphed linear equations by plotting points, using intercepts, recognizing horizontal and vertical lines, and using the point–slope method. Once we see how an equation in slope–intercept form and its graph are related, we’ll have one more method we can use to graph lines.

See (Figure). Let’s find the slope of this line.

The red lines show us the rise is 1 and the run is 2. Substituting into the slope formula:

$\begin{array}{ccc}\hfill m& =\hfill & \dfrac{\text{rise}}{\text{run}}\hfill \\ \hfill m& =\hfill & \dfrac{1}{2}\hfill \end{array}$

What is the y-intercept of the line? The y-intercept is where the line crosses the y-axis, so y-intercept is $\left(0,3\right)$. The equation of this line is:

Notice, the line has:

When a linear equation is solved for $y$, the coefficient of the $x$ term is the slope and the constant term is the y-coordinate of the y-intercept. We say that the equation $y=\dfrac{1}{2}x+3$ is in slope–intercept form.

Slope-intercept form of an equation of a line

The slope–intercept form of an equation of a line with slope $m$ and y-intercept, $\left(0,b\right)$ is,

$y=mx+b$

Sometimes the slope–intercept form is called the “y-form.”

EXAMPLE 1

Use the graph to find the slope and y-intercept of the line, $y=2x+1$.

Compare these values to the equation$y=mx+b$.

Solution

To find the slope of the line, we need to choose two points on the line. We’ll use the points $\left(0,1\right)$ and $\left(1,3\right)$.


Find the rise and run.


Find the y-intercept of the line.	The y-intercept is the point (0, 1).

The slope is the same as the coefficient of $x$ and the y-coordinate of the y-intercept is the same as the constant term.

TRY IT 1

Use the graph to find the slope and y-intercept of the line $y=\dfrac{2}{3}x-1$. Compare these values to the equation $y=mx+b$.

Show answer

2.1 Solve Linear Equations

jgray — Wed, 12 Aug 2020 18:25:20 +0000

Learning Objectives

By the end of this section it is expected that you will be able to:

Verify a solution of an equation
Solve equations using the Subtraction and Addition Properties of Equality
Solve equations using the Division and Multiplication Properties of Equality
Solve an equation with variables and constants on both sides

Verify a Solution of an Equation

Solving an equation is like discovering the answer to a puzzle. The purpose in solving an equation is to find the value or values of the variable that make each side of the equation the same – so that we end up with a true statement. Any value of the variable that makes the equation true is called a solution to the equation. It is the answer to the puzzle!

Solution of an equation

A solution of an equation is a value of a variable that makes a true statement when substituted into the equation.

HOW TO: Determine whether a number is a solution to an equation

Substitute the number in for the variable in the equation.
Simplify the expressions on both sides of the equation.
Determine whether the resulting equation is true (the left side is equal to the right side).
- If it is true, the number is a solution.
- If it is not true, the number is not a solution.

EXAMPLE 1

Determine whether $x=-3$ is a solution of $4x+5=-7$.

Solution

Since a solution to an equation is a value of the variable that makes the equation true, begin by substituting the value of the solution for the variable.

Substitute $x=-3$ for x	$4(-3)+5=-7$
Multiply.	$-12+5=-7$
Simplify.	$-7=-7$

Since $x=-3$ results in a true statement (-7 is in fact equal to -7), -3 is a solution to the equation $4x+5=-7$.

TRY IT 1

Is $y=2$ a solution of $9y-2=6$?

Show answer

2.2 Use a General Strategy to Solve Linear Equations

jgray — Wed, 12 Aug 2020 18:26:04 +0000

Learning Objectives

By the end of this section it is expected that you will be able to:

Solve equations using a general strategy
Classify equations

Solve Equations Using the General Strategy

Until now we have dealt with solving one specific form of a linear equation. It is time now to lay out one overall strategy that can be used to solve any linear equation. Some equations we solve will not require all these steps to solve, but many will.

Beginning by simplifying each side of the equation makes the remaining steps easier.

EXAMPLE 1

How to Solve Linear Equations Using the General Strategy

Solve: $-6\left(x+3\right)=24.$

Solution

TRY IT 1

Solve: $5\left(x+3\right)=35.$

Show answer

2.4 Solve a Formula for a Specific Variable

jgray — Wed, 12 Aug 2020 18:26:29 +0000

Learning Objectives

By the end of this section it is expected that you will be able to:

Solve a formula for a specific variable
Use formulas to solve applications

Solve a Formula for Specific Variable

In the last section we have worked with some geometry formulas when solving problems. A formula is a mathematical description of the relationship between variables. Formulas are also used in the sciences, such as chemistry, physics, and biology. In medicine they are used for calculations for dispensing medicine or determining body mass index. Spreadsheet programs rely on formulas to make calculations. It is important to be familiar with formulas and be able to manipulate them easily.

To solve a formula for a specific variable means to isolate that variable on one side of the equals sign with a coefficient of 1. All other variables and constants are on the other side of the equals sign. To see how to solve a formula for a specific variable, we will start with the distance, rate and time formula.

EXAMPLE 1

Solve the formula $d=rt$ for $t$:

when $d=520$ and $r=65$
in general

Solution

We will write the solutions side-by-side to demonstrate that solving a formula in general uses the same steps as when we have numbers to substitute.

a) when $d=520$ and $r=65$		b) in general
Write the formula.	$\phantom{\rule{1em}{0ex}}d=rt$	Write the formula.	$d=rt$
Substitute.	$520=65t$
Divide, to isolate $t$.	$\frac{520}{65}=\frac{65t}{65}$	Divide, to isolate $t$.	$\frac{d}{r}=\frac{rt}{r}$
Simplify.	$\phantom{\rule{1.2em}{0ex}}8=t$	Simplify.	$\frac{d}{r}=t$

We say the formula $t=\frac{d}{r}$ is solved for $t$.

TRY IT 1

Solve the formula $d=rt$ for $r$:

a) when $d=180\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}t=4$ b) in general

Show answer

2.3 Use a Problem Solving Strategy

imazur — Sat, 01 May 2021 01:10:20 +0000

Learning Objectives

By the end of this section it is expected that you will be able to:

Translate to an equation and solve
Translate and solve applications

Translate to an Equation and Solve

To solve applications algebraically, we will begin by translating from English sentences into equations. Our first step is to look for the word (or words) that would translate to the equals sign. In the next few examples, we will translate sentences into equations and then solve the equations.

EXAMPLE 1

Translate and solve: Eleven more than x is equal to 54.

Solution

Translate.
Subtract 11 from both sides.
Simplify.
Check: Is 54 eleven more than 43? $\begin{array}{ccc}\hfill 43+11& \stackrel{?}{=}\hfill & 54\hfill \\ \hfill 54& =\hfill & 54✓\hfill \end{array}$

TRY IT 1

Translate and solve: Ten more than x is equal to 41.

Show answer

5.1 Use Properties of Angles, Triangles, and the Pythagorean Theorem

imazur — Thu, 20 May 2021 05:57:01 +0000

[latexpage]

Learning Objectives

By the end of this section it is expected that you will be able to

Use the properties of angles
Use the properties of triangles
Use the Pythagorean Theorem

Use the Properties of Angles

Are you familiar with the phrase ‘do a $180\text{’?}$ It means to make a full turn so that you face the opposite direction. It comes from the fact that the measure of an angle that makes a straight line is $180$ degrees. See (Figure 1).

[caption id="" align="aligncenter" width="152"] Figure 1[/caption]

An angle is formed by two rays that share a common endpoint. Each ray is called a side of the angle and the common endpoint is called the vertex. An angle is named by its vertex. In (Figure 2), $\angle A$ is the angle with vertex at point $A$. The measure of $\angle A$ is written $m\angle A$.

$\angle A$ is the angle with vertex at $\text{point}\phantom{\rule{0.2em}{0ex}}A$.

[caption id="" align="aligncenter" width="201"] Figure 2[/caption]

We measure angles in degrees, and use the symbol ° to represent degrees. We use the abbreviation $m$ to for the measure of an angle. So if $\angle A$ is 27°, we would write $m\angle A=27$.

If the sum of the measures of two angles is $\text{180}$°, then they are called supplementary angles. In (Figure 3), each pair of angles is supplementary because their measures add to $\text{180}$°. Each angle is the supplement of the other.

The sum of the measures of supplementary angles is $\text{180}$°.

[caption id="" align="aligncenter" width="574"] Figure 3[/caption]

If the sum of the measures of two angles is $\text{90}$°, then the angles are complementary angles. In (Figure 4), each pair of angles is complementary, because their measures add to $\text{90}$°. Each angle is the complement of the other.

The sum of the measures of complementary angles is $\text{90}$°.

[caption id="" align="aligncenter" width="286"] Figure 4[/caption]

Supplementary and Complementary Angles

If the sum of the measures of two angles is $\text{180}$°, then the angles are supplementary.

If $\angle A$ and $\angle B$ are supplementary, then $m\angle A+m\angle B=\text{180}$°.

If the sum of the measures of two angles is $\text{90}$°, then the angles are complementary.

If $\angle A$ and $\angle B$ are complementary, then $m\angle A+m\angle B=\text{90}$°.

In this section and the next, you will be introduced to some common geometry formulas. We will adapt our Problem Solving Strategy for Geometry Applications. The geometry formula will name the variables and give us the equation to solve.

In addition, since these applications will all involve geometric shapes, it will be helpful to draw a figure and then label it with the information from the problem. We will include this step in the Problem Solving Strategy for Geometry Applications.

HOW TO: Use a Problem Solving Strategy for Geometry Applications

Read the problem and make sure you understand all the words and ideas. Draw a figure and label it with the given information.
Identify what you are looking for.
Name what you are looking for and choose a variable to represent it.
Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.
Solve the equation using good algebra techniques.
Check the answer in the problem and make sure it makes sense.
Answer the question with a complete sentence.

The next example will show how you can use the Problem Solving Strategy for Geometry Applications to answer questions about supplementary and complementary angles.

EXAMPLE 1

An angle measures $\text{40}$°. Find a) its supplement, and b) its complement.

Solution

a)
Step 1. Read the problem. Draw the figure and label it with the given information.
Step 2. Identify what you are looking for.
Step 3. Name. Choose a variable to represent it.
Step 4. Translate. Write the appropriate formula for the situation and substitute in the given information.
Step 5. Solve the equation.
Step 6. Check:
Step 7. Answer the question.

b)
Step 1. Read the problem. Draw the figure and label it with the given information.
Step 2. Identify what you are looking for.
Step 3. Name. Choose a variable to represent it.
Step 4. Translate. Write the appropriate formula for the situation and substitute in the given information.
Step 5. Solve the equation.
Step 6. Check:
Step 7. Answer the question.

TRY IT 1

An angle measures $\text{25}$°. Find its: a) supplement b) complement.

Show answer

5.2 Solve Applications: Sine, Cosine and Tangent Ratios.

imazur — Thu, 20 May 2021 06:03:17 +0000

[latexpage]

Learning Objectives

By the end of this section it is expected that you will be able to

Find missing side of a right triangle using sine, cosine, or tangent ratios
Find missing angle of a right triangle using sine, cosine, or tangent ratios
Solve applications using right angle trigonometry

Now, that we know the fundamentals of algebra and geometry associated with a right triangle, we can start exploring trigonometry. Many real life problems can be represented and solved using right angle trigonometry.

Sine, Cosine, and Tangent Ratios

We know that any right triangle has three sides and a right angle. The side opposite to the right angle is called the hypotenuse. The other two angles in a right triangle are acute angles (with a measure less than 90 degrees). One of those angles we call reference angle and we use θ (theta) to represent it. The hypotenuse is always the longest side of a right triangle. The other two sides are called opposite side and adjacent side. The names of those sides depends on which of the two acute angles is being used as a reference angle.

[caption id="attachment_9622" align="aligncenter" width="420"] Figure 1.[/caption]

In the right triangle each side is labeled with a lowercase letter to match the uppercase letter of the opposite vertex.

EXAMPLE 1

Label the sides of the triangle and find the hypotenuse, opposite, and adjacent. Solution We labeled the sides with a lowercase letter to match the uppercase letter of the opposite vertex. c is hypotenuse a is opposite b is adjacent

TRY IT 1

Label the sides of the triangle and find the hypotenuse, opposite and adjacent.

Show answer

1.6 Roots and Radicals

imazur — Wed, 02 Jun 2021 05:24:26 +0000

Learning Objectives

By the end of this section it is expected that you will be able to:

Simplify expressions with roots
Estimate and approximate roots
Use radicals in applications

Simplify Expressions with Square Roots

Remember that when a number n is multiplied by itself, we write ${n}^{2}$ and read it “n squared.” The result is called the square of n. For example,

$\begin{array}{cccc}{8}^{2}\hfill & & & \text{read}\phantom{\rule{0.2em}{0ex}}\text{‘}8\phantom{\rule{0.2em}{0ex}}\text{squared’}\hfill \\ 64\hfill & & & 64\phantom{\rule{0.2em}{0ex}}\text{is called the}\phantom{\rule{0.2em}{0ex}}\mathit{\text{square}}\phantom{\rule{0.2em}{0ex}}\text{of}\phantom{\rule{0.2em}{0ex}}8.\hfill \end{array}$

Similarly, 121 is the square of 11, because ${11}^{2}$ is 121.

Square of a Number

If ${n}^{2}=m,$ then m is the square of n.

Complete the following table to show the squares of the counting numbers 1 through 15.

The numbers in the second row are called perfect square numbers. It will be helpful to learn to recognize the perfect square numbers.

The squares of the counting numbers are positive numbers. What about the squares of negative numbers? We know that when the signs of two numbers are the same, their product is positive. So the square of any negative number is also positive.

$\begin{array}{cccccccccc}{\left(-3\right)}^{2}=9\hfill & & & {\left(-8\right)}^{2}=64\hfill & & & {\left(-11\right)}^{2}=121\hfill & & & {\left(-15\right)}^{2}=225\hfill \end{array}$

Did you notice that these squares are the same as the squares of the positive numbers?

Sometimes we will need to look at the relationship between numbers and their squares in reverse. Because ${10}^{2}=100,$ we say 100 is the square of 10. We also say that 10 is a square root of 100. A number whose square is $m$ is called a square root of m.

Square Root of a Number

If ${n}^{2}=m,$ then n is a square root of m.

Notice ${\left(-10\right)}^{2}=100$ also, so $-10$ is also a square root of 100. Therefore, both 10 and $-10$ are square roots of 100.

So, every positive number has two square roots—one positive and one negative. What if we only wanted the positive square root of a positive number? The radical sign, $\sqrt{m},$ denotes the positive square root. The positive square root is called the principal square root. When we use the radical sign that always means we want the principal square root.

We also use the radical sign for the square root of zero. Because ${0}^{2}=0,$ $\sqrt{0}=0.$ Notice that zero has only one square root.

Square Root Notation

$\sqrt{m}$ is read “the square root of m”

If $m={n}^{2},$ then $\sqrt{m}=n,$ for $n\ge 0.$

The square root of m, $\sqrt{m},$ is the positive number whose square is m.

Since 10 is the principal square root of 100, we write $\sqrt{100}=10.$ You may want to complete the following table to help you recognize square roots.

EXAMPLE 1

Simplify: a) $\sqrt{25}$ b) $\sqrt{121}.$

Solution

a) Since ${5}^{2}=25$	$\begin{array}{c}\sqrt{25}\\ 5\end{array}$
b) Since ${11}^{2}=121$	$\begin{array}{c}\sqrt{121}\\ 11\end{array}$

TRY IT 1

Simplify: a) $\sqrt{36}$ b) $\sqrt{169}.$

Show answer

1.1 Algebraic Expressions

imazur — Wed, 02 Jun 2021 05:50:44 +0000

Learning Objectives

By the end of this section it is expected that you will be able to:

Use variables and algebraic symbols
Identify expressions and equations
Simplify expressions with exponents
Simplify expressions using the order of operations
Evaluate algebraic expressions

Use Variables and Algebraic Symbols

In algebra, letters of the alphabet are used to represent variables.

Letters often used for variables are $x,y,a,b,\text{and}\phantom{\rule{0.2em}{0ex}}c$.

Variables and Constants

A variable is a letter that represents a number or quantity whose value may change.

A constant is a number whose value always stays the same.

To write algebraically, we need some symbols as well as numbers and variables. There are several types of symbols we will be using. There are four basic arithmetic operations: addition, subtraction, multiplication, and division. We will summarize them here, along with words we use for the operations and the result.

Operation	Notation	Say:	The result is…
Addition	$a+b$	$a\phantom{\rule{0.2em}{0ex}}\text{plus}\phantom{\rule{0.2em}{0ex}}b$	the sum of $a$ and $b$
Subtraction	$a-b$	$a\phantom{\rule{0.2em}{0ex}}\text{minus}\phantom{\rule{0.2em}{0ex}}b$	the difference of $a$ and $b$
Multiplication	$a\cdot b,\left(a\right)\left(b\right),\left(a\right)b,a\left(b\right)$	$a\phantom{\rule{0.2em}{0ex}}\text{times}\phantom{\rule{0.2em}{0ex}}b$	The product of $a$ and $b$
Division	$a\div b,a/b,\phantom{\rule{0.2em}{0ex}}\frac{a}{b},b\overline{)a}$	$a$ divided by $b$	The quotient of $a$ and $b$

In algebra, the cross symbol, ×, is not used to show multiplication because that symbol may cause confusion. Does $3xy$ mean $3 \phantom{\rule{0.2em}{0ex}} \times \phantom{\rule{0.2em}{0ex}}y$ (three times $y$) or $3 \cdot x \cdot y$ (three times $x\phantom{\rule{0.2em}{0ex}}\text{times}\phantom{\rule{0.2em}{0ex}}y$)? To make it clear, use • or parentheses for multiplication.When two quantities have the same value, we say they are equal and connect them with an equal sign.

Equality Symbol

$a=b\phantom{\rule{0.2em}{0ex}}\text{is read}\phantom{\rule{0.2em}{0ex}}a\phantom{\rule{0.2em}{0ex}}\text{is equal to}\phantom{\rule{0.2em}{0ex}}b$

The symbol $=$ is called the equal sign.

An inequality is used in algebra to compare two quantities that may have different values. The number line can help you understand inequalities. Remember that on the number line the numbers get larger as they go from left to right. So if we know that $b$ is greater than $a$, it means that $b$ is to the right of $a$ on the number line. We use the symbols < and > for inequalities.

Inequality

$a$ < $b$ is read $a$ is less than $b$

$a$ is to the left of $b$ on the number line

$a$ > $b$ is read $a$ is greater than $b$

$a$ is to the right of $b$ on the number line

The expressions $a$ < $b\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}a$ > $\phantom{\rule{0.2em}{0ex}}b$ can be read from left-to-right or right-to-left, though in English we usually read from left-to-right. In general,

When we write an inequality symbol with a line under it, such as $a\le b$, it means \(a

We summarize the symbols of equality and inequality in the table below.

Algebraic Notation	Say
\(a=b\)	\(a\) is equal to \(b\)
\(a\ne b\)	\(a\) is not equal to \(b\)
\(a\) < \(b\)	\(a\) is less than \(b\)
\(a\) > \(b\)	\(a\) is greater than \(b\)
\(a\le b\)	\(a\) is less than or equal to \(b\)
\(a\ge b\)	\(a\) is greater than or equal to \(b\)

Algebraic Notation

Say

$a=b$

$a$ is equal to $b$

$a\ne b$

$a$ is not equal to $b$

$a$ < $b$

$a$ is less than $b$

$a$ > $b$

$a$ is greater than $b$

$a\le b$

$a$ is less than or equal to $b$

$a\ge b$

$a$ is greater than or equal to $b$

Symbols < and >

The symbols < and > each have a smaller side and a larger side.

smaller side < larger side larger side > smaller side

The smaller side of the symbol faces the smaller number and the larger faces the larger number.

Grouping symbols in algebra are much like the commas, colons, and other punctuation marks in written language. They indicate which expressions are to be kept together and separate from other expressions. The table below lists three of the most commonly used grouping symbols in algebra.

Common Grouping Symbols
Name	Symbol
parentheses	$\left(\phantom{\rule{0.5em}{0ex}}\right)$
brackets	$\left[\phantom{\rule{0.5em}{0ex}}\right]$
braces	$\left\{\phantom{\rule{0.5em}{0ex}}\right\}$

Here are some examples of expressions that include grouping symbols. We will simplify expressions like these later in this section.

$8\left(14-8\right)\phantom{\rule{4em}{0ex}}21-3\left[2+4\left(9-8\right)\right]\phantom{\rule{4em}{0ex}}24\div \left\{13-2\left[1\left(6-5\right)+4\right]\right\}$

Identify Expressions and Equations

What is the difference in English between a phrase and a sentence? A phrase expresses a single thought that is incomplete by itself, but a sentence makes a complete statement. “Running very fast” is a phrase, but “The football player was running very fast” is a sentence. A sentence has a subject and a verb.

In algebra, we have expressions and equations. An expression is like a phrase. Here are some examples of expressions and how they relate to word phrases:

Expression	Words	Phrase
$3+5$	$3\phantom{\rule{0.2em}{0ex}}\text{plus}\phantom{\rule{0.2em}{0ex}}5$	the sum of three and five
$n-1$	$n$ minus one	the difference of $n$ and one
$6\cdot 7$	$6\phantom{\rule{0.2em}{0ex}}\text{times}\phantom{\rule{0.2em}{0ex}}7$	the product of six and seven
$\frac{x}{y}$	$x$ divided by $y$	the quotient of $x$ and $y$

Notice that the phrases do not form a complete sentence because the phrase does not have a verb. An equation is two expressions linked with an equal sign. When you read the words the symbols represent in an equation, you have a complete sentence in English. The equal sign gives the verb. Here are some examples of equations:

Equation	Sentence
$3+5=8$	The sum of three and five is equal to eight.
$n-1=14$	$n$ minus one equals fourteen.
$6\cdot 7=42$	The product of six and seven is equal to forty-two.
$x=53$	$x$ is equal to fifty-three.
$y+9=2y-3$	$y$ plus nine is equal to two $y$ minus three.

Expressions and Equations

An expression is a number, a variable, or a combination of numbers and variables and operation symbols.

An equation is made up of two expressions connected by an equal sign.

EXAMPLE 1

Determine if each is an expression or an equation:

$\phantom{\rule{0.2em}{0ex}}16-6=10$
$\phantom{\rule{0.2em}{0ex}}4\cdot 2+1$
$\phantom{\rule{0.2em}{0ex}}x\div 25$
$\phantom{\rule{0.2em}{0ex}}y+8=40$

Solution

a. $\phantom{\rule{0.2em}{0ex}}16-6=10$	This is an equation—two expressions are connected with an equal sign.
b. $\phantom{\rule{0.2em}{0ex}}4\cdot 2+1$	This is an expression—no equal sign.
c. $\phantom{\rule{0.2em}{0ex}}x\div 25$	This is an expression—no equal sign.
d. $\phantom{\rule{0.2em}{0ex}}y+8=40$	This is an equation—two expressions are connected with an equal sign.

TRY IT 1

Determine if each is an expression or an equation:

$\phantom{\rule{0.2em}{0ex}}23+6=29\phantom{\rule{0.4em}{0ex}}$
$\phantom{\rule{0.2em}{0ex}}7\cdot 3-7$

Show Answer

Introduction

jgray — Mon, 18 Mar 2019 17:38:55 +0000

This is where you can write your introduction.

absolute frequency

kmoshenko — Tue, 13 Aug 2019 21:47:10 +0000

the number of occurrences of a particular value or characteristic

relative frequency

kmoshenko — Tue, 13 Aug 2019 21:48:27 +0000

the number of times a particular value occurs in relation to the total number of occurences

frequency distribution table

kmoshenko — Thu, 15 Aug 2019 22:56:49 +0000

shows the number of occurrences of a characteristic or outcome

slope

kmoshenko — Thu, 29 Aug 2019 03:44:45 +0000

indicates the steepness of a line

extrapolating

kmoshenko — Thu, 29 Aug 2019 04:16:11 +0000

estimating or drawing conclusions based on trends in the data

median

kmoshenko — Tue, 20 Apr 2021 21:00:35 +0000

The median is the data item in the middle of each set of ranked, or ordered, data. The median separates the upper half and the lower half of a data set.

mean

kmoshenko — Thu, 22 Apr 2021 22:12:34 +0000

the sum of the data values divided by the number of data values

outlier

kmoshenko — Thu, 22 Apr 2021 22:32:41 +0000

a data point that is substantially different from the other data values

sample

kmoshenko — Wed, 28 Apr 2021 18:42:07 +0000

a subset of the larger population under study

range

kmoshenko — Wed, 28 Apr 2021 20:56:55 +0000

the difference between the highest and lowest values in a set of data

standard deviation

kmoshenko — Wed, 28 Apr 2021 20:58:12 +0000

a measure of dispersion of all data values around the mean of a data set

1.2 Integers

imazur — Wed, 02 Jun 2021 05:52:21 +0000

Learning Objectives

By the end of this section it is expected that you will be able to:

Use negatives and opposites
Simplify: expressions with absolute value
Add ans subtract integers
Multiply and divide integers
Simplify Expressions with Integers

Use Negatives and Opposites

If you have ever experienced a temperature below zero or accidentally overdrawn your checking account, you are already familiar with negative numbers. Negative numbers are numbers less than $0$. The negative numbers are to the left of zero on the number line. See Figure 1.

[caption id="" align="aligncenter" width="389"] Figure 1 The number line shows the location of positive and negative numbers.[/caption] The arrows on the ends of the number line indicate that the numbers keep going forever. There is no biggest positive number, and there is no smallest negative number.

Is zero a positive or a negative number? Numbers larger than zero are positive, and numbers smaller than zero are negative. Zero is neither positive nor negative.

Consider how numbers are ordered on the number line. Going from left to right, the numbers increase in value. Going from right to left, the numbers decrease in value. See Figure 2.

[caption id="" align="aligncenter" width="440"] Figure 2 The numbers on a number line increase in value going from left to right and decrease in value going from right to left.[/caption]

Remember that we use the notation:

a < b (read “a is less than b”) when a is to the left of b on the number line.

a > b (read “a is greater than b”) when a is to the right of b on the number line.

Now we need to extend the number line which showed the whole numbers to include negative numbers, too. The numbers marked by points in Figure 3 are called the integers. The integers are the numbers $\text{…}-3,-2,-1,0,1,2,3\text{…}$

[caption id="" align="aligncenter" width="401"] Figure 3 All the marked numbers are called integers.[/caption]

EXAMPLE 1

Order each of the following pairs of numbers, using < or >: a) $14$___$6$ b) $-1$___$9$ c) $-1$___$-4$ d) $2$___$-20$.

Solution

It may be helpful to refer to the number line shown.

a) 14 is to the right of 6 on the number line.	$14$___$6$ $14$ > $6$
b) −1 is to the left of 9 on the number line.	$-1$___$9$ $-1<9$
c) −1 is to the right of −4 on the number line.	$-1$___$-4$
d) 2 is to the right of −20 on the number line.	$2$___$-20$ $2$ > $-20$

TRY IT 1

Order each of the following pairs of numbers, using < or > $\text{:}$ a) $15$___$7$ b) $-2$___$5$ c) $-3$___$-7$ d) $5$___$-17$.

Show answer

6.1 Measurement; Health Applications

imazur — Fri, 04 Jun 2021 20:17:27 +0000

Learning Objectives

By the end of this section it is expected that you will be able to:

Make unit conversions in the imperial system
Make unit conversions in the metric system
Convert between the imperial and the metric systems of measurement
Convert between Fahrenheit and Celsius temperatures

Make Unit Conversions in the Imperial System

There are two systems of measurement commonly used around the world. Metric system used by most countries and imperial system used by the United States. Even though since 1970, Canada tried to switch to the metric system, the imperial system of measurements is still used in Canada. We will look at the imperial system first.

The imperial system of measurement uses units of inch, foot, yard, and mile to measure length and pound and ton to measure weight. For capacity, the units used are cup, pint, quart, and gallons. Both the imperial system and the metric system measure time in seconds, minutes, and hours.

The equivalencies of measurements are shown in the table below. The table also shows, in parentheses, the common abbreviations for each measurement.

Imperial System of Measurement
$\begin{array}{ccc}\mathbf{\text{Length}}\hfill & & \begin{array}{ccc}1\phantom{\rule{0.2em}{0ex}}\text{foot}\phantom{\rule{0.2em}{0ex}}\text{(ft.)}\hfill & =\hfill & 12\phantom{\rule{0.2em}{0ex}}\text{inches}\phantom{\rule{0.2em}{0ex}}\text{(in.)}\hfill \\ 1\phantom{\rule{0.2em}{0ex}}\text{yard}\phantom{\rule{0.2em}{0ex}}\text{(yd.)}\hfill & =\hfill & 3\phantom{\rule{0.2em}{0ex}}\text{feet}\phantom{\rule{0.2em}{0ex}}\text{(ft.)}\hfill \\ 1\phantom{\rule{0.2em}{0ex}}\text{mile}\phantom{\rule{0.2em}{0ex}}\text{(mi.)}\hfill & =\hfill & 5,280\phantom{\rule{0.2em}{0ex}}\text{feet}\phantom{\rule{0.2em}{0ex}}\text{(ft.)}\hfill \end{array}\hfill \end{array}$	$\begin{array}{ccc}\mathbf{\text{Volume}}\hfill & & \begin{array}{ccc}3\phantom{\rule{0.2em}{0ex}}\text{teaspoons}\phantom{\rule{0.2em}{0ex}}\text{(t)}\hfill & =\hfill & 1\phantom{\rule{0.2em}{0ex}}\text{tablespoon}\phantom{\rule{0.2em}{0ex}}\text{(T)}\hfill \\ \text{16 tablespoons}\phantom{\rule{0.2em}{0ex}}\text{(T)}\hfill & =\hfill & \text{1 cup}\phantom{\rule{0.2em}{0ex}}\text{(C)}\hfill \\ \text{1 cup}\phantom{\rule{0.2em}{0ex}}\text{(C)}\hfill & =\hfill & \text{8 fluid ounces}\phantom{\rule{0.2em}{0ex}}\text{(fl. oz.)}\hfill \\ \text{1 pint}\phantom{\rule{0.2em}{0ex}}\text{(pt.)}\hfill & =\hfill & \text{2 cups}\phantom{\rule{0.2em}{0ex}}\text{(C)}\hfill \\ \text{1 quart}\phantom{\rule{0.2em}{0ex}}\text{(qt.)}\hfill & =\hfill & \text{2 pints}\phantom{\rule{0.2em}{0ex}}\text{(pt.)}\hfill \\ \text{1 gallon}\phantom{\rule{0.2em}{0ex}}\text{(gal)}\hfill & =\hfill & \text{4 quarts}\phantom{\rule{0.2em}{0ex}}\text{(qt.)}\hfill \end{array}\hfill \end{array}$
$\begin{array}{ccc}\mathbf{\text{Weight}}\hfill & & \begin{array}{ccc}\text{1 pound}\phantom{\rule{0.2em}{0ex}}\text{(lb.)}\hfill & =\hfill & \text{16 ounces}\phantom{\rule{0.2em}{0ex}}\text{(oz.)}\hfill \\ \text{1 ton}\hfill & =\hfill & \text{2000 pounds}\phantom{\rule{0.2em}{0ex}}\text{(lb.)}\hfill \end{array}\hfill \end{array}$	$\begin{array}{ccc}\mathbf{\text{Time}}\hfill & & \phantom{\rule{1em}{0ex}}\begin{array}{ccc}\text{1 minute}\phantom{\rule{0.2em}{0ex}}\text{(min)}\hfill & =\hfill & \text{60 seconds}\phantom{\rule{0.2em}{0ex}}\text{(sec)}\hfill \\ \text{1 hour}\phantom{\rule{0.2em}{0ex}}\text{(hr)}\hfill & =\hfill & \text{60 minutes}\phantom{\rule{0.2em}{0ex}}\text{(min)}\hfill \\ \text{1 day}\hfill & =\hfill & \text{24 hours}\phantom{\rule{0.2em}{0ex}}\text{(hr)}\hfill \\ \text{1 week}\phantom{\rule{0.2em}{0ex}}\text{(wk)}\hfill & =\hfill & \text{7 days}\hfill \\ \text{1 year}\phantom{\rule{0.2em}{0ex}}\text{(yr)}\hfill & =\hfill & \text{365 days}\hfill \end{array}\hfill \end{array}$

In many medical applications, we need to convert between units of measurement, such as feet and inches, minutes and seconds, pounds and ounces, etc. We will use the identity property of multiplication to do these conversions. We’ll restate the identity property of multiplication here for easy reference.

Identity Property of Multiplication

For any real number $a$: $a\cdot1=a$ $1\cdot a=a$ 1 is the multiplicative identity.

To use the identity property of multiplication, we write 1 in a form that will help us convert the units. For example, suppose we want to change inches to feet. We know that 1 foot is equal to 12 inches, so we will write 1 as the fraction $\dfrac{1\text{ foot}}{12\text{ inches}}$. When we multiply by this fraction we do not change the value, but just change the units.

But $\dfrac{12\text{ inches}}{1\text{ foot}}$ also equals 1. How do we decide whether to multiply by $\dfrac{1\text{ foot}}{12\text{ inches}}$ or $\dfrac{12\text{ inches}}{1\text{ foot}}$? We choose the fraction that will make the units we want to convert from divide out. Treat the unit words like factors and “divide out” common units like we do common factors. If we want to convert $66$ inches to feet, which multiplication will eliminate the inches?

The inches divide out and leave only feet. The second form does not have any units that will divide out and so will not help us.

EXAMPLE 1

MaryAnne is 66 inches tall. Convert her height into feet.

Solution

TRY IT 1

Lexie is 30 inches tall. Convert her height to feet.

Show answer

6.2 Ratio, Rate, and Percent; Health Applications

imazur — Fri, 04 Jun 2021 20:24:00 +0000

Learning Objectives

By the end of this section it is expected that you will be able to:

Write a ratio as a fraction
Application of ratio
Write ratio as a fraction
Find unit rate
Use the definition of percent
Convert percents to fractions and decimals
Convert decimals and fractions to percents

Write a Ratio as a Fraction

Ratios

A ratio compares two numbers or two quantities that are measured with the same unit. The ratio of $a$ to $b$ is written $a\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}b,\phantom{\rule{0.2em}{0ex}}\frac{a}{b},\phantom{\rule{0.2em}{0ex}}\text{or}\phantom{\rule{0.2em}{0ex}}\mathit{\text{a}}\text{:}\mathit{\text{b}}\text{.}$

In this section, we will use the fraction notation. When a ratio is written in fraction form, the fraction should be simplified. If it is an improper fraction, we do not change it to a mixed number. Because a ratio compares two quantities, we would leave a ratio as $\frac{4}{1}$ instead of simplifying it to $4$ so that we can see the two parts of the ratio.

EXAMPLE 1

Write each ratio as a fraction: a) $\phantom{\rule{0.2em}{0ex}}15\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}27\phantom{\rule{0.2em}{0ex}}$ b) $\phantom{\rule{0.2em}{0ex}}45\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}18$.

Solution

	$\text{15 to 27}$
Write as a fraction with the first number in the numerator and the second in the denominator.	$\frac{15}{27}$
Simplify the fraction.	$\frac{5}{9}$

We leave the ratio in b) as an improper fraction.

	$\text{45 to 18}$
Write as a fraction with the first number in the numerator and the second in the denominator.	$\frac{45}{18}$
Simplify.	$\frac{5}{2}$

TRY IT 1

Write each ratio as a fraction: a) $\phantom{\rule{0.2em}{0ex}}21\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}56\phantom{\rule{0.2em}{0ex}}$ b) $\phantom{\rule{0.2em}{0ex}}48\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}32$.

Show answer

6.3 Proportions; Health Applications

imazur — Fri, 04 Jun 2021 20:26:45 +0000

Learning Objectives

By the end of this section it is expected that you will be able to:

Use the definition of proportion
Solve proportions
Solve applications using proportions
Write percent equations as proportions
Translate and solve percent proportions

Use the Definition of Proportion

When two ratios or rates are equal, the equation relating them is called a proportion.

Proportion

A proportion is an equation of the form $\frac{a}{b}=\frac{c}{d}$, where $b\ne 0,d\ne 0$.

The proportion states two ratios or rates are equal. The proportion is read $\text{“}a$ is to $b$, as $c$ is to $d\text{”.}$

The equation $\frac{1}{2}=\frac{4}{8}$ is a proportion because the two fractions are equal. The proportion $\frac{1}{2}=\frac{4}{8}$ is read $\text{“}1$ is to $2$ as $4$ is to $8\text{”.}$

If we compare quantities with units, we have to be sure we are comparing them in the right order.

EXAMPLE 1

Write the sentence as a proportion:

72 heartbeats in 1 minute is the same as 216 heartbeats in 3 minutes.

Solution

	72 is to 1 as 216 is to 3.
Write as a proportion.	$\frac{\text{72 heartbeats}} {\text{1 minute}}=\frac{\text{216 heartbeats}}{\text{3 minutes}}$

TRY IT 1

Write the sentence as a proportion:

$\phantom{\rule{0.2em}{0ex}}5$ is to $9$ as $20$ is to $36$.

Show answer

1.3 Fractions

imazur — Wed, 23 Jun 2021 04:53:49 +0000

Learning Objectives

By the end of this section it is expected that you will be able to:

Multiply and divide fractions
Simplifying expressions with fraction bar
Add or subtract fractions with a common denominator
Add or subtract fractions with different denominators
Use the order of operations to simplify complex fractions

Multiply Fractions

Many people find multiplying and dividing fractions easier than adding and subtracting fractions. So we will start with fraction multiplication.

We’ll use a model to show you how to multiply two fractions and to help you remember the procedure. Let’s start with $\frac{3}{4}$.

Now we’ll take $\frac{1}{2}$ of $\frac{3}{4}$.

Notice that now, the whole is divided into 8 equal parts. So $\frac{1}{2}\cdot\frac{3}{4}=\frac{3}{8}$.

To multiply fractions, we multiply the numerators and multiply the denominators.

Fraction Multiplication

If $a,b,c\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}d$ are numbers where $b\ne 0\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}d\ne 0$, then

$\frac{a}{b}\cdot\frac{c}{d}=\frac{ac}{bd}$ To multiply fractions, multiply the numerators and multiply the denominators.

When multiplying fractions, the properties of positive and negative numbers still apply, of course. It is a good idea to determine the sign of the product as the first step. In Example 1, we will multiply negative and a positive, so the product will be negative.

EXAMPLE 1

Multiply: $-\phantom{\rule{0.2em}{0ex}}\frac{11}{12}\cdot \frac{5}{7}$.

Solution

The first step is to find the sign of the product. Since the signs are the different, the product is negative.

	$-\phantom{\rule{0.2em}{0ex}}\frac{11}{12}\cdot\frac{5}{7}$
Determine the sign of the product; multiply.	$-\phantom{\rule{0.2em}{0ex}}\frac{11\cdot5}{12\cdot7}$
Are there any common factors in the numerator and the demoninator? No.	$-\phantom{\rule{0.2em}{0ex}}\frac{55}{84}$

TRY IT 1

Multiply: $-\phantom{\rule{0.2em}{0ex}}\frac{10}{28}\cdot\frac{8}{15}$.

Show answer

1.4 Decimals

imazur — Wed, 23 Jun 2021 04:55:25 +0000

Learning Objectives

By the end of this section it is expected that you will be able to:

Round decimals
Add and subtract decimals
Multiply and divide decimals
Convert decimals, fractions, and percent

Decimals are another way of writing fractions whose denominators are powers of 10.

$\begin{array}{cccccccc}\hfill 0.1& =\hfill & \frac{1}{10}\hfill & & & & & 0.1\phantom{\rule{0.2em}{0ex}}\text{is “one tenth”}\hfill \\ \hfill 0.01& =\hfill & \frac{1}{100}\hfill & & & & & 0.01\phantom{\rule{0.2em}{0ex}}\text{is “one hundredth”}\hfill \\ \hfill 0.001& =\hfill & \frac{1}{1,000}\hfill & & & & & \text{0.001 is “one thousandth”}\hfill \\ \hfill 0.0001& =\hfill & \frac{1}{10,000}\hfill & & & & & \text{0.0001 is “one ten-thousandth”}\hfill \end{array}$

Notice that “ten thousand” is a number larger than one, but “one ten-thousandth” is a number smaller than one. The “th” at the end of the name tells you that the number is smaller than one.

When we name a whole number, the name corresponds to the place value based on the powers of ten. We read 10,000 as “ten thousand” and 10,000,000 as “ten million.” Likewise, the names of the decimal places correspond to their fraction values. Figure 1 shows the names of the place values to the left and right of the decimal point.

Place value of decimal numbers are shown to the left and right of the decimal point.

[caption id="" align="aligncenter" width="286"] Figure 1[/caption]

Round Decimals

Rounding decimals is very much like rounding whole numbers. We will round decimals with a method based on the one we used to round whole numbers.

EXAMPLE 1

Round 18.379 to the nearest hundredth.

Solution

TRY IT 1

Round to the nearest hundredth: $1.047$.

Show answer

1.5. Exponents and Scientific Notation

imazur — Wed, 23 Jun 2021 05:01:04 +0000

Learning Objectives

By the end of this section it is expected that you will be able to:

Simplify expressions with exponents
Simplify expressions with zero exponents
Use the definition of a negative exponent
Use formulas with exponents in applications
Convert from decimal notation to scientific notation
Convert scientific notation to decimal form
Multiply and divide using scientific notation

Simplify Expressions with Exponents

Remember that an exponent indicates repeated multiplication of the same quantity. For example, ${2}^{4}$ means to multiply 2 by itself 4 times, so ${2}^{4}$ means 2 · 2 · 2 · 2

Let’s review the vocabulary for expressions with exponents.

Exponential Notation (Power)

This is read $a$ to the ${m}^{th}$ power.

In the expression ${a}^{m}$, the exponent $m$ tells us how many times we use the base $a$ as a factor.

Before we begin working with expressions containing exponents, let’s simplify a few expressions involving only numbers.

EXAMPLE 1

Simplify: a) ${4}^{3}$ b) ${7}^{1}$ c) ${\left(\dfrac{5}{6}\right)}^{2}$ d) ${\left(0.63\right)}^{2}$.

Solution

a)	${4}^{3}$
Multiply three factors of 4.	4 · 4 · 4
Simplify.	$64$
b)	${7}^{1}$
Multiply one factor of 7.	$7$
c)	${\left(\dfrac{5}{6}\right)}^{2}$
Multiply two factors.	$\left(\dfrac{5}{6}\right)\left(\dfrac{5}{6}\right)$
Simplify.	$\dfrac{25}{36}$
d)	${\left(0.63\right)}^{2}$
Multiply two factors.	$\left(0.63\right)\left(0.63\right)$
Simplify.	$0.3969$

TRY IT 1

Simplify: a) ${6}^{3}$ b) ${15}^{1}$ c) ${\left(\dfrac{3}{7}\right)}^{2}$ d) ${\left(0.43\right)}^{2}$.

Show answer

1.7 The Real Numbers

imazur — Fri, 02 Jul 2021 02:07:38 +0000

Learning Objectives

By the end of this section it is expected that you will be able to:

Identify integers, rational numbers, irrational numbers, and real numbers
Locate fractions on the number line
Locate decimals on the number line

Identify Integers, Rational Numbers, Irrational Numbers, and Real Numbers

We have already described numbers as counting numbers, whole numbers, and integers. What is the difference between these types of numbers?

$\begin{array}{cccccc}\text{Counting numbers}\hfill & & & & & 1,2,3,4,\text{…}\hfill \\ \text{Whole numbers}\hfill & & & & & 0,1,2,3,4,\text{…}\hfill \\ \text{Integers}\hfill & & & & & \text{…}-3,-2,-1,0,1,2,3,\text{…}\hfill \end{array}$

What type of numbers would we get if we started with all the integers and then included all the fractions? The numbers we would have form the set of rational numbers. A rational number is a number that can be written as a ratio of two integers.

Rational Number

A rational number is a number of the form $\frac{p}{q}$, where p and q are integers and $q\ne 0$. A rational number can be written as the ratio of two integers.

All signed fractions, such as $\frac{4}{5},-\phantom{\rule{0.2em}{0ex}}\frac{7}{8},\frac{13}{4},-\phantom{\rule{0.2em}{0ex}}\frac{20}{3}$ are rational numbers. Each numerator and each denominator is an integer.

Are integers rational numbers? To decide if an integer is a rational number, we try to write it as a ratio of two integers. Each integer can be written as a ratio of integers in many ways. For example, 3 is equivalent to $\frac{3}{1},\frac{6}{2},\frac{9}{3},\frac{12}{4},\frac{15}{5}\text{…}$

An easy way to write an integer as a ratio of integers is to write it as a fraction with denominator one.

$\begin{array}{ccccccc}\hfill 3=\frac{3}{1}\hfill & & & \hfill -8=-\phantom{\rule{0.2em}{0ex}}\frac{8}{1}\hfill & & & \hfill 0=\frac{0}{1}\hfill \end{array}$

Since any integer can be written as the ratio of two integers, all integers are rational numbers! Remember that the counting numbers and the whole numbers are also integers, and so they, too, are rational.

What about decimals? Are they rational? Let’s look at a few to see if we can write each of them as the ratio of two integers.

We’ve already seen that integers are rational numbers. The integer $-8$ could be written as the decimal $-8.0$. So, clearly, some decimals are rational.

Think about the decimal 7.3. Can we write it as a ratio of two integers? Because 7.3 means $7\frac{3}{10}$, we can write it as an improper fraction, $\frac{73}{10}$. So 7.3 is the ratio of the integers 73 and 10. It is a rational number.

In general, any decimal that ends after a number of digits (such as 7.3 or $-1.2684$ is a rational number. We can use the place value of the last digit as the denominator when writing the decimal as a fraction.

EXAMPLE 1

Write as the ratio of two integers: a) $-27$ b) 7.31

Solution

a) Write it as a fraction with denominator 1.	$\begin{array}{c}-27\\ \frac{-27}{1}\end{array}$
b) Write it as a mixed number. Remember, 7 is the whole number and the decimal part, 0.31, indicates hundredths. Convert to an improper fraction.	$\begin{array}{c}7.31\\ 7\frac{31}{100}\\ \frac{731}{100}\end{array}$

So we see that $-27$ and 7.31 are both rational numbers, since they can be written as the ratio of two integers.

TRY IT 1

Write as the ratio of two integers: a) $-24$ b) 3.57

Show answer

Authors

jgray — Wed, 31 Jul 2019 22:51:04 +0000

Cover

jgray — Wed, 31 Jul 2019 22:51:04 +0000

About

jgray — Wed, 31 Jul 2019 22:51:04 +0000

Buy

jgray — Wed, 31 Jul 2019 22:51:04 +0000

Access Denied

jgray — Wed, 31 Jul 2019 22:51:04 +0000

Book Information

jgray — Wed, 31 Jul 2019 22:51:04 +0000

4. Systems of Equations

jgray — Wed, 24 Jul 2019 22:10:36 +0000

7. Data Analysis I

kmoshenko — Wed, 07 Aug 2019 18:01:38 +0000

Data is collected daily from a variety of sources for the purpose of providing information. Once you have collected data, what will you do with it? Data can be described and presented in many different formats. For example, suppose you are interested in buying a house in a particular area. You may have no clue about the house prices, so you might ask your real estate agent to give you a sample data set of prices. Looking at all the prices in the sample often is overwhelming. A better way might be to look at the median price and the variation of prices. The median and variation are just two ways that you will learn to describe data. Your agent might also provide you with a graph of the data. In this chapter, you will study numerical and graphical ways to describe and display your data. You will learn how to calculate, and even more importantly, how to interpret these measurements and graphs. A table can be used to collect and organize data which can then be more easily analyzed to determine patterns or trends. Frequency distributions and stem-and-leaf plots provide a tabular view that can be more revealing than a basic table. A graph is a tool that helps you learn about the shape or distribution of a sample or a population. A graph can be a more effective way of presenting data than a mass of numbers because we can see where data clusters and where there are only a few data values. Newspapers and the Internet use graphs to show trends and to enable readers to compare facts and figures quickly. Statisticians often graph data first to get a picture of the data. Then, more formal tools may be applied. Some of the types of graphs that are used to summarize and organize data are the bar graph, the histogram, the frequency polygon (a type of broken line graph), the pie chart, and the box plot. In this chapter, we will look at ways to collect, present and describe data. We will also consider how data can be presented in misleading ways.

Learning Objectives

By the end of the chapter the student should be able to:

Present and analyze data using frequency distributions, stem-and-leaf plots, pictographs, bar graphs, line graphs, and pie charts.
Describe and calculate the central measures of tendency: mean, median and mode
Design a statistical experiment, collect the data and analyze the results.

1. Operations with Real Numbers

jgray — Tue, 11 Dec 2018 18:22:19 +0000

The chart below shows us how the number sets fit together. In this chapter we will work with rational numbers, but you will be also introduced to irrational numbers. The set of rational numbers together with the set of irrational numbers make up the set of real numbers.

H5P listing

bpayne — Mon, 01 Jun 2020 21:38:21 +0000

standard deviation

kmoshenko — Wed, 28 Apr 2021 20:59:42 +0000

a measure of the spread of all data values around the mean of a set of data

histogram

kmoshenko — Wed, 28 Apr 2021 21:59:35 +0000

a diagram that illustrates the frequency distribution of data that has been grouped into intervals

frequency

kmoshenko — Thu, 29 Apr 2021 21:29:11 +0000

the number of times that a particular data value occurs

symmetrical

kmoshenko — Tue, 11 May 2021 18:08:24 +0000

one half of an object is the mirror image of the other half

principal

kmoshenko — Wed, 12 May 2021 21:09:01 +0000

the initial amount of money that is loaned or borrowed

simple interest

kmoshenko — Wed, 12 May 2021 21:12:03 +0000

is calculated only once on the amount invested or borrowed

compound amount

kmoshenko — Thu, 13 May 2021 20:41:38 +0000

is the total amount of principal and accumulated interest at the end of a loan or investment period.

compound interest

kmoshenko — Thu, 13 May 2021 20:43:44 +0000

interest on a loan or deposit is calculated based on both the initial principal and any accumulated interest from previous periods.

solution of an equation

jgray — Fri, 14 May 2021 20:10:33 +0000

A value of a variable that makes a true statement when substituted into the equation.

8 Data Analysis 2

kmoshenko — Thu, 04 Jun 2020 18:16:38 +0000

9 Financial Mathematics

kmoshenko — Thu, 04 Jun 2020 20:50:08 +0000

3. Equations and their Graphs

jgray — Fri, 07 Aug 2020 23:03:03 +0000

[latexpage] Graphs are found in all areas of our lives. In this chapter, we will study the rectangular coordinate system, which is the basis for most graphs. We will look at linear graphs, slopes of lines, equations of lines, and linear inequalities.

2. Solving Linear Equations and Inequalities

jgray — Wed, 12 Aug 2020 18:25:09 +0000

[latexpage]

The rocks in this formation must remain perfectly balanced around the centre for the formation to hold its shape.

If we carefully placed more rocks of equal weight on both sides of this formation, it would still balance. Similarly, the expressions in an equation remain balanced when we add the same quantity to both sides of the equation. In this chapter, we will solve equations, remembering that what we do to one side of the equation, we must also do to the other side.

Attributions

This chapter has been adapted from the “Introduction” in Chapter 2 of Elementary Algebra (OpenStax) by Lynn Marecek and MaryAnne Anthony-Smith, which is under a CC BY 4.0 Licence. Adapted by Izabela Mazur. See the Adaptation Statement for more information.

5. Trigonometry

imazur — Thu, 20 May 2021 05:54:41 +0000

Trigonometry is a part of geometry that takes its origin in the ancient study of the relationship of the sides and angles of a right triangle. "Trigon" from Greek means triangle and "metron" means measure. Applications of trigonometry are essential to many disciplines like carpentry, engineering, surveying, and astronomy, just to name a few. How tall is the Riverpole? Do we have to climb the pole to find out? Fortunately, with the knowledge of trigonometry, we can find out the measurements of tall objects without too much hassle. In this chapter we will explore the basic properties of angles and triangles, and the applications of the Pythagorean Theorem and trigonometric ratios.

6. Health Option

imazur — Fri, 04 Jun 2021 20:22:07 +0000

Mathematics is a tool frequently used to solve real-life problems. [caption id="attachment_6943" align="aligncenter" width="425"] Young smiling doctor standing in hospital with colleagues talking to patient in background.[/caption] In this chapter, understanding of measurement conversion, ratio, rate, proportion, and percent will help us to solve some health related problems.

\(x\)	\(y\)	\(\left(x,y\right)\)
0	3	\(\left(0,3\right)\)
2	0	\(\left(2,0\right)\)
1	\(\dfrac{3}{2}\)	\(\left(1,\dfrac{3}{2}\right)\)

Substitute \(x=-3\) for x	\(4(-3)+5=-7\)
Multiply.	\(-12+5=-7\)
Simplify.	\(-7=-7\)

a) when \(d=520\) and \(r=65\)		b) in general
Write the formula.	\(\phantom{\rule{1em}{0ex}}d=rt\)	Write the formula.	\(d=rt\)
Substitute.	\(520=65t\)
Divide, to isolate \(t\).	\(\frac{520}{65}=\frac{65t}{65}\)	Divide, to isolate \(t\).	\(\frac{d}{r}=\frac{rt}{r}\)
Simplify.	\(\phantom{\rule{1.2em}{0ex}}8=t\)	Simplify.	\(\frac{d}{r}=t\)

a) Since \({5}^{2}=25\)	\(\begin{array}{c}\sqrt{25}\\ 5\end{array}\)
b) Since \({11}^{2}=121\)	\(\begin{array}{c}\sqrt{121}\\ 11\end{array}\)

Operation	Notation	Say:	The result is…
Addition	\(a+b\)	\(a\phantom{\rule{0.2em}{0ex}}\text{plus}\phantom{\rule{0.2em}{0ex}}b\)	the sum of \(a\) and \(b\)
Subtraction	\(a-b\)	\(a\phantom{\rule{0.2em}{0ex}}\text{minus}\phantom{\rule{0.2em}{0ex}}b\)	the difference of \(a\) and \(b\)
Multiplication	\(a\cdot b,\left(a\right)\left(b\right),\left(a\right)b,a\left(b\right)\)	\(a\phantom{\rule{0.2em}{0ex}}\text{times}\phantom{\rule{0.2em}{0ex}}b\)	The product of \(a\) and \(b\)
Division	\(a\div b,a/b,\phantom{\rule{0.2em}{0ex}}\frac{a}{b},b\overline{)a}\)	\(a\) divided by \(b\)	The quotient of \(a\) and \(b\)

39	54	59	65	75	79	84	92
42	54	60	67	76	80	86	92
43	55	60	69	76	80	88	94
48	57	60	69	77	82	88	95
51	57	63	72	77	83	89	96
51	59	65	72	78	83	91	97

72	65	85	52	61	49	65	82
55	99	58	79	98	79	58	93
88	48	97	74	65	85	71	75
99	39	60	96	80	70	54	77

Name	Symbol
parentheses	\(\left(\phantom{\rule{0.5em}{0ex}}\right)\)
brackets	\(\left[\phantom{\rule{0.5em}{0ex}}\right]\)
braces	\(\left\{\phantom{\rule{0.5em}{0ex}}\right\}\)

Expression	Words	Phrase
\(3+5\)	\(3\phantom{\rule{0.2em}{0ex}}\text{plus}\phantom{\rule{0.2em}{0ex}}5\)	the sum of three and five
\(n-1\)	\(n\) minus one	the difference of \(n\) and one
\(6\cdot 7\)	\(6\phantom{\rule{0.2em}{0ex}}\text{times}\phantom{\rule{0.2em}{0ex}}7\)	the product of six and seven
\(\frac{x}{y}\)	\(x\) divided by \(y\)	the quotient of \(x\) and \(y\)

Equation	Sentence
\(3+5=8\)	The sum of three and five is equal to eight.
\(n-1=14\)	\(n\) minus one equals fourteen.
\(6\cdot 7=42\)	The product of six and seven is equal to forty-two.
\(x=53\)	\(x\) is equal to fifty-three.
\(y+9=2y-3\)	\(y\) plus nine is equal to two \(y\) minus three.

a. \(\phantom{\rule{0.2em}{0ex}}16-6=10\)	This is an equation—two expressions are connected with an equal sign.
b. \(\phantom{\rule{0.2em}{0ex}}4\cdot 2+1\)	This is an expression—no equal sign.
c. \(\phantom{\rule{0.2em}{0ex}}x\div 25\)	This is an expression—no equal sign.
d. \(\phantom{\rule{0.2em}{0ex}}y+8=40\)	This is an equation—two expressions are connected with an equal sign.

a) 14 is to the right of 6 on the number line.	\(14\)___\(6\) \(14\) > \(6\)
b) −1 is to the left of 9 on the number line.	\(-1\)___\(9\) \(-1<9\)
c) −1 is to the right of −4 on the number line.	\(-1\)___\(-4\)
d) 2 is to the right of −20 on the number line.	\(2\)___\(-20\) \(2\) > \(-20\)

\(\begin{array}{ccc}\mathbf{\text{Length}}\hfill & & \begin{array}{ccc}1\phantom{\rule{0.2em}{0ex}}\text{foot}\phantom{\rule{0.2em}{0ex}}\text{(ft.)}\hfill & =\hfill & 12\phantom{\rule{0.2em}{0ex}}\text{inches}\phantom{\rule{0.2em}{0ex}}\text{(in.)}\hfill \\ 1\phantom{\rule{0.2em}{0ex}}\text{yard}\phantom{\rule{0.2em}{0ex}}\text{(yd.)}\hfill & =\hfill & 3\phantom{\rule{0.2em}{0ex}}\text{feet}\phantom{\rule{0.2em}{0ex}}\text{(ft.)}\hfill \\ 1\phantom{\rule{0.2em}{0ex}}\text{mile}\phantom{\rule{0.2em}{0ex}}\text{(mi.)}\hfill & =\hfill & 5,280\phantom{\rule{0.2em}{0ex}}\text{feet}\phantom{\rule{0.2em}{0ex}}\text{(ft.)}\hfill \end{array}\hfill \end{array}\)	\(\begin{array}{ccc}\mathbf{\text{Volume}}\hfill & & \begin{array}{ccc}3\phantom{\rule{0.2em}{0ex}}\text{teaspoons}\phantom{\rule{0.2em}{0ex}}\text{(t)}\hfill & =\hfill & 1\phantom{\rule{0.2em}{0ex}}\text{tablespoon}\phantom{\rule{0.2em}{0ex}}\text{(T)}\hfill \\ \text{16 tablespoons}\phantom{\rule{0.2em}{0ex}}\text{(T)}\hfill & =\hfill & \text{1 cup}\phantom{\rule{0.2em}{0ex}}\text{(C)}\hfill \\ \text{1 cup}\phantom{\rule{0.2em}{0ex}}\text{(C)}\hfill & =\hfill & \text{8 fluid ounces}\phantom{\rule{0.2em}{0ex}}\text{(fl. oz.)}\hfill \\ \text{1 pint}\phantom{\rule{0.2em}{0ex}}\text{(pt.)}\hfill & =\hfill & \text{2 cups}\phantom{\rule{0.2em}{0ex}}\text{(C)}\hfill \\ \text{1 quart}\phantom{\rule{0.2em}{0ex}}\text{(qt.)}\hfill & =\hfill & \text{2 pints}\phantom{\rule{0.2em}{0ex}}\text{(pt.)}\hfill \\ \text{1 gallon}\phantom{\rule{0.2em}{0ex}}\text{(gal)}\hfill & =\hfill & \text{4 quarts}\phantom{\rule{0.2em}{0ex}}\text{(qt.)}\hfill \end{array}\hfill \end{array}\)
\(\begin{array}{ccc}\mathbf{\text{Weight}}\hfill & & \begin{array}{ccc}\text{1 pound}\phantom{\rule{0.2em}{0ex}}\text{(lb.)}\hfill & =\hfill & \text{16 ounces}\phantom{\rule{0.2em}{0ex}}\text{(oz.)}\hfill \\ \text{1 ton}\hfill & =\hfill & \text{2000 pounds}\phantom{\rule{0.2em}{0ex}}\text{(lb.)}\hfill \end{array}\hfill \end{array}\)	\(\begin{array}{ccc}\mathbf{\text{Time}}\hfill & & \phantom{\rule{1em}{0ex}}\begin{array}{ccc}\text{1 minute}\phantom{\rule{0.2em}{0ex}}\text{(min)}\hfill & =\hfill & \text{60 seconds}\phantom{\rule{0.2em}{0ex}}\text{(sec)}\hfill \\ \text{1 hour}\phantom{\rule{0.2em}{0ex}}\text{(hr)}\hfill & =\hfill & \text{60 minutes}\phantom{\rule{0.2em}{0ex}}\text{(min)}\hfill \\ \text{1 day}\hfill & =\hfill & \text{24 hours}\phantom{\rule{0.2em}{0ex}}\text{(hr)}\hfill \\ \text{1 week}\phantom{\rule{0.2em}{0ex}}\text{(wk)}\hfill & =\hfill & \text{7 days}\hfill \\ \text{1 year}\phantom{\rule{0.2em}{0ex}}\text{(yr)}\hfill & =\hfill & \text{365 days}\hfill \end{array}\hfill \end{array}\)

	\(\text{15 to 27}\)
Write as a fraction with the first number in the numerator and the second in the denominator.	\(\frac{15}{27}\)
Simplify the fraction.	\(\frac{5}{9}\)

	\(\text{45 to 18}\)
Write as a fraction with the first number in the numerator and the second in the denominator.	\(\frac{45}{18}\)
Simplify.	\(\frac{5}{2}\)

39	54	59	65	75	79	84	92
42	54	60	67	76	80	86	92
43	55	60	69	76	80	88	94
48	57	60	69	77	82	88	95
51	57	63	72	77	83	89	96
51	59	65	72	78	83	91	97

72	65	85	52	61	49	65	82
55	99	58	79	98	79	58	93
88	48	97	74	65	85	71	75
99	39	60	96	80	70	54	77

	72 is to 1 as 216 is to 3.
Write as a proportion.	\(\frac{\text{72 heartbeats}} {\text{1 minute}}=\frac{\text{216 heartbeats}}{\text{3 minutes}}\)

	\(-\phantom{\rule{0.2em}{0ex}}\frac{11}{12}\cdot\frac{5}{7}\)
Determine the sign of the product; multiply.	\(-\phantom{\rule{0.2em}{0ex}}\frac{11\cdot5}{12\cdot7}\)
Are there any common factors in the numerator and the demoninator? No.	\(-\phantom{\rule{0.2em}{0ex}}\frac{55}{84}\)

a)	\({4}^{3}\)
Multiply three factors of 4.	4 · 4 · 4
Simplify.	\(64\)
b)	\({7}^{1}\)
Multiply one factor of 7.	\(7\)
c)	\({\left(\dfrac{5}{6}\right)}^{2}\)
Multiply two factors.	\(\left(\dfrac{5}{6}\right)\left(\dfrac{5}{6}\right)\)
Simplify.	\(\dfrac{25}{36}\)
d)	\({\left(0.63\right)}^{2}\)
Multiply two factors.	\(\left(0.63\right)\left(0.63\right)\)
Simplify.	\(0.3969\)

a) Write it as a fraction with denominator 1.	\(\begin{array}{c}-27\\ \frac{-27}{1}\end{array}\)
b) Write it as a mixed number. Remember, 7 is the whole number and the decimal part, 0.31, indicates hundredths. Convert to an improper fraction.	\(\begin{array}{c}7.31\\ 7\frac{31}{100}\\ \frac{731}{100}\end{array}\)

39	54	59	65	75	79	84	92
42	54	60	67	76	80	86	92
43	55	60	69	76	80	88	94
48	57	60	69	77	82	88	95
51	57	63	72	77	83	89	96
51	59	65	72	78	83	91	97

72	65	85	52	61	49	65	82
55	99	58	79	98	79	58	93
88	48	97	74	65	85	71	75
99	39	60	96	80	70	54	77