Solution Manual for Pathways to College Mathematics, 2nd Edition

INSTRUCTOR’S SOLUTIONS MANUAL DANIEL S. MILLER Niagara County Community College P ATHWAYS TO C OLLEGE M ATHEMATICS SECOND EDITION Robert Blitzer Miami Dade College The author and publisher of this book have used their best efforts in preparing this book. These efforts include the development, research, and testing of the theories and programs to determine their effectiveness. The author and publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation contained in this book. The author and publisher shall not be liable in any event for incidental or consequential damages in connection with, or arising out of, the furnishing, performance, or use of these programs. Reproduced by Pearson from electronic files supplied by the author. Copyright © 2020, 2016 by Pearson Education, Inc. 221 River Street, Hoboken, NJ 07030. All rights reserved. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher. Printed in the United States of America. ISBN-13: 978-0-13-527708-9 ISBN-10: 0-13-527708-6 Instructor’s Solutions Manual Pathways to College Mathematics 2e Table of Contents Chapter P Prealgebra Pathways ………………………………………………………………………………….1 Chapter 1 Numerical Pathways ………………………………………………………………………………..77 Chapter 2 Algebraic Pathways: Equations and Inequalities ………………………………………………………………………143 Chapter 3 Algebraic Pathways: Graphs, Functions, Linear Functions, and Linear Systems ………………………….215 Chapter 4 Algebraic Pathways: Polynomials, Quadratic Equations, and Quadratic Functions ………………………295 Chapter 5 Geometric Pathways: Measurement …………………………………………………………………………………………373 Chapter 6 Geometric Pathways ………………………………………………………………………………393 Chapter 7 Pathways to Probability: Counting Methods and Probability Theory………………………………………………..435 Chapter 8 Statistical Pathways ……………………………………………………………………………….491 Copyright © 2020 Pearson Education Inc. i Chapter P Prealgebra Pathways Check Points P.1 1. a. millions place b. ones place c. hundreds place d. millions place 2. Work from left to right. Write the name of the three-digit number in each period, followed by the name of the period and a comma. Do not write the name of the last period, “ones.” a. twenty-seven thousand, one hundred forty-three b. five hundred twenty-one million, six hundred thirty thousand, fifty-seven 3. a. Begin by noting how to write the number within each period. 53  406   fifty-three thousand, four hundred six Write the digits for the number in each period, followed by a comma. The standard form for the number is 53,406. b. Begin by noting how to write the number within each period. 016  204   932    two hundred four million, nine hundred thirty-two thousand, sixteen Write the digits for the number in each period, followed by a comma. The standard form for the number is 204,932,016. Copyright © 2020 Pearson Education, Inc. 1 Chapter P Prealgebra Pathways 4. a. The place value chart shows that 704,663 contains 7 hundred-thousands, 0 ten-thousands, 4 thousands, 6 hundreds, 6 tens, and 3 ones. Thus 704,663 is written in expanded form as follows: 704,663  700, 000  4000  600  60  3. b. The place value chart shows that 49,063,400 contains 4 ten-millions, 9 millions, 0 hundred-thousands, 6 tenthousands, 3 thousands, 4 hundreds, 0 tens, and 0 ones. Thus 49,063,400 is written in expanded form as follows: 49, 063, 400  40, 000, 000  9, 000, 000  60, 000  3000  400. 5. a. 14  5 because 14 is to the right of 5 on the number line. b. 0  16 because 0 is to the left of 16 on the number line. 6. a. The digit to the right of the thousands digit is 4, which is less than 5. This indicates to leave the thousands digit the same. Replace all digits to the right with zeros. 57, 498  57, 000 b. The digit to the right of the hundred-thousands digit is 5. This indicates to add one to the hundred-thousands digit. Replace all digits to the right with zeros. 4,856,902  4,900, 000 c. The digit to the right of the thousands digit is 6, which is greater than 5. This implies to add one to the thousands digit. Replace all digits to the right with zeros. 9602  10, 000 d. The digit to the right of the millions digit is 2, which is less than 5. This implies to leave the millions digit the same. Replace all digits to the right with zeros. 684, 236, 042  684, 000, 000 7. a. The digit to the right of the billions digit is 5. This implies to add one to the billions digit. Replace all digits to the right with zeros. 7,599, 445,183  8, 000, 000, 000 b. The digit to the right of the ten-thousands digit is 5. This implies to add one to the ten-thousands digit. Replace all digits to the right with zeros. 7,599, 445,183  7,599, 450, 000 c. The digit to the right of the thousands digit is 1, which is less than 5. This implies to leave the thousands digit the same. Replace all digits to the right with zeros. 7,599, 445,183  7,599, 445, 000 8. a. The cost of a coronary bypass in the United States is $67,583 b. The country with the least amount in the CT scan column is India. The average cost for this procedure in India is $43. c. The charge for an appendectomy in Chile is $5509. The countries in which an appendectomy costs more than in Chile are Canada, Switzerland, and United States. 9. a. We begin with the number of marriages between an African-American husband and a white wife in 2010. Look at the bars labeled with the year 2010. The yellow bar to the right represents the number of marriages between an African-American husband and a white wife. The number above this bar is 390, representing 390 thousand. Thus, in 2010, there were 390,000 marriages between an African-American husband and a white wife. b. Look for the red bar labeled 61 (for 61 thousand, or 61,000). This is the bar to the left for the year labeled 1990. Thus, in 1990, there were 61,000 marriages between a white husband and an African-American wife. 2 Copyright © 2020 Pearson Education, Inc. Section P.1 Whole Numbers Concept and Vocabulary Check P.1 1. whole; 0 11. hundred-millions 2. standard 12. hundred-millions 3. periods 13. two hundred fifty-eight 4. millions; hundred-thousands; thousands; tens 14. three hundred twenty-four 5. millions; forty-two; nine 15. eight thousand, three hundred seventy-six 6. expanded; 5000; 60; 8 16. six thousand, two hundred twenty-six 7. number line 17. thirty-six thousand, eight hundred eighty 8. 19. seven million, five hundred sixty-six thousand 10. 8; 5; add 1; 9,000,000 20. four million, three hundred two thousand 11. 2; 3; do not change; 8,542,000 21. thirty-five million, two hundred sixty thousand, three hundred seventy-five Exercise Set P.1 22. fifty-seven million, forty-four thousand, two hundred eight Note that exercises #1 – 22 use the following table: 23. The standard form is 3468. 24. The standard form is 5283. 25. The standard form is 86,500. 26. The standard form is 58,004. 27. The standard form is 16,402,012. 28. The standard form is 14,204,015. 1. hundreds 2. hundreds 3. ones 4. ones 5. hundred-thousands 6. hundred-thousands 7. millions 8. millions 9. ten-millions 10. ten-millions 29. The standard form is 9,000,009. 30. The standard form is 5,000,005. 31. The standard form is 26,034,203. 32. The standard form is 52,028,706. 33. The standard form is 620,595. 34. The standard form is 430,696. 35. The expanded form is 600  40  3. 36. The expanded form is 500  70  2. 37. The expanded form is 5000  40  6. 38. The expanded form is 3000  50  7. Copyright © 2020 Pearson Education, Inc. 3 Chapter P Prealgebra Pathways 39. The expanded form is 80, 000  1000  300  60  4. 40. The expanded form is 70, 000  2000  500  40  6. 41. The expanded form is 50, 000  5000  30  8. 42. The expanded form is 40, 000  4000  20  9. 43. The expanded form is 20, 000, 000  8, 000, 000  600, 000  40, 000. 44. The expanded form is 50, 000, 000  6, 000, 000  300, 000  7000  30  2. 45. 9  3 because 9 is to the right of 3 on the number line. 46. 7  2 because 7 is to the right of 2 on the number line. 47. 0  14 because 0 is to the left of 14 on the number line. 48. 0  16 because 0 is to the left of 16 on the number line. 49. 3600  4500 because 36000 is to the left of 4500 on the number line. 50. 2300  3200 because 2300 is to the left of 3200 on the number line. 51. 200, 000  20, 000 because 200,000 is to the right of 20,000 on the number line. 52. 300, 000  30, 000 because 300,000 is to the right of 30,000 on the number line. 53. 624 rounded to the nearest ten is 620. 54. 372 rounded to the nearest ten is 370. 55. 627 rounded to the nearest ten is 630. 56. 378 rounded to the nearest ten is 380. 57. 4891 rounded to the nearest hundred is 4900. 58. 5482 rounded to the nearest hundred is 5500. 59. 4831 rounded to the nearest hundred is 4800. 60. 5432 rounded to the nearest hundred is 5400. 61. 61,529 rounded to the nearest thousand is 62,000. 62. 72,503 rounded to the nearest thousand is 73,000. 63. 61,129 rounded to the nearest thousand is 61,000. 64. 72,103 rounded to the nearest thousand is 72,000. 65. 24,628 rounded to the nearest ten-thousand is 20,000. 66. 34,628 rounded to the nearest ten-thousand is 30,000. 67. 345,207 rounded to the nearest ten-thousand is 350,000. 4 Copyright © 2020 Pearson Education, Inc. Section P.1 Whole Numbers 68. 645,308 rounded to the nearest ten-thousand is 650,000. 69. 86,609,100 rounded to the nearest million is 87,000,000. 70. 75,809,100 rounded to the nearest million is 76,000,000. 71. 86,409,100 rounded to the nearest million is 86,000,000. 72. 75,309,100 rounded to the nearest million is 75,000,000. 73. 86,609,100 rounded to the nearest ten-million is 90,000,000. 74. 75,809,100 rounded to the nearest million is 80,000,000. 75. ninety-two quadrillion, two hundred thirty-three trillion, seven hundred twenty billion, three hundred sixty-eight million, five hundred forty-seven thousand, eight hundred. 76. ten-quadrillions 77. 700, 000, 000, 000  20, 000, 000, 000 78. 90, 000, 000, 000, 000, 000  2, 000, 000, 000, 000, 000 79. 92,233,720,368,547,800 rounded to the nearest ten-quadrillion is 90,000,000,000,000,000. The word name is ninety quadrillion. 80. 92,233,720,368,547,800 rounded to the nearest quadrillion is 92,000,000,000,000,000. The word name is ninety-two quadrillion. 81. The greatest yearly earnings are for males with a Doctorate which is $131,569. In word form this is one hundred thirtyone thousand, five hundred sixty-nine dollars. 82. The least yearly earnings are for females with a 9th to 12th grade, non-graduate which is $16,812. In word form this is sixteen thousand, eight hundred twelve dollars. 83. Women with bachelor’s degree earns $50,856. In word form this is fifty thousand, eight hundred fifty-six dollars. 84. Men with associate degree earns $51,865. In word form this is fifty-one thousand, eight hundred sixty-five dollars. 85. Men with bachelor’s degree will earns $79,927 which would round to $80,000. In word form this is seventy-nine thousand, nine hundred twenty-seven dollars. 86. Women with master’s degree will earns $64,834 which would round to $60,000. In word form this is sixty-four thousand, eight hundred thirty-four dollars. 87. In 1960 there were only 9,700,000 immigrants. 88. In 2016 there were 43,700,000 immigrants. 89. In 1900 there were only 10,300,000 immigrants. In 1940 there were only 11,600,000 immigrants. In 1960 there were only 9,700,000 immigrants. 90. In 1980 there were 14,100,000 immigrants. In 2000 there were 31,100,000 immigrants. In 2016 there were 43,700,000 immigrants. 91. In 1980 there were 14,100,000 immigrants. Copyright © 2020 Pearson Education, Inc. 5 Chapter P Prealgebra Pathways 92. In 2000 there were 31,100,000 immigrants. 93. 2,376,206; two million, three hundred seventy-six thousand, two hundred six 94. 1,857,160; one million; eight hundred fifty-seven thousand, one hundred sixty 95. Williams 96. Brown and Jones 97. 1,857,160 rounded to the nearest hundred-thousand is 1,900,000. 98. 2,376,206 rounded to the nearest hundred-thousand is 2,400,000. 99. The 3 is in the ten-thousands place. 100. The 8 is in the ten-thousands place. 101. 1,380,145 1,380,145 103. two thousand, four hundred fifty-three 104. two hundred two thousand, twenty-two 105. 102,063 106. 12,042 107. – 117. Answers will vary. 118. does not make sense; Explanations will vary. Sample explanation: Adding one to this number would create a bigger number. 119. makes sense 120. makes sense 121. makes sense 122. true 123. false; Changes to make the statement true will vary. A sample change is: The number 32,864 is written in standard form. 124. false; Changes to make the statement true will vary. A sample change is: When rounding whole numbers, the digit to be rounded either stays the same or increases by 1. 125. false; Changes to make the statement true will vary. A sample change is: When comparing numbers of various items, tables are just as effective as bar graphs. 126. The whole numbers from 10 to 40 would be rounded to 10 or 20 or 30 or 40. So there are four different rounded numbers. 6 Copyright © 2020 Pearson Education, Inc. Section P.2 Fractions and Mixed Numbers Check Points P.2 1. a. 3 Numerator is 3. Denominator is 10. 10 b. 10 Numerator is 10. Denominator is 3. 3 c.   180 Numerator is  . Denominator is 180. 2. a. There are 7 parts shaded out of a total 10 equal parts. Thus, the fraction 7 (seven-tenths) represents the shaded 10 portion of the figure. b. There are 2 parts shaded out of a total 5 equal parts. Thus, the fraction 2 (two-fifths) represents the shaded portion 5 of the figure. c. There are 13 parts shaded out of a total 16 equal parts. Thus, the fraction 13 (thirteen-sixteenths) represents the 16 shaded portion of the figure. number of fatal gun accidents with children between 15 and 34 3. Fraction of fatal gun accidents with children between 15 and 34 is . total number of fatal gun accidents 145  107 606 252  606  4. a. 3 3 is a proper fraction. Because the numerator is less than the denominator, 10 10 b. 10 10 is an improper fraction. Because the numerator and denominator are equal, 10 10 c. 10 10 Because the numerator is greater than the denominator, is an improper fraction. 3 3 5. a. Improper Fraction: Each whole object is divided into 3 equal parts, or thirds. A total of 5 of the thirds are shaded. 5 The improper fraction represents the shaded portion of the group of figures. 3 2 2 of a second object. The mixed number 1 Mixed Number: The shaded portions include 1 whole object and 3 3 represents the shaded portion of the group of figures. Copyright © 2020 Pearson Education, Inc. 7 Chapter P Prealgebra Pathways b. Improper Fraction: Each whole object is divided into 4 equal parts, or fourths. A total of 9 of the fourths are shaded. 9 represents the shaded portion of the group of figures. The improper fraction 4 1 1 of a second object. The mixed number 2 Mixed Number: The shaded portions include 2 whole object and 4 4 represents the shaded portion of the group of figures. 6. a. 2 5 8  2  5 16  5 21    8 8 8 8 b. 12 7. a. 7 16 12  7 192  7 199    16 16 16 16 1 35 3 2 5 3 Write the mixed number using quotient b. 12 7 86 7 16 14 22 86 7 Write the mixed number using quotient c. 5 2 remainder . Thus,  1 . 3 3 original denominator 513 19 86 2 remainder  12 . . Thus, 7 7 original denominator 27 19 513 38 133 133 200 Write the mixed number using quotient 513 remainder  27. . Thus, 19 original denominator Concept and Vocabulary Check P.2 1. numerator; denominator 2. 3; 8; 3 8 3. proper; improper 4. mixed 5. 5; 3; 2; 5 8 Copyright © 2020 Pearson Education, Inc. Section P.2 Fractions and Mixed Numbers 6. 2; 1 Exercise Set P.2 1. 5 Numerator is 5. Denominator is 13. 13 2. 7 Numerator is 7. Denominator is 19. 19 3. 13 Numerator is 13. Denominator is 5. 5 4. 19 Numerator is 19. Denominator is 7. 7 5. 6.  60  180 Numerator is  . Denominator is 60. Numerator is  . Denominator is 180. 7. There are 3 parts shaded out of a total 5 equal parts. Thus, the fraction 3 represents the shaded portion of the figure. 5 8. There are 2 parts shaded out of a total 5 equal parts. Thus, the fraction 2 represents the shaded portion of the figure. 5 9. There is 1 part shaded out of a total 6 equal parts. Thus, the fraction 1 represents the shaded portion of the figure. 6 10. There are 2 parts shaded out of a total 6 equal parts. Thus, the fraction 2 represents the shaded portion of the figure. 6 11. There are 7 parts shaded out of a total 16 equal parts. Thus, the fraction 7 represents the shaded portion of the figure. 16 12. There are 11 parts shaded out of a total 16 equal parts. Thus, the fraction 13. a. Probability of landing on a red region  11 represents the shaded portion of the figure. 16 red region 3  . total regions 10 b. Probability of landing on a blue region  blue region 2  . total regions 10 c. Red is more likely. Copyright © 2020 Pearson Education, Inc. 9 Chapter P Prealgebra Pathways 14. a. Probability of landing on a brown region  brown region 3  . total regions 10 b. Probability of landing on a yellow region  yellow region 2  . total regions 10 c. Brown is more likely. 15. 9 9 Because the numerator is less than the denominator, is a proper fraction. 11 11 16. 8 8 is a proper fraction. Because the numerator is less than the denominator, 13 13 17. c. 11 11 is an improper fraction. Because the numerator is greater than the denominator, 9 9 18. c. 13 13 Because the numerator is greater than the denominator, is an improper fraction. 8 8 19. 1 2 is a mixed number. 9 20. 1 5 is a mixed number. 8 21. 6 6 Because the numerator and denominator are equal, is an improper fraction. 6 6 22. 8 8 Because the numerator and denominator are equal, is an improper fraction. 8 8 23. Improper Fraction: 4 . 3 1 Mixed Number: 1 . 3 24. Improper Fraction: 5 . 3 2 Mixed Number: 1 . 3 11 . 4 3 Mixed Number: 2 . 4 25. Improper Fraction: 10 Copyright © 2020 Pearson Education, Inc. Section P.2 Fractions and Mixed Numbers 26. Improper Fraction: 5 . 2 1 Mixed Number: 2 . 2 17 . 5 2 Mixed Number: 3 . 5 27. Improper Fraction: 28. Improper Fraction: 18 . 5 3 Mixed Number: 3 . 5 29. Improper Fraction: 25 . 7 4 Mixed Number: 3 . 7 30. Improper Fraction: 26 . 7 5 Mixed Number: 3 . 7 3 8  2  3 19 31. 2   8 8 8 32. 2 7 9  2  7 25   9 9 9 33. 7 3 5  7  3 38   5 5 5 34. 6 2 5  6  2 32   5 5 5 35. 8 7 16  8  7 135   16 16 16 36. 9 5 16  9  5 149   16 16 16 37. 12 18 25 12  18 318   25 25 25 38. 15 21 25 15  21 396   25 25 25 Copyright © 2020 Pearson Education, Inc. 11 Chapter P Prealgebra Pathways 39. 23 3 4 5 5 40. 47 7 5 8 8 41. 76 4 8 9 9 42. 59 5 6 9 9 43. 711 11  35 20 20 44. 788 13  31 25 25 45. 1247  29 43 46. 1665  45 37 47. 48. 49. 50. 51. a. Shark attacks  24  14  3  2  1  1  1  46 b. Florida had the greatest number of attacks, 24. Thus, the fraction of attacks that occurred in Florida is 24 . 46 c. Alabama, North Carolina, and Texas had one shark attack each. Thus, the fraction of attacks that occurred in these 3 . three states is 46 52. a. Shark attacks  24  14  3  2  1  1  1  46 12 Copyright © 2020 Pearson Education, Inc. Section P.3 Simplifying Fractions; Operations with Fractions b. Hawaii had the second greatest number of attacks, 14. Thus, the fraction of attacks that occurred in Hawaii is 14 . 46 c. South Carolina and California had a combined 5 shark attacks. Thus, the fraction of attacks that occurred in these 5 . two states is 46 53. – 58. Answers will vary. 59. does not make sense; Explanations will vary. Sample explanation: Since 3 is an improper fraction the price would be 2 higher than the original price. 60. makes sense 61. does not make sense; Explanations will vary. Sample explanation: There are 8 total people, so the denominator would be 8. 62. true 63. true 64. false; Changes to make the statement true will vary. A sample change is: A mixed number has a whole number and a 5 is not a proper fraction. proper fraction. 4 65. false; Changes to make the statement true will vary. A sample change is: 17 2 3 . 5 5 Check Points P.3 1. a. 300  3 100  3 10 10  3 2 5 25  2  2 355 b. 36  6  6  2 3 2 3  2  2 33 2. a. 10 2  5 2   15 3  5 3 Copyright © 2020 Pearson Education, Inc. 13 Chapter P Prealgebra Pathways b. c. 13 13 ; Because 13 and 15 share no common factors (other than 1), is already reduced to its lowest terms. 15 15 d. 9 1 9 1   45 5  9 5 3. a. b. c. d. 4. a. 4 2 42 8    11 3 11  3 33 6 3 6 3 18 3    3 5 1 5 5 5 3 2 3 2 6 2  3 2      7 3 7  3 21 7  3 7 Remember that you can divide numerators and denominators by common factors before performing multiplication. 3 2 3 2 2     7 3 7 3 7 1  2   1  17 3 51  3 5  1 2   5  2  10  5 10    5 3 5 8 5 4  2 10 1  3      4 8 4 3 4 3 3 3 b. 2 2 3 2 1 2 3      3 3 1 3 3 9 c. 3 1 27 9 27 4 9  3 4 3 1 3 2        1 8 4 8 4 8 9 4 2 9 2 2 5. a. 2 3 5   11 11 11 b. 5 1 4 2    6 6 6 3 c. 3 1 27 9 18 9 1 3 1     2 8 8 8 8 8 4 4 6. 14 42 2  3 7 7   24 2  2  2  3 4 When reducing fractions, it may not be necessary to write prime factorizations. We can use the greatest common factor to reduce this fraction. 42 7  6 7   24 4  6 4 2 2  7 14   3 3  7 21 Copyright © 2020 Pearson Education, Inc. Section P.3 Simplifying Fractions; Operations with Fractions 7. a. b. c. 8. a. 1 1 3 1 5 3  2 5 6 11 or 1       2 5 2  5 5  2 10 10 10 10 4 3 4  4 3  3 16 9 7       3 4 3  4 4  3 12 12 12 1 11 19 23 19  2 23    3 1  6 12 6 12 6  2 12 38 23 15    12 12 12 5 1  or 1 4 4 3 7 2 10 7 17 27 b. 3 16 7 20 16 15 35 c. d. 5 10 2 5  35  35 8 16 2 8 7 12 10 2 12 17 5 5 18  18  1  19 12 12 12 16 19 23 5 16 23 24 8 e. 5 7 14 23 19 20 4  2 20 18 16 3 15 5 3  16  16 4 20 5 4 Copyright © 2020 Pearson Education, Inc. 15 Chapter P Prealgebra Pathways 9. To subtract the fraction parts, we need to borrow from the number 8 because This is done as follows: 8 5 7 is less than . 9 9 5 5 14 14  7 1  7  7 . 9 9 9 8 9 5 14  7 9 9 7 7 3 3 9 9 8 4 10. a. 7 9 Using inspection, the least common denominator for the fraction parts is 35. 2 25 10 4  4  4 7 7 5 35 11 11 11  9 9 9 35 35 35 13 21 7 3 3  13  13 35 5 7 5 b. Using inspection, the least common denominator for the fraction parts is 10. 1 1 5 5 26  26  26 2 25 10 3 3 2 6  10   10   10 5 5 2 10 36 c. 11 1 1  36  1  37 10 10 10 Using inspection, the least common denominator for the fraction parts is 12. 2 24 8 32  32  32 3 3 4 12 5 5 5  9 9 9 12 12 12 23 3 3 1 1  23  23 12 4 3 4 d. The least common denominator for the fraction is 35. 2 25 10 14  14  14 7 75 35 3 37 21  6 6 6 5 57 35 We will need to borrow from the number 14 in order to do the subtraction. 14 16 10 10 45 45  13  1  13   13 . 35 35 35 35 Copyright © 2020 Pearson Education, Inc. Section P.3 Simplifying Fractions; Operations with Fractions Now perform the subtraction. 10 45  13 35 35 21 21 6 6 35 35 14 7 e. 24 35 We need to borrow 1 from the whole number 23 and create a fraction to perform the subtraction. We will choose 1 11 to be . 11 11 11 4 4  17   17 11 11 7 5 11 23 11. a.  22 We are interested in the fraction of jobs that will not require college education. This includes the fraction that will require less than high school education plus the fraction that will require a high school graduate education. 11 6 11 64    100 25 100 25  4 11 24   100 100 35  100 7  20 7 of all jobs will not require any college education. By 2020, 20 b. We have seen that the phrase “how much greater” implies subtraction. To determine how much greater the fraction that will require a college degree is than the fraction that will require some college, we find the difference of these fractions. 7 3 7 3 2    20 10 20 10  2 7 6   20 20 1  20 1 By 2020, the fraction of jobs that will require a college degree will be greater than the fraction that will require 20 some college. Copyright © 2020 Pearson Education, Inc. 17 Chapter P Prealgebra Pathways Concept and Vocabulary Check P.3 9. composite; 140  10 14  25 27  2 257 1. natural 2. prime 3. factors; product 4. 5. 10. composite; 110  10 11  2  5 11 a b 11. 79 has no factors other than 1 and 79, so 79 is prime. ac 12. 83 has no factors other than 1 and 83, so 83 is prime. bd 6. reciprocals 7. 8. 13. composite; 81  9  9  3333 d c 14. composite; 64  8  8  2  4  2  4  222222 ac b 15. composite; 240  10  24  2  5  2  12  25 23 4  25 23 2 2  2 2 2 235 9. least common denominator Exercise Set P.3 1. composite; 22  2 11 16. composite; 360  10  36  2566  2 5 2 3 2 3  2  2  2 335 2. composite; 15  3  5 3. composite; 20  4  5  225 18 4. composite; 75  3  25  355 17. 10 2  5 5   16 2  8 8 5. 37 has no factors other than 1 and 37, so 37 is prime. 18. 8 2 4 4   14 2  7 7 6. 23 has no factors other than 1 and 23, so 23 is prime. 19. 15 3  5 5   18 3  6 6 7. composite; 36  4  9  2 2 33 20. 18 9  2 2   45 9  5 5 8. composite; 200  10  20  2  5  2 10  25 2 25 21. 35 5 7 7   50 5 10 10 22. 45 5 9 9   50 5 10 10 Copyright © 2020 Pearson Education, Inc. Section P.3 Simplifying Fractions; Operations with Fractions 23. 32 16  2 2   80 16  5 5  3   3  15 8 120 20  6 39.  3  1      6 20 1  4   5  4 5 20 24. 75 5 15 15   80 5 16 16 40.  4   1  14 5 70 10  7   2  1     5 4 5 4 20 10  2  44 2  22 22 25.   50 2  25 25 7 1 or 3 2 2 38 2 19 19 26.   50 2  25 25 41. 5 4 5 3 5  3 15      4 3 4 4 4  4 16 120 2  60 60 27.   86 2  43 43 42. 7 2 7 3 7  3 21 5      or 1 8 3 8 2 8  2 16 16 116 2  58 58 28.   86 2  43 43 43. 29. 2 1 2 1 2    5 3 5  3 15 44. 18 18 1 2   5 5 2 18 1 18 2  9 9 4     or 1 5  2 10 2  5 5 5 12 12 1 3   7 7 3 12 1 12 3  4 4     7  3 21 3  7 7 30. 3 1 3 1 3    7 4 7  4 28 31. 3 7 3  7 21    8 11 8 11 88 45. 2  18 2 5 10 2  5 5      5 1 18 18 2  9 9 32. 5 3 5  3 15    8 11 8 11 88 46. 3  12 3 7 21 3  7 7 3      or 1 7 1 12 12 3  4 4 4 33. 9  4 9 4 9  4 36 1 or 5     7 1 7 1 7 7 7 47. 3 1 3 4 3  4 12      3 4 4 4 1 4 1 4 34. 8  3 8 3 8  3 24 3     or 3 7 1 7 1 7 7 7 48. 3 1 3 7 3  7 21      3 7 7 7 1 7 1 7 35. 1 5 1 5 5 5 1 1      10 6 10  6 60 5 12 12 49. 7 5 7 3 7  3 21 3 7 7        6 3 6 5 6  5 30 3 10 10 36. 1 2 1 2 2 2 1 1      8 3 8  3 24 2 12 12 50. 37. 5 6 5  6 30 2 15 15 1      or 1 4 7 4  7 28 2 14 14 14 7 6 7  6 42 2  21 21 38.      4 11 4 11 44 2  22 22 7 3 7 8 7  8 56 4 14       4 8 4 3 4  3 12 4  3 14 2  or 4 3 3 51. 1 1 1 7 7 7 1 1       14 7 14 1 14 7  2 2 52. 1 1 1 4 4 4 1 1       8 4 8 1 8 4  2 2 Copyright © 2020 Pearson Education, Inc. 19 Chapter P Prealgebra Pathways 3 1 33 11  53. 6  1  5 10 5 10 33 10 11  3 5  2 6      6 5 11 1 5 11 3 5 7 21 54. 1  2   4 8 4 8 7 8 56 28  2 2      4 21 84 28  3 3 55. 56. 57. 66. 3 67. 2 4 24 6    11 11 11 11 5 2 52 7    13 13 13 13 68. 5 1 6 2 3 3     58. 16 16 16 2  8 8 5 5 10 2  5 5 1     or 1 8 8 8 2 4 4 4 60. 3 3 6 2 3 3     8 8 8 2 4 4 61. 7 5 2 2 1 1     12 12 12 2  6 6 62. 13 5 8 2 4 4     18 18 18 2  9 9 63. 16 2 14 7  2    2 7 7 7 7 1 64. 17 2 15   3 5 5 5 65. 2 5 1 2 5 20 70. 71. 72. 3 5 3 28 or 5 5 3 8 1 2 8 73. 2 1 13  3 or 8 4 4 3 10 1 1 10 4 3 69. 6 3 15  3 or 8 4 4 5 3 7 1 8 4 2 2     12 12 12 4  3 3 59. 5 8 1 1 8 2 2 1 16  3 or 10 5 5 1 1 1 5 1 2      2 5 2 5 5 2 5 2 52 7     10 10 10 10 1 1 1 5 1 3      3 5 3 5 5 3 5 3 53 8     15 15 15 15 3 3 3 5 3     4 20 4 5 20 15 3   20 20 18 2 9 9    20 2 10 10 2 2 2 3 2     5 15 5 3 15 6 2 8    15 15 15 3 5 3 3 5 2      8 12 8 3 12 2 9 10 19    24 24 24 Copyright © 2020 Pearson Education, Inc. Section P.3 Simplifying Fractions; Operations with Fractions 74. 75. 2 1 11 5 82. 3  2   3 2 3 2 11 2 5 3     3 2 2 3 22 15 7 1    or 1 6 6 6 6 3 2 3 3 2 2      10 15 10 3 15 2 9 4 13    30 30 30 11 2 11 2 2 11 4 7        18 9 18 9 2 18 18 18 17 4 17 4 2 17 8 9        76. 18 9 18 9 2 18 18 18 9 1 1   9 2 2 77. 78. 79. 80. 4 3 4 4 3 3      3 4 3 4 4 3 16 9 7    12 12 12 83. 84. 3 2 3 3 2 2      2 3 2 3 3 2 9 4 5    6 6 6 85. 3 3 2 6  3  3 7 72 14 1 1 1  2  2 2 14 14 14 7 1 11 5  5 or 14 2 2 3 86. 2 4 1  1 3 6 1 3  4 4 2 6 5 7 1 37  6 or 6 6 6 3 1 15 7 81. 3  2   4 3 4 3 15 3 7 4     4 3 3 4 45 28 17 5    or 1 12 12 12 12 87. 2 21 7 3 21 9 3 9 9 21 7 6 3 14 4  14 14 7 1 23  23 14 2 9 11 14 7  20 14 18 4 2 25  26  26 14 14 7 5 13 14 7  30 14 20 6 3 37  38  38 14 14 7 7 8 15 2  2 15 14 12 88. 6 2  12 15 5 5 6 1  2 6 13 4 2 11  11 6 3 Copyright © 2020 Pearson Education, Inc. 21 Chapter P Prealgebra Pathways 89. 2 11  6 9 9 5 5 4 4 9 9 2 90. 2 9  5 7 7 5 5 3 3 7 7 14 1 2 15  15 3 6 1 1  12   12 6 6 3 6  18 5 10 3 3  22   22 10 10 18 9 10 4 16  87 5 20 3 15  72   72 4 20 87 4 16  55 9 36 3 27  63   63 4 36 43 7  119 36 36 5 25  56 6 30 7 21  18   18 10 30 56 38 22 31 11  160 20 20 55 118 95. 98. 5 6 1 2 22  29  28 10 20 20 1 5 5  5  5 5 4 20 20 29 3 1  27 6 2 159 94. 1 6 1 2 8 19  19  18 3 6 6 1 3 3  4  4 4 2 6 6 4 7 40 93. 25 97. 27 92. 2 4 51  51 3 6 1 3  26   26 2 6 6 2 2 9 3 6 2 91. 96. 7 23 17 20 1 1 5 3 2 5 99.           2 3 8 6 6 8 1 5   6 8 1 8 8 2 4 4      6 5 30 2 15 15 1 1 1 1 2 1 3 2 100.                 2 4  2 3  4 4  6 6 3 5   4 6 3 6   4 5 18 2 9 9    20 2 10 10 2 4 1 3 2 3  2 9  101.                 3 3   6 4   3 4   12 12  1 11   2 12 6 11   12 12 17 5  1 12 12 4 2  38 30 15 Copyright © 2020 Pearson Education, Inc. Section P.3 Simplifying Fractions; Operations with Fractions  3 9   7 8   3 10   21 16  102.                 5 10   10 15   5 9   30 30  2 5   3 30 2 1   3 6 4 1   6 6 5  6 103. a. 10 6 16 8    50 50 50 25 b. The phrase “how much greater” implies we should subtract fractions. 10 6 10  6 4 2     50 50 50 50 25 104. a. 7 5 12 6    50 50 50 25 b. The phrase “how much greater” implies we should subtract fractions. 7 5 75 2 1     50 50 50 50 25 105. a. 11 b. 11 106. a. 10 b. 10 107. a. 1 9 10  6  17  18 hours 10 10 10 1 9 11 9 2 1  6  10  6  4  4 hours 10 10 10 10 10 5 4 7 11 1  7  17  18 hours 10 10 10 10 4 7 14 7 7 7  9 7  2 10 10 10 10 10 8 8 64   9 9 81 b. 3 8 24 12  2 2     4 9 36 12  3 3 c. There are black keys to the left of the keys for the notes D, E, G, A, and B. 108. The numbers are the prime numbers between 2 and 97, inclusive. 109. – 116. Answers will vary. 117. makes sense 118. makes sense Copyright © 2020 Pearson Education, Inc. 23 Chapter P Prealgebra Pathways 119. makes sense 120. does not make sense; Explanations will vary. Sample explanation: When adding fractions, we need to have a common denominator before adding the fractions. 121. does not make sense; Explanations will vary. Sample explanation: Although that method will work, it is easier to leave them as mixed numbers and perform the addition. 122. makes sense 123. 1 1 1 1 1 1 1 1 1 1         2  2 1 2 2  3 3  4 4  5 5  6 2 2  3 3  2 2 5 5 23 1 2  3  5 1 2  5 1 5 1 3 1 2     2  2 2  2 35 2 3 2 5 3 2 5 2 53 5 2 3 2 30 10 5 3 2      60 60 60 60 60 50  60 5  6 124. The notes in each measure add to 3 . 4 Check Points P.4 1. a. The underlined digit of 3.258 is in the hundredths place. b. The underlined digit of 6.9347 is in the tenths place. c. The underlined digit of 2056.31479 is in the thousandths place. d. The underlined digit of 0.002576 is in the hundred-thousandths place. e. The underlined digit of 22.005124689 is in the millionths place. 2. Work from left to right. Write the whole-number part in words, followed by the word “and” for the decimal point. Then write the decimal part in words as though it were a whole number and follow it by the place value of the last digit. a. nineteen and twenty-four hundredths b. eight hundred twenty-six and three hundred seventy-five thousandths c. fifty-two ten-thousandths 24 Copyright © 2020 Pearson Education, Inc. Section P.4 Decimals 3. a. Write the digits for the whole-number part, which comes before the word “and.” Write a decimal point for the word “and.” Then write the digits for the decimal part as if it were a whole number. 45   32    forty-five and thirty-two hundredths Make sure the last digit is in the hundredths place. The standard form for the number is 45.32. b. Write the digits for the whole-number part, which comes before the word “and.” Write a decimal point for the word “and.” Then write the digits for the decimal part as if it were a whole number. 11   400      four hundred and eleven thousandths Make sure the last digit is in the thousandths place by inserting a 0 in the tenths place. The standard form for the number is 400.011. c. The word form does not contain the word “and” thus, the whole-number part is zero.  411   four hundred eleven thousandths Make sure the last digit is in the thousandths place. The standard form for the number is 0.411. d. Write the digits for the whole-number part, which comes before the word “and.” Write a decimal point for the word “and.” Then write the digits for the decimal part as if it were a whole number. 52 8      eight and fifty-two ten-thousandths Make sure the last digit is in the ten-thousandths place by inserting a 0 in the tenths place and the hundredths place. The standard form for the number is 8.0052. 4. a. 0.7 has 1 decimal place so the denominator will contain 1 zero, or 10. 7 0.7  10 b. 0.59 has 2 decimal places so the denominator will contain 2 zeros, or 100. 59 0.59  100 c. 0.012 has 3 decimal places so the denominator will contain 3 zeros, or 1000. 12 4 3 3 0.012    1000 4  250 250 5. a. 72.6 has 1 decimal place so the denominator will contain 1 zero, or 10. 6 2 3 3 72.6  72  72  72 10 5 2 5 b. 184.083 has 3 decimal places so the denominator will contain 3 zeros, or 1000. 83 184.083  184 1000 c. 5.0044 has 4 decimal places so the denominator will contain 4 zeros, or 10,000. 44 4 11 11 5.0044  5 5 5 10, 000 2500 4  2500 Copyright © 2020 Pearson Education, Inc. 25 Chapter P Prealgebra Pathways 6. a. The digit to the right of the tenths digit is 3, which is less than 5. Therefore do not change the digit to be rounded. Drop all digits to the right of the tenths place. 23.436  23.4 b. The digit to the right of the hundredths digit is 6, which is greater than 5. Therefore we add 1 to the hundredths digit. Drop all digits to the right of the hundredths place. 392.7869  392.79 c. The digit to the right of the thousandths digit is 5. Therefore we add 1 to the thousandths digit. Drop all digits to the right of the thousandths place. 1400.61558  1400.616 d. The digit to the right of the ones digit is 6, which is greater than 5. Therefore we add 1 to the ones digit. Drop all digits to the right of the ones place. 298.603  299 7. The digit to the right of the hundredths digit is 6, which is greater than 5. Therefore we add 1 to the hundredths digit: 9  1  10. Record 0 in the hundredths place and carry the 1 to the tenths place. 38.496  38.50 Concept and Vocabulary Check P.4 1. whole-number; decimal 2. tens; ones; tenths; hundredths 3. and; thousandths 4. 10; 100; 1000 5. 6; 7; add 1; 12.37 Exercise Set P.4 1. hundredths 2. hundredths 3. thousandths 4. thousandths 5. tens 6. tens 7. ten-thousandths 8. ten-thousandths 9. millionths 10. millionths 11. hundred-thousandths 26 Copyright © 2020 Pearson Education, Inc. Section P.4 Decimals 12. hundred-thousandths 39. 0.3  3 10 40. 0.9  9 10 13. seven and sixty-three hundredths 14. eight and fifty-seven hundredths 15. sixteen and six tenths 16. eighteen and eight tenths 41. 0.39  39 100 17. eight hundred sixty-five and two hundred five thousandths 42. 0.37  37 100 18. six hundred eighty-five and three hundred seven thousandths 43. 0.8  8 2 4 4   10 2  5 5 44. 0.6  6 2 3 3   10 2  5 5 19. twenty-three ten-thousandths 20. thirty-two ten-thousandths 21. six thousand and three thousandths 22. three thousand and six thousandths 45. 0.64  64 4 16 16   100 4  25 25 46. 0.75  75 25  3 3   100 25  4 4 23. two hundred seven and twenty-seven hundredthousandths 24. three hundred nine and thirty-nine hundredthousandths 25. The standard form is 5.9. 47. 3.005  3 5 5 1 1 3 3 1000 200 5  200 48. 7.005  7 5 5 1 1 7 7 1000 200 5  200 26. The standard form is 4.7. 27. The standard form is 17.37. 28. The standard form is 19.29. 29. The standard form is 800.08. 49. 876.32  876 32 4 8 8  876  876 100 25 4  25 50. 678.34  678 34 2 17 17  678  678 100 50 2  50 30. The standard form is 900.09. 31. The standard form is 306.535. 32. The standard form is 507.543. 33. The standard form is 0.292. 34. The standard form is 0.529. 51. 0.0009  9 10, 000 52. 0.0007  7 10, 000 35. The standard form is 200.092. 36. The standard form is 500.029. 53. 17.268  17 268 4  67 67  17  17 1000 250 4  250 54. 19.375  19 375 125  3 3  19  19 1000 8 125  8 37. The standard form is 0.0075. 38. The standard form is 0.0057. Copyright © 2020 Pearson Education, Inc. 27 Chapter P Prealgebra Pathways 55. 0.4006  4006 2  2003 2003   10, 000 2  5000 5000 56. 0.6004  6004 4 1501 1501   10, 000 4  2500 2500 57. 0.38 rounded to the nearest tenth is 0.4. 58. 0.27 rounded to the nearest tenth is 0.3. 59. 0.8647 rounded to the nearest hundredth is 0.86. 60. 0.7843 rounded to the nearest hundredth is 0.78. 61. 24.23651 rounded to the nearest thousandth is 24.237. 62. 37.32851 rounded to the nearest thousandth is 37.329. 63. 63.498 rounded to the nearest hundredth is 63.50. 64. 71.396 rounded to the nearest hundredth is 71.40. 65. 32.98601 rounded to the nearest tenth is 33.0. 66. 43.97801 rounded to the nearest tenth is 44.0. 67. 32.98601 rounded to the nearest ten is 30. 68. 43.97801 rounded to the nearest ten is 40. 69. 0.041652 rounded to the nearest ten-thousandth is 0.0417. 70. 0.032751 rounded to the nearest ten-thousandth is 0.0328. 71. 318.489 rounded to the nearest whole number is 318. 72. 617.498 rounded to the nearest whole number is 617. 73. one thousand, five and one hundred five ten-billionths 74. two thousand, seven and two hundred seven ten-billionths 75. 11.00011101  11 11,101 100, 000, 000 76. 13.00011103  13 11,103 100, 000, 000 77. In 2014 the average age of cars on U.S. roads was at a maximum of 11.4  11 78. In 2008 the average age of cars on U.S. roads was at a minimum of 10.1  10 4 2  11 years. 10 5 1 years. 10 79. In 2010 the average vehicle age was between 10.5 and 11.0 years at ten and sixth tenths years. 28 Copyright © 2020 Pearson Education, Inc. Section P.4 Decimals 80. In 2012 the average vehicle age was between 11.0 and 11.4 years at eleven and two tenths years. 81. In 2008 the average age of vehicles on the U.S. roads was 10.1 years , which rounded to the nearest whole number equals 10 years. 82. In 2010, 2012, and 2014 the average age of vehicles on the U.S. roads when rounded to the nearest whole number equals 11 years. 1 2 1 83. In 2012 the average age of cars on U.S. roads was 11 years because 11.2  11  11 years. 5 10 5 84. In 2010 the average age of cars on U.S. roads was 10 3 6 3 years because 10.6  10  10 years. 5 10 5 85. Check to Safeway: 86. Check to Dr. Jill Cantor: 87. Cobra: nine thousandths 88. Krait, five tenths 89. cobra, mamba, brown snake, taipan, krait because 0.009  0.02  0.05  0.11  0.5. 90. krait, taipan, brown snake, mamba, cobra because 0.009  0.02  0.05  0.11  0.5. 91. a. $4.539 rounded to the nearest cent is $4.54. b. $4.539 rounded to the nearest whole dollar is $5. 92. a. $4.239 rounded to the nearest cent is $4.24. b. $4.239 rounded to the nearest whole dollar is $4. 93. 2.718281828459045 rounded to the nearest hundredth is 2.72. 94. 2.718281828459045 rounded to the nearest thousandth is 2.718. 95. 2.718281828459045 rounded to the nearest ten-thousandth is 2.7183. Copyright © 2020 Pearson Education, Inc. 29 Chapter P Prealgebra Pathways 96. 2.718281828459045 rounded to the nearest millionth is 2.718282. 97. – 101. Answers will vary. 102. does not make sense; Explanations will vary. Sample explanation: The place values to the immediate left of a decimal start with ones. 103. does not make sense; Explanations will vary. Sample explanation: 0.17 in words is seventeen hundredths. 104. does not make sense; Explanations will vary. Sample explanation: Adding zeroes before the first digit in a number’s decimal part will change the value of the number. Adding zeroes after the last digit in a number’s decimal part will not change the value of the number. 105. Answers will vary; an example is 18.6001. 106. Answers will vary; an example is 23.5399. 107. Answers will vary; an example is 62.35. 108. 9 3 3 4 12 9 9  9.12 25 25  4 100 Mid-Chapter P Check Point 1. eight billion, sixty-three million, five hundred sixty-one thousand, four 2. 54,302,628 3. a. nearest hundred: 64,517  64,500 b. nearest thousand: 64,517  65,000 4. 18  0 5. 18  19 6. a. The James Bond franchise had the greatest number of movies. The total world gross for this franchise was $5,116,147,171. b. three billion, two hundred eighty-seven million, two hundred eighty-five thousand, five dollars c. Star Wars had 7 movies. $4, 279, 632, 749  $4, 000, 000, 000 d. Shrek and Lord of the Rings each had total world gross less than $3,500,000,000. 7. a. 61 (1980)  28 (Ancient Greece and Rome) 33 People born in 1980 are expected to live 33 years more than people born in ancient Greece and Rome. b. Average life expectancy was 48 years in 1950. 30 Copyright © 2020 Pearson Education, Inc. Mid-Chapter P Check Point c. The following life expectancies round to 30. Stone Age: 25  30 Ancient Greece and Rome: 28  30 Middle Ages: 30  30 1900: 31  30 d. life expectancy 1950: 48 life expectancy Middle Ages: 30 2 1 14 7 14 5 14 5 2 17.       15 5 15 7 15 7 3 3 1 5 2 ? 2 48  2  30  12 3 1 2 25 20 25 9 25 9 15 7 18. 4  2        1 6 9 6 9 6 20 8 8 6 20 4 2 10 15 10 14 10 14 4 19.       21 14 21 15 21 15 9 ? 48  60  12 3 3 ? 48  48 true 20. 8. There are 2 parts shaded out of a total 5 equal parts. 2 Thus, the fraction (two-fifths) represents the 5 shaded portion of the figure. 9. Improper Fraction: 7 . 3 10. 4 11. 14 7  30 15 21. 1 Mixed Number: 2 . 3 13 26  18 36 5 15    12 36 7 10  4  7 47   10 10 10 11 36 85 1  14 6 6 22. 12. 150  10 15  2 535 13. 1 5  6 30 3 9    10 30 4 12 4  3 3   28 4  7 7 23. 1 1 5 8 5 8 1 14.     24 15 24 15 9 3 2 16 1  1 5 40 7 35  3  3 8 40 1 3 9  5  4 2 6 6 2 4 4  2  2  2 3 6 6 5 3 2 1 15. 6  51 11 5 40 40 7 6 7 7 1    3 12 1 12 2 2 24. 35  15  20 questions left. 20 4  of the test left. 35 7 2 3 3  2  1  5 9 5 9 15 16. 1  2       3 4 4  3  4  3 4 3 4 1 5 6 25. hundredths 26. twenty-three and two hundred three thousandths 27. The standard form is 3003.03. Copyright © 2020 Pearson Education, Inc. 31