Chapter 9 Solutions - IAmMEA.org
Chapter 9 Solutions - IAmMEA.org
Chapter 9 Solutions - IAmMEA.org
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ch09_p355_387.qxd 3/17/10 12:44 PM Page 356356 <strong>Chapter</strong> 9 Discrete Mathematics27.28.29.25C 3 =52C 5 =48C 3 =30. Choose 7 positions from the 20:20!20C 7 =7!120 - 72! = 20!7!13! = 77,52031. Choose A♠ and K♠, and 11 cards from the other 50:2C 2# 50 C 11 = 1 #50!50 C 11 =11!150 - 112! = 50!11!39!=37,353,738,80032. 8C 3 =33. We either have 3, 2, or 1 student(s) nominated:6C 3 + 6 C 2 + 6 C 1 =20+15+6=4134. We either have 3, 2, or 1 appetizer(s) represented:5C 3 + 5 C 2 + 5 C 1 =10+10+5=2535. Each of the 5 dice have 6 possible outcomes: 6 5 =777620!36. 20C 8 =8!120 - 82! = 20! = 125, 9708!12!37. 2 9 -1=51138. 3*4*3*2 6 =230439. Since each topping can be included or left off, the totalnumber of possibilities with n toppings is 2 n . Since2 11 =2048 is less than 4000 but 2 12 =4096 is greaterthan 4000, Luigi offers at least 12 toppings.40. There are 2 n subsets, of which 2 n -2 are proper subsets.41. 2 10 =102442. 5 10 =9,765,62543. True.25!3!125 - 32! = 25!3!22! = 2300 (d) Number of 5-digit numbers that can be formed usingonly the digits 1 and 2.52!5!152 - 52! = 52!5!47! = 2,598,96048!3!148 - 32! = 48!3!45! = 17,2968!3!18 - 32! = 8!3!5! = 56a n a b = n!a!1n - a2! = n!a!b! = n!1n - b2!b! = an b b44. False. For example, a5 is greater than a 5 2 b = 104 b = 5.45. There are a 6 different combinations of vegetables.2 b = 15The total number of entrée-vegetable-dessert4 # 15 # 6 = 360variations is . The answer is D.46. 10P 5 = 30,240. The answer is D.n!47. nP n =The answer is B.1n - n2! = n!48. There are as many ways to vote as there are subsets of aset with 5 members. That is, there are 2 5 ways to fill outthe ballot. The answer is C.49. Answers will vary. Here are some possible answers:(a) Number of 3-card hands that can be dealt from adeck of 52 cards(b) Number of ways to choose 3 chocolates from a boxof 12 chocolates(c) Number of ways to choose a starting soccer teamfrom a roster of 25 players (where position matters)(e) Number of possible pizzas that can be ordered at aplace that offers 3 different sizes and up to 10different toppings.50. Counting the number of ways to choose the two eggs youare going to have for breakfast is equivalent to countingthe number of ways to choose the ten eggs you are notgoing to have for breakfast.51. (a) Twelve(b) Every 0 represents a factor of 10, or a factor of5 multiplied by a factor of 2. In the product50 # 49 # 48 # Á # 2 # 1, the factors 5, 10, 15, 20, 30, 35,40, and 45 each contain 5 as a factor once, and 25 and50 each contain 5 twice, for a total of twelve occurrences.Since there are 47 factors of 2 to pair up withthe twelve factors of 5, 10 is a factor of 50! twelvetimes.52. (a) Each combination of the n vertices taken two at atime determines a segment that is either an edge or adiagonal. There are n C 2 such combinations.(b) Subtracting the n edges from the answers in (a),n!we find that nC 2 - n =2!1n - 22! - nn1n - 12 n 2 - 3n= - 2n = .2 2 253. In the nth week, 5 n copies of the letter are sent. In the lastweek of the year, that’s 5 52 ≠2.22*10 36 copies of theletter. This exceeds the population of the world, which isabout 6*10 9 , so someone (several people, actually) hashad to receive a second copy of the letter.54. Six. No matter where the first person sits, there are the3!=6 ways to sit the others in different positions relativeto the first person.55. Three. This is equivalent to the round table problem(Exercise 42), except that the necklace can be turnedupside-down. Thus, each different necklace accounts fortwo of the six different orderings.56. The chart on the left is more reasonable. Each pair ofactresses will require about the same amount of time tointerview. If we make a chart showing n (the number ofactresses) and n C 2 (the number of pairings), we can seethat chart 1 allows approximately 3 minutes per pairthroughout, while chart 2 allows less and less time perpair as n gets larger.Number Number of Time per Pair Time per Pairn Pairs n C 2 Chart 1 Chart 23 3 3.33 3.336 15 3 29 36 3.06 1.6712 66 3.03 1.5215 105 3.05 1.43Copyright © 2011 Pearson Education, Inc. Publishing as Addison-Wesley.
ch09_p355_387.qxd 3/17/10 12:44 PM Page 357Section 9.2 The Binomial Theorem 35757. There are 52 C 13 =635,013,559,600 distinct bridge hands.Every day has 60 # 60 # 24 = 86,400 seconds; a year has365.24 days, which is 31,556,736 seconds. Therefore it will635,013,559,600take aboutL 20,123 years. (Using 365 days31,556,736per year, the computation gives about 20,136 years.)58. Each team can choose 5 players in 12 C 5 =792 ways, sothere are 792 2 =627,264 ways total.■ Section 9.2 The Binomial TheoremExploration 11. 3C 0 = 3!0!3! = 1, 3C 1 = 3!1!2! = 3,3C 2 = 3!These are (in order) the2!1! = 3, 3C 3 = 3!3!0! = 1.coefficients in the expansion of (a+b) 3 .2. {1 4 6 4 1}. These are (in order) the coefficients in theexpansion of (a+b) 4 .3. {1 5 10 10 5 1}. These are (in order) the coefficients inthe expansion of (a+b) 5 .Quick Review 9.21. x 2 +2xy+y 22. a 2 +2ab+b 23. 25x 2 -10xy+y 24. a 2 -6ab+9b 25. 9s 2 +12st+4t 26. 9p 2 -24pq+16q 27. u 3 +3u 2 v+3uv 2 +v 38. b 3 -3b 2 c+3bc 2 -c 39. 8x 3 -36x 2 y+54xy 2 -27y 310. 64m 3 +144m 2 n+108mn 2 +27n 3Section 9.2 Exercises1.2.3.4.1a + b2 4 = a 4 + a 4 = a 4 + 4a 3 b + 6a 2 b 2 + 4ab 3 + b 44 ba0 b 40 ba4 b 0 + a 4 1 ba3 b 1 + a 4 2 ba2 b 2 + a 4 3 ba1 b 31a + b2 6 = a 6 0 ba6 b 0 + a 6 1 ba5 b 1 + a 6 2 ba4 b 2 + a 6 3 ba3 b 3+ a 6 4 ba2 b 4 + a 6 5 ba1 b 5 + a 6 6 ba0 b 6= a 6 + 6a 5 b + 15a 4 b 2 + 20a 3 b 3 + 15a 2 b 4 + 6ab 5 + b 61x + y2 7 = a 7 + a 7 + a 7 6 bx1 y 6 + a 7 7 bx0 y 74 bx3 y 4 + a 7 5 bx2 y 50 bx7 y 0 + a 7 1 bx6 y 1 + a 7 2 bx5 y 2 + a 7 3 bx4 y 3= x 7 + 7x 6 y + 21x 5 y 2 + 35x 4 y 3 + 35x 3 y 4+ 21x 2 y 5 + 7xy 6 + y 71x + y2 10 = a 10 + a 10 + a 10 5 bx5 y 5 + a 106 bx4 y 63 bx7 y 3 + a 10 4 bx6 y 40 bx10 y 0 + a 10 1 bx9 y 1 + a 10 2 bx8 y 2+ a 10 + a 10 9 bx1 y 9 + a 1010 bx0 y 107 bx3 y 7 + a 10 8 bx2 y 8= x 10 + 10x 9 y + 45x 8 y 2 + 120x 7 y 3 + 210x 6 y 4+ 252x 5 y 5 + 210x 4 y 6 + 120x 3 y 7+ 45x 2 y 8 + 10xy 9 + y 105. Use the entries in row 3 as coefficients:(x+y) 3 =x 3 +3x 2 y+3xy 2 +y 311. a 166166 b = 166!166!0! = 16. Use the entries in row 5 as coefficients:(x+y) 5 =x 5 +5x 4 y+10x 3 y 2 +10x 2 y 312. a 1660 b = 166!0!166! = 1+5xy 4 +y 57. Use the entries in row 8 as coefficients:13. a 14 3 b = a 1411 b = 364(p+q) 8 =p 8 +8p 7 q+28p 6 q 2 +56p 5 q 3 +70p 4 q 4+56p 3 q 5 +28p 2 q 6 +8pq 7 +q 814. a 13 8 b = a 135 b = 12878. Use the entries in row 9 as coefficients:(p+q) 9 =p 9 +9p 8 q+36p 7 q 2 +84p 6 q 3 +126p 5 q 4 +1 15. = 1-22 8 a 12 =126,72026p 4 q 5 +84p 3 q 6 +36p 2 q 7 +9pq 8 +q 91-228 a 12 8 b4 b9. a 9 2 b = 9!2!7! = 9 # 82 # 1 = 36 16. = 1-32 4 a 111-324 a 11 =26,7304 b 7 b10. a 1511 b = 15!11!4! = 15 # 14 # 13 # 124 # 3 # 2 # 1= 136517. f(x)=(x-2) 5 =x 5 +5x 4 (–2)+10x 3 (–2) 2 +10x 2 (–2) 3 +5x(–2) 4 +(–2) 5 =x 5 -10x 4 +40x 3 -80x 2 +80x-3218. g(x)=(x+3) 6 =x 6 +6x # 5 3 + 15x 4 # 3 2 + 20x 3 # 3 3 +15x # 2 3 4 + 6x # 3 5 + 3 6=x 6 +18x 5 +135x 4 +540x 3 +1215x 2 +1458x+72919. h(x)=(2x-1) 7=(2x) 7 +7(2x) 6 (–1)+21(2x) 5 (–1) 2 +35(2x) 4 (–1) 3 +35(2x) 3 (–1) 4 +21(2x) 2 (–1) 5 +7(2x)(–1) 6 +(–1) 7=128x 7 -448x 6 +672x 5 -560x 4 +280x 3 -84x 2 +14x-1Copyright © 2011 Pearson Education, Inc. Publishing as Addison-Wesley.
ch09_p355_387.qxd 3/17/10 12:44 PM Page 358358 <strong>Chapter</strong> 9 Discrete Mathematics20. f(x)=(3x+4) 5 =(3x) 5 +5(3x) 4 +10(3x) 2=243x 5 +1620x 4 +4320x 3 +5760x 2 +3840x+102421. (2x+y) 4 =(2x) 4 +4(2x) 3 y+6(2x) 2 y 2 +4(2x)y 3 +y 4 =16x 4 +32x 3 y+24x 2 y 2 +8xy 3 +y 422. (2y-3x) 5 =(2y) 5 +5(2y) 4 (–3x)+10(2y) 3 (–3x) 2 +10(2y) 2 (–3x) 3 +5(2y)(–3x) 4 +(–3x) 5=32y 5 -240y 4 x+720y 3 x 2 -1080y 2 x 3 +810yx 4 -243x 523.1- 1y2 + 1512# x24 1- 1y2 2 + 2011x2 3 1- 1y2 3 + 1511x2 2 1- 2 y24+ 611x21- 1y2 5 + 1- 1y2 6 =x 3 -6x 5/2 y 1/2 +15x 2 y-20x 3/2 y 3/2 +15xy 2 -6x 1/2 y 5/2 +y 324. 11x + 132 4 = 11x2 4 + 411x2 3 1132 + 611x2 2 # 1132 2 + 411x21132 3 + 1132 411x - 1y2 6 = 11x2 6 + 611x2 5 # 4 3 + 513x2 # 4 4 + 4 5= x 2 + 4x13x + 18x + 1213x + 9# 3 + 101x -2 2 3 # 3 2 + 101x -2 2 2 # 3 3 + 51x -2 2 # 3 4 + 3 5= x -10 + 15x -825. (x –2 +3) 5 =(x –2 ) 5 +5(x –2 ) 4# 4 + 1013x2 3 # 4 2+90x –6 +270x –4 +405x –2 +24326. (a-b –3 ) 7 =a 7 +7a 6 (–b –3 )+21a 5 (–b –3 ) 2 +35a 4 (–b –3 ) 3 +35a 3 (–b –3 ) 4 +21a 2 (–b –3 ) 5 +7a(–b –3 ) 6 +(–b –3 ) 7=a 7 -7a 6 b –3 +21a 5 b –6 -35a 4 b –9 +35a 3 b –12 -21a 2 b –15 +7ab –18 -b –2127. Answers will vary.28. Answers will vary.29. a n 1 b = n!1!1n - 12! = n = n!1n - 12!1!n!=1n - 12!3n - 1n - 124! = a nn - 1 b30. a n r b = n!r!1n - r2! = n!1n - r2!r!n!=1n - r2!3n - 1n - r24! = a nn - r b31. a n - 1r - 1 b + an - 1 br1n - 12!=1r - 12!31n - 12 - 1r - 124! + 1n - 12!r!1n - 1 - r2!r1n - 12!=r1r - 12!1n - r2! +1n - 12!1n - r2r!1n - r21n - r - 12!r1n - 12! 1n - r21n - 12!= +r!1n - r2! r!1n - r2!1r + n - r21n - 12!=r!1n - r2!n!=r!1n - r2!= a n r b32. (a)Any pair (n, m) of nonnegative integers — except for(1, 1) — provides a counterexample. For example, n=2and m=3: (2+3)!=5!=120, but2!+3!=2+6=8.(b) Any pair (n, m) of nonnegative integers — except for(0, 0) or any pair (1, m) or (n, 1) — provides a counterexample.For example, n=2 and m=3:12 # 32! = 6! = 720, but, 2! # 3! = 2 # 6 = 12.n! 1n + 12!33. a n +2 b + a n + 1 b =2 2!1n - 22! 2!1n - 12!n1n - 12 1n + 121n2= +22= n2 - n + n 2 + n= n 2234.nn!an - 2 b + an + 1n - 1 b = 1n - 22!3n - 1n - 224!1n + 12!+1n - 12!31n + 12 - 1n - 124!n! 1n + 12!=+1n - 22!2! 1n - 12!2!n1n - 12 1n + 12n= +22= n2 - n + n 2 + n2=n 235. True. The signs of the coefficients are determined by thepowers of the (–y) terms, which alternate between oddand even.36. True. In fact, the sum of every row is a power of 2.37. The fifth term of the expansion isa 8 The answer is C.4 b12x24 112 4 = 1120x 4 .38. The two smallest numbers in row 10 are 1 and 10. Theanswer is B.39. The sum of the coefficients of (3x-2y) 10 is the same asthe value of (3x-2y) 10 when x=1 and y=1.The answer is A.40. The even-numbered terms in the two expressions areopposite-signed and cancel out, while the odd-numberedterms are identical and add together. The answer is D.41. (a) 1, 3, 6, 10, 15, 21, 28, 36, 45, 55(b) They appear diagonally down the triangle, startingwith either of the 1’s in row 2.(c) Since n and n+1 represent the sides of the givenrectangle, then n(n+1) represents its area. The triangularnumber is 1/2 of the given area. Therefore, then1n + 12triangular number is .2Copyright © 2011 Pearson Education, Inc. Publishing as Addison-Wesley.
ch09_p355_387.qxd 3/17/10 12:44 PM Page 359Section 9.3 Probability 359(d) From part (c), we observe that the nth triangularn1n + 12number can be written as . We know that2binomial coefficients are the values of a n forr br=0,1,2,3,...,n. We can show thatn1n + 12= a n + 1 b as follows:22n1n + 12 1n + 12n1n - 12!=221n - 12!1n + 12!=2!1n - 12!1n + 12!=2!11n + 12 - 22!= a n + 1 b.2So, to find the fourth triangular number, for example,5 # 4 #compute a4 + 15!3!b= a 5 = =2 2 b 2!15 - 22! 2!3!5 # 4= =10.242. (a) 2 (Every other number appears at least twice.)(b) 1(c) No (They all appear in order down the seconddiagonal.)(d) 0 (See Exercise 44 for a proof.)(e) All are divisible by p.(f) Rows that are powers of 2: 2, 4, 8, 16, etc.(g) Rows that are 1 less than a power of 2: 1, 3, 7, 15, etc.(h) For any prime numbered row, or row where the firstelement is a prime number, all the numbers in thatrow (excluding the 1’s) are divisible by the prime. Forexample, in the seventh row (1 7 21 35 35 217 1) 7, 21, and 35 are all divisible by 7.43. The sum of the entries in the nth row equals the sum ofthe coefficients in the expansion of (x+y) n . But thissum, in turn, is equal to the value of (x+y) n when x=1and y=1:2 n = 11 + 12 n= a n 0 b1n 1 0 + a n 1 b1n-1 1 1 + a n 2 b1n-2 1 2= a n 0 b + an 1 b + an 2 b + Á + ann b44. 0 = 11 - 12 n+ Á + ann b10 1 n= a n 0 b1n + a n 1 b1n - 1 1-12+ a n 2 b1n - 2 1-12 2+ Án + a (–1) nn b= a n 0 b - an 1 b + an 2 b + Á + 1-12 n a n n b45. 3 n =(1+2) n■ Section 9.3 ProbabilityExploration 11.2.3.= a n 0 b1n + a n 1 b1n - 1 2 + a n 2 b1n - 2 2 2= a n 0 b + 2an 1 b + 4an 2 b + Á + 2 n a n n bp = 0.994p = 0.994p = 0.994p = 0.006p = 0.006p = 0.0064. P(±)=0.00598+0.01491=0.020890.005980.02089 L 0.286 (A little more than 1 chance in 4.)P1antibody present and +25. P(antibody present |±)=P1+2Quick Review 9.31. 22. 63. 2 3 =84. 6 3 =2165. 52 C 5 =2,598,9606. 10 C 2 =457. 5!=120AntibodiespresentAntibodiesabsentAntibodiespresentAntibodiesabsentAntibodiespresentAntibodiesabsentp = 0.997p = 0.003p = 0.015p = 0.985p = 0.997p = 0.003p = 0.015p = 0.985+–+–+–+–+ Á + ann b2np = 0.00598p = 0.00002p = 0.01491p = 0.97909Copyright © 2011 Pearson Education, Inc. Publishing as Addison-Wesley.
ch09_p355_387.qxd 3/17/10 12:44 PM Page 360360 <strong>Chapter</strong> 9 Discrete Mathematics8. 5 P 3 =609.10.5C 310C 3=5C 210C 2=5!3!2!10!3!7!5!2!3!10!2!8!= 112= 2 9Section 9.3 ExercisesFor #1–8, consider ordered pairs (a, b) where a is the value ofthe red die and b is the value of the green die.41. E={(3, 6), (4, 5), (5, 4), (6, 3)}: P(E)=36 = 1 92. E={both dice even, both dice odd}:3 # 3 + 3 # 3P(E)== 1836 36 = 1 23. E={(2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2),(5, 3), (5, 4), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)}:15P(E)=36 = 5124. E={(1, 1), (1, 2),…,(6, 2), (6, 3)}30P(E)=36 = 5 63 # 35. P(E)=36 = 1 43 # 36. P(E)=36 = 1 47. E={(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2),(3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5)}15P(E)=36 = 5128. E={(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5)}:8P(E)=36 = 2 99. (a) No. 0.25+0.20+0.35+0.30=1.1. The numbers donot add up to 1.(b) There is a problem with Alrik’s reasoning. Since thegerbil must always be in exactly one of the fourrooms, the proportions must add up to 1, just like aprobability function.10. Since 4+3+2+1=10, we can divide each number inthe ratio by 10 and get the proportions relative to thewhole. The table then becomesCompartment A B C DProportion 0.4 0.3 0.2 0.1Yes, this is a valid probability function.11. P(B or T)=P(B)+P(T)=0.3+0.1=0.412. P(R or G or O)=P(R)+P(G)+P(O)=0.2+0.1+0.1=0.413. P(R)=0.214. P(not R)=1-P(R)=1-0.2=0.815. P[not (O or Y)]=1-P(O or Y)=1-(0.2+0.1)=0.716. P[not (B or T)]=1-P(B or T)=1-(0.3+0.1)=0.617. P(B 1 and B 2 )=P(B 1 ) # P(B 2 )=(0.3)(0.3)=0.0918. P(O 1 and O 2 )=P(O 1 ) # P(O 2 )=(0.1)(0.1)=0.0119. P[(R 1 and G 2 ) or (G 1 and R 2 )]=P(R 1 ) # P(G 2 )+P(G 1 ) # P(R 2 )=(0.2)(0.2)+(0.2)(0.2)=0.0820. P(B 1 and Y 2 )=P(B 1 ) # P(Y 2 )=(0.3)(0.2)=0.0621. P(neither is yellow)=P(not Y 1 and not Y 2 )=P(not Y 1 )#P(not Y 2 )=(0.8)(0.8)=0.6422. P(not R 1 and not O 2 )=P(not R 1 )#P(not O 2 )=(0.8)(0.9)=0.7223. There are 24 C 6 =134,596 possible hands; of these, only 11consists of all spades, so the probability is .134,59624. Of the 24 C 6 =134,596 possible hands, one consists of allspades, one consists of all clubs, one consists of all hearts,and one consists of all diamonds, so the probability is4134,596 = 133,649 .25. Of the 24 C 6 =134,596 possible hands, there are4C 4# 20 C 2 =190 hands with all the aces, so the probability190is134,596 = 53542 .26. There are 2C 2# 22 C 4 = 7315 ways to get both black jacksand 4 “other” cards. Similarly, there are 7315 ways to getboth red jacks. These two numbers together count twicethe 2C 2# 2 C 2# 20 C 2 = 190 ways to get all four jacks.Therefore, altogether we have 2 # 7315 - 190 = 14,440distinct ways to have both bowers, so the probability is14,440134,596 = 1901771 .27. (a)0.3 0.20.3A(b) 0.3(c) 0.2(d) 0.2P1A and B2(e) Yes. P(A |B)=P1B228. (a)0.5 0.20.2ABB0.20.1= 0.30.5= 0.6 = P1A2(b) 0.5(c) 0.2(d) 0.1P1A and B2(e) No. P(A |B)== 0.2 = 0.5 Z P1A2P1B2 0.429. P(John will practice)=(0.6)(0.8)+(0.4)(0.4)=0.64Copyright © 2011 Pearson Education, Inc. Publishing as Addison-Wesley.
ch09_p355_387.qxd 3/17/10 12:44 PM Page 361130. (a) P(meatloaf is served)= = 0.2014C 840. (a) = 30035 20C 8 125,970 = 773230(b) P(meatloaf and peas are served)=(0.20)(0.70)14C 5# 6 C 3=0.14(b) = 40,04020C 8 125,970 = 308969(c) P(peas are served)=(0.20)(0.70)+(0.80)(0.30)=0.38(c)20C 831. If all precalculus students were put on a single list and aname then randomly chosen, the probability P(from45,045 + 20,592 + 300312Mr. Abel’s class | girl) would be . But when one of22 = 6=125,970111the two classes is selected at random, and then a student 41. = 19C 2 36from this class is selected, P(from Mr. Abel’s class | girl)P1girl from Mr. Abel’s class2=P1girl232. Within each box, any of the coins is equally likely to bechosen and either side is equally likely to be shown. But ahead in the 2-coin box is more likely to be displayed thana head in the 3-coin box.P1H from 2-coin box2P(from 2-coin box | H)=P1H2= 320C 233. = 19025C 2 300 = 1953034. P(none are defective)= 4 C 0# (0.037) 0 (0.963) 4=(0.963) 4 ≠0.86035. (a) P(cardiovascular disease or cancer)=0.45+0.22=0.67(b) P(other cause of death)=1-0.67=0.3336. P(Yahtzee)= 6 # a 1 56 b 1=129637. The sum of the probabilities is greater than 1 – an impossibility,since the events are mutually exclusive.38. P(at least one false positive)=1-P(no false positives)=1-(0.993) 60 ≠0.344.17239. (a) P(a woman)=254 = 86127(b) P(went to graduate school)=124 + 58254(c) P(a woman who went to graduate school)= 124254 = 62127=== 3 5a 1 2 ba12 20 ba 1 2 ba12 20 b + a 1 2 ba10 25 ba 1 2 ba3 4 ba 1 2 ba3 4 b + a 1 2 ba3 6 b= 91127Section 9.3 Probability 36142. This cannot be true. Let A be the event that it is cloudyall day, B be the event that there is at least 1 hour of sunshine,and C be the event that there is some sunshine, butless than 1 hour. Then A, B, and C are mutually exclusiveevents, so P(A or B or C)=P(A)+P(B)+P(C)=0.22+0.78+P(C)=1+P(C). Then it must be thecase that P(C)=0. This is absurd; there must be someprobability of having more than 0 but less than1 hour of sunshine.43. P(HTTTTTTTTT)= a 1 102 b44. P(THTTTTHTTT)= a 1 102 b =45. P(HHHHHHHHHH)= a 1 102 b =46. P(9 H and 1 T)= 10C 1# a 1 102 b47. P(2 H and 8 T)= 10C 2# a 1 102 b48. P(3 H and 7 T)= 10C 3# a 1 102 b49. P(at least one H)=1-P(0 H)= 1 - 10 C 0# a 1 102 b= 1 -14C 6# 6 C 2 + 14 C 7# 6 C 1 + 14 C 8# 6 C 011024 = 10231024= 68,640125,970 = 176323= 12 10 = 110241102411024= 101024 = 5512= 451024= 1201024 = 1512850. P(at least two H)=1-P(0 H)-P(1 H)=1 - 10 C 0# a 1 102 b - 10 C 1# a 1 102 b = 1 - 111024 = 1013102451. False. A sample space consists of outcomes, which are notnecessarily equally likely.52. False. All probabilities are between 0 and 1, inclusive.53. Of the 36 different, equally-likely ways the dice can land,4 ways have a total of 5. So the probability is 4/36=1/9.The answer is D.54. A probability must always be between 0 and 1, inclusive.The answer is E.55. P(B and A)=P(B) P(A|B), and for independentevents, P(B and A)=P(B)P(A). It follows that theanswer is A.56. A specific sequence of one “heads” and two “tails” hasprobability (1/2) 3 =1/8. There are three such sequences.The answer is C.Copyright © 2011 Pearson Education, Inc. Publishing as Addison-Wesley.
ch09_p355_387.qxd 3/17/10 12:44 PM Page 364364 <strong>Chapter</strong> 9 Discrete Mathematics39. The numbers of seats in each row form a finite arithmeticsequence with a 1 =7, d=2, and n=25. The totalnumber of seats is25.(b) Notice that after n-2 months, there are a n–1 pairs,of which a n–2 (the number of pairs present onemonth earlier) are fertile. Therefore, after n-12 32172 + 125 - 121224 = 775 months, the number of pairs will be a n =a n–1+a n–2 : to last month’s total, we add the number ofnew pairs born. Thus a 4 =3, a 5 =5, a 6 =8, a 7 =13,a 8 =21, a 9 =34, a 10 =55, a 11 =89, a 12 =144,a 13 =233.40. The numbers of tiles in each row form a finite arithmeticsequence with a 1 =15, a n =30, and n=16. The totalnumber of tiles is15 + 3016a b = 360 .241. The ten-digit numbers will vary; thus the sequences willvary. The end result will, however, be the same. Each limitwill be 9. One example is:Five random digits: 1, 4, 6, 8, 9Five random digits: 2, 3, 4, 5, 6List: 1, 2, 3, 4, 4, 5, 6, 6, 8, 9Ten-digit number: 2, 416, 345, 689Ten-digit number: 9, 643, 128, 564a 1 =positive difference of the ten-digit numbers=7, 226, 782, 875a n+1 = sum of the digits of a n ,soa 2 =sum of the digits of a 1 =54a 3 =sum of the digits of a 2 =9.All successive sums of digits will be 9, so the sequenceconverges and the limit is 9.42. Everyone should end up at the word “all.”43. True. Since two successive terms are negative, the commonratio r must be positive, and so the sign of the first termdetermines the sign of every number in the sequence.44. False. For example, consider the sequence 5, 1, –3, –7, …45. a 1 =2 and a 2 =8 implies d=8-2=6c=a 1 -d so c=2-6=–4a 4 =6#4+(–4)=20.The answer is A.(c) Since a 1 is the initial number of pairs, and a 2 is thenumber of pairs after one month, we see that a 13 is thenumber of pairs after 12 months.50. Use a calculator: a 1 =1, a 2 =1, a 3 =2, a 4 =3, a 5 =5,a 6 =8, a 7 =13. These are the first seven terms of theFibonacci sequence.51. (a) For a polygon with n sides, let A be the vertex inquadrant I at the top of the vertical segment, and letB be the point on the x-axis directly below A.Together with (0, 0), these two points form a right triangle;the acute angle at the origin has measureu = 2p , since there are 2n such triangles making2n = p nup the polygon. The length of the side opposite thisangle issin u = sin p , and there are 2n such sides making upnpthe perimeter of the polygon, so sin orn = a n2n ,a n =2n sin(∏/n).(b) a 10 ≠6.1803, a 100 ≠6.2822, a 1000 ≠6.2832,a 10,000 ≠6.2832≠2∏. It appears that a n : 2∏ asn : q , which makes sense since the perimeter ofthe polygon should approach the circumference ofthe circle.y46. lim 2n = lim so the sequence diverges.n: q n: q n1>2 = q,The answer is B.47. r = a 2= 6 a 1 2 = 3a # 3 5 6 = a 1 r 5 = 2 = 486 and a 2 = 6, soa 6= 486 = 81.a 2 6The answer is E.48. The geometric sequence will be defined by a n+1 =a n ÷3for n Ú 1 and a 1 Z 0.a 2 = a 13a 3 = a 23 = a 1>3= a 13 9a 4 = a 33 = a 1>9= a 13 27a n = a 13 n - 1, which represents a geometric sequence.The answer is C.49. (a) a 1 =1 because there is initially one male-female pair(this is the number of pairs after 0 months). a 2 =1because after one month, the original pair has onlyjust become fertile. a 3 =2 because after two months,the original pair produces a new male-female pair.θ1a n2n52. P 1 =525,000; P n =1.0175P n-1 , n Ú 253. The difference of successive terms in {log (a n )} will be ofthe form log (a n1 )-log (a n )=log a a n + 1b. Since {a n }isa nageometric, is constant. This makes log a a n + 1n + 1b constant,so {log (a n )} is a sequence with a constanta n a ndifference(arithmetic).54. The ratios of successive terms in {10 bn } will be of the form10 bn±1 ÷10 bn =10 bn±1 –bn . Since {b n } is arithmetic, b n1 -b nis constant. This makes 10 bn±1-bn constant, so {10 bn } is asequence with a common ratio (geometric).xCopyright © 2011 Pearson Education, Inc. Publishing as Addison-Wesley.
ch09_p355_387.qxd 3/17/10 12:44 PM Page 365Section 9.5 Series 36555. a 1 =[1 1], a 2 =[1 2], a 3 =[2 3], a 4 =[3 5],5. aa 5 =[5 8], a 6 =[8 13], a 7 =[13 21]. The entries in1 =1 and a 2 =2 yields r = 2 1 = 2the terms of this sequence are successive pairs of termsa 10 =1#2 9 =512from the Fibonacci sequence.a 10 =51256. a 1 =[1 a] r = c 1 d0 1 d 6. a 4 =1 and a 4 =a 1# r 3 ; a 6 =2 and a 6 =a 1# r 5a 1# r 5a 2 =a 1# r=[1 a+d]aa 3 =a 2#1# r 3 = 2 1r=[1 d+a+d]=[1 a+2d]#a 4 =a 3 r=[1 d+a+2d]=[1 a+3d]a n =[1 a+(n-1)d].So, the second entries of this geometric sequence ofmatrices form an arithmetic sequence with the first term aand common difference d.■ Section 9.5 SeriesExploration 11. 3 + 6 + 9 + 12 + 15 = 452. 5 2 + 6 2 + 7 2 + 8 2 = 25 + 36 + 49 + 64 = 1743. cos(0)+cos(∏)+…+cos(11∏)+cos(12∏)=1-1+1+…-1+1=14. sin(0)+sin(∏)+…+sin(k∏)+...=0+0+…+0+...=035. …31+…=10 + 3100 + 31,000 + 1,000,000 + 3Exploration 21. 1+ 2+ 3+…+99+1002. 100+ 99+ 98+…+ 2+ 13. 101+101+101+…+101+1014. 100(101)=10,1005. The sum in 4 involves two copies of the same progression,so it doubles the sum of the progression. The answer thatGauss gave was 5050.Quick Review 9.51. a 1 =4; d=2 soa 10 =a 1 +(n-1)da 10 =4+(10-1)2=4+18=22a 10 =222. a 1 =3; a 2 =1sod=1-3=–2a 10 =a 1 +(n-1)da 10 =3+(10-1)(–2)=3-18=–15a 10 =–153. a 3 =6 and a 8 =21a 3 =a 1 +2d and a 8 =a 1 +7d(a 1 +7d)-(a 1 +2d)=21-6 so5d=15 Q d=3.6=a 1 +2(3) so a 1 =0a 10 =0+9(3)=27a 10 =274. a 5 =3, and a n+1 =a n +5 for n Ú 1 Q a 6 =3+5=8a 5 =3 and a 6 =8 Q d=5a 5 =a 1 +4d so 3=a 1 +4(5) Q a 1 =–17a 10 =–17+9(5)=28a 10 =28r 2 =2a a 10 = 524 1-229 = -25601 = 5= -4064 ; 247. a 7 =5 and r=–2 Q 5=a 1 (–2) 6a 10 = -408. a 8 =10 and a 8 =a 1 r 7 ; a 12 =40 Q a 12 =a 1 r 119.10.r 4 = 4; so r = 142 1>4Section 9.5 Exercises1.2.3.4.5.6.1 = a 1 1222 3 ;1a 10 =222 12229 = 1622222 = 8a 10 = 8a 1# r 11a 1# r 7 = 401010 = a 1 1142 1>4 2 7;a 10 = 10 2 9 = 10149>4 2= 1014 2>4 2 = 10 # 2 = 204 7>4141>4 4 7>4a 10 = 205a n 2 = 1 + 4 + 9 + 16 + 25 = 55n = 15a 12n - 12 = 1 + 3 + 5 + 7 + 9 = 25n = 111a 16k - 132k = 110a 13k - 12k = 1n + 1a k 2k = 1n + 1a k 3k = 1qa 61-22 kk = 0qa 51-32 kk = 0For #7–12, use one of the formulas S n =na a 1 + a nb or2nS n = [2a 1 +(n-1)d]. In most cases, the first of these is2easier (since the last term a n is given); note thata n - a 1n= + 1.d7. 6 # aQr = 22a 1 =a 1 = 104 7>4-7 + 13b = 6 # 3 = 182#11222 = 13 222#Copyright © 2011 Pearson Education, Inc. Publishing as Addison-Wesley.
ch09_p355_387.qxd 3/17/10 12:44 PM Page 366366 <strong>Chapter</strong> 9 Discrete Mathematics8. 6 # a9.10.-8 + 27b = 3 # 19 = 57280 # a 1 + 80 b = 40 # 81 = 3240235 # a 2 + 70 b = 35 # 36 = 1260211. 13 #117 + 33a b = 13 # 75 = 9752ƒ` `111 + 2712. 29a b = 29 # 69 = 20012a 1 11 - r n 2For #13–16, use the formula S n = . Note that1 - ra nn=1+log |r| = 1 + ln ƒa n>a 1a 1 ln ƒrƒ.311 - 2 13 213.1 - 2511 - 3 10 2= 24,57314.1 - 3= 147,6204231 - 11>62 9 415.1 - 11>62= 50.411 - 6 -9 2 L 50.44231 - 1-1>62 10 416.1 - 1-1>62= 3611 - 6 -10 2 = 36 - 6 -8 L 36nFor #17–22, use one of the formulas S n = [2a 1 +(n-1)d]a 1 11 - r n 22or S n = .1 - r1017. Arithmetic with d=3: [2 # 2+(10-1)(3)]2=5 # 31=155918. Arithmetic with d=–6: [2 # 14+(9-1)(–6)]2=9 # (–10)=–90431 - 1-1>22 12 419. Geometric with r= - 1 :2 1 - 1-1>228= 11 - 2 -12 2 L 2.666320. Geometric with r= - 1 2 : 631 - 1-1>2211 41 - 1-1>22= 4 # 11 + 2 -11 2 L 4.00221. Geometric with r= -11: -131 - 1-1129 41 - 1-112= - 1 # 11 + 11 9 2 = -196,495,6411222. Geometric with r= -12: -231 - 1-1228 41 - 1-122= - 2 # 11 - 12 8 2 = 66,151,0301323. (a) The first six partial sums are {0.3, 0.33, 0.333, 0.3333,0.33333, 0.333333}. The numbers appear to be approachinga limit of 0.3 = 1>3. The series is convergent.(b) The first six partial sums are {1, –1, 2, –2, 3, –3}. Thenumbers approach no limit. The series is divergent.24. (a) The first six partial sums are {–2, 0, –2, 0, –2, 0}. Thenumbers approach no limit. The series is divergent.(b) The first six partial sums are {1, 0.3, 0.23, 0.223,0.2223, 0.22223}. The numbers appear to be approachinga limit of 0.2 = 2>9. The series is convergent.25. r = 1 6, so it converges to S=21 - 11>22 = 12.26. r = 1 4, so it converges to S=31 - 11>32 = 6.27. r=2, so it diverges.28. r=3, so it diverges.29. r = 1 3>4, so it converges to S=41 - 11>42 = 1.30. r = 2 10>3, so it converges to S=31 - 12>32 = 10.31. 7 + 1499 = 69399 + 1499 = 7079932. 5 + 9399 = 5 + 3133 = 1963333. -17 - 268999 = -17,251 99934. -12 - 876 292= -12 -999 333 = -4288 33335. (a) The ratio of any two successive account balances isr=1.1. That is,$22,000$20,000 = $24,200$22,000 = $26,620$24,200 = $29,282$26,620 = 1.1.(b) Each year, the balance is 1.1 times as large as the yearbefore. So, n years after the balance is $20,000, it willbe $20,000 (1.1) n .(c) The sum of the eleven terms of the geometric$20,00011 - 1.1 11 2sequence is = $370,623.34.1 - 1.136. (a) The difference of any two successive account balancesis d=$2016. That is $20,016-$18,000=$22,032-$20,016=$24,048-$22,032=$26,064-$24,048=$2016.(b) Each year, the balance is $2016 more than the yearbefore. So, n years after the balance is $18,000, it willbe $18,000+$2016n.(c) The sum of the eleven terms of the arithmeticsequence is11321$18,0002 + 11021$201624 = $308,880.237. (a) The first term, 120(1+0.07/12) 0 , simplifies to 120.The common ratio of terms, r, equals 1+0.07/12.(b) The sum of the 120 terms is120 31 - 11 + 0.07>122 120 4= $20,770.18.1 - 11 + 0.07>12238. (a) The first term, 100(1+0.08/12) 0 , simplifies to 100.The common ratio of terms, r, equals 1+0.08/12.Copyright © 2011 Pearson Education, Inc. Publishing as Addison-Wesley.
ch09_p355_387.qxd 3/17/10 12:44 PM Page 367Section 9.5 Series 367(b) The sum of the 120 terms is100 31 - 11 + 0.08>122 120 4= $18,294.60.1 - 11 + 0.08>12239. 2+2 # [2(0.9)+(2(0.9) 2 +2(0.9) 3 +…+2(0.9) 9 ]9=2+ a 410.92 k = 2 + 3.611 - 0.99 2=k = 11 - 0.938-36(0.9 9 )≠24.05 m: The ball travels down 2 m.Between the first and second bounces, it travels 2(0.9) mup, then the same distance back down. Between thesecond and third bounces, it travels 2(0.9) 2 m up, then thesame distance back down, etc.40. f(x) does have a horizontal asymptote to the left (sinceas x : – q, f(x) : –40), but as x : q, f(x) : q.Thegiven series is geometric with r=1.05; the sum of the firstn terms in the series is f(n). Observing that f(x) : q asx : q is confirmation that the geometric series diverges.41. False. The series might diverge. For example, examine theseries 1+2+3+4+5 where all of the terms are positive.Consider the limit of the sequence of partial sums.The first five partial sums are {1, 3, 6, 10, 15}. These numbersincrease without bound and do not approach a limit.Therefore, the series diverges and has no sum.42. False. Justifications will vary. One example is to examineqa nn = 1andqa 1-n2.n = 1qqBoth of these diverge, but a 1n + 1-n22 = a 0 = 0.n = 1n = 1So the sum of the two divergent series converges.43. 3 –1 +3 –2 +3 –3 +…+3 –n +…=13 + 1 9 + 127 + 181 + 1243 + Á + 1 3 n + ÁThe first five partial sums are e 1 These3 , 4 9 , 1327 , 4081 , 121243 f.appear to be approaching a limit of 1/2, which wouldsuggest that the series converges to 1/2. The answer is A.q44. If a x n = 4, then x=0.8.n = 1qa 0.8 n = 0.8+0.64+0.512+0.4096+0.32768n = 1+0.262144+…The first six partial sums are {0.8, 1.44, 1.952, 2.3616,2.68928, 2.951424}. It appears from this sequence ofpartial sums that the series is converging. If the sequence ofpartial sums were extended to the 40th partial sum, youwould see that the series converges to 4. The answer is D.45. The common ratio is 0.75/3=0.25, so the sum of the infiniteseries is 3/(1-0.25)=4. The answer is C.46. The sum is an infinite geometric series with ƒrƒ = 5>3 7 1.The answer is E.47. (a) Heartland: 19,237,759 people. Heartland: Southeast:Southeast: 42,614,977 people. Iowa ≠52.00 Alabama ≠86.01Kansas ≠32.68 Arkansas ≠50.26(b) Heartland: 517,825 mi 2 . Minnesota ≠58.29 Florida ≠272.53Southeast: 348,999 mi 2 . Missouri ≠80 .28 Ge<strong>org</strong>ia ≠138.97Nebraska ≠22.12 Louisiana ≠93.5919,237,759(c) Heartland:517,825L 37.15 people/mi 2 . N. Dakota ≠9.08 Mississippi ≠59.65S. Dakota ≠9.79 S. Carolina ≠128.9542,614,977Southeast: L 122.11 people/mi 2 .348,999Total ≠264.24 Total ≠829.96Average ≠37.75 Average ≠118.57(d) The table is shown on the right; the answer differsbecause the overall population densityg population1is generally not the same as the average of the population densities,The larger statesn g a population b.g areaareawithin each group have a greater effect on the overall mean density. In a similar way, if a student’s grades are based on a 100-point test and four 10-point quizzes, her overall average grade depends more on the test grade than on the four quiz grades.848. a 1k 2 - 22k = 1Copyright © 2011 Pearson Education, Inc. Publishing as Addison-Wesley.
ch09_p355_387.qxd 3/17/10 12:44 PM Page 368368 <strong>Chapter</strong> 9 Discrete Mathematicsn49. The table suggests that S n = a F k = F n + 2 - 1.n F n S n F n+2 -11 1 1 12 1 2 23 2 4 44 3 7 75 5 12 126 8 20 207 13 33 338 21 54 549 34 88 8850. The nth triangular number is simply the sum of the first nconsecutive positive integers:1+2+3+…+n= na 1 + n n1n + 12b =221n - 12n n1n + 1251. Algebraically: T n1 +T n = +22n 2 - n + n 2 + n== n 22Geometrically: The array of black dots in the figurerepresents T n =1+2+3+…+n (that is, there areT n dots in the array). The array of gray dots representsT n1 =1+2+3+…+(n-1). The two triangulararrays fit together to form an n*n square array, whichhas n 2 dots.n – 1nn – 1nn152. If a ln n for all n, then the sum diverges since ask = 1k Ún : q, ln n : q.[–1, 10] by [–2, 4]k = 1■ Section 9.6 Mathematical InductionExploration 11. Start with the rightmost peg if n is odd and the middlepeg if n is even. From that point on, the first move formoving any smaller stack to a destination peg should bedirectly to the destination peg if the smaller stack’s size nis odd and to the other available peg if n is even. The factthat the winning strategy follows such predictable rules iswhat makes it so interesting to students of computer programming.Exploration 21. 43, 47, 53, 61, 71, 83, 97, 113, 131, 151. Yes.2. 173, 197, 223, 251, 281, 313, 347, 383, 421, 461. Yes.3. 503, 547, 593, 641, 691, 743, 797, 853, 911, 971. Yes.Inductive thinking might lead to the conjecture thatn¤+n+41 is prime for all n, but we have no proofas yet!4. The next 9 numbers are all prime, but 40¤+40+41is not. Quite obviously, neither is the number41¤+41+41.Quick Review 9.61. n 2 +5n2. n 2 -n-63. k 3 +3k 2 +2k4. (n+3)(n-1)5. (k+1) 36. (n-1) 37. f(1)=1+4=5, f(t)=t+4,f(t+1)=t+1+4=t+518. f(1)=1 + 1 = 1 2 , f1k2 = kk + 1 ,k + 1f(k+1)=k + 1 + 1 = k + 1k + 22 # 19. P(1)=3 # 1 + 1 = 1 2 ,P1k2 =2k; P(k + 1)3k + 110. P(1)=2(1) 2 -1-3=–2, P(k)=2k 2 -k-3,P(k+1)=2(k+1) 2 -(k+1)-3=2k 2 +3k-2Section 9.6 Exercises1. P n : 2+4+6+…+2n=n 2 +n.P 1 is true: 2(1)=1 2 +1.Now assume P k is true: 2+4+6+ … +2k=k 2 +k. Add 2(k+1) to both sides:2+4+6+…+2k+2(k+1)=k 2 +k+2(k+1)=k 2 +3k+2=k 2 +2k+1+k+1=(k+1) 2 +(k+1), soP k+1 is true. Therefore, P n is true for all n Ú 1.2. P n : 8+10+12+…+(2n+6)=n 2 +7n.P 1 is true: 2(1)+6=1 2 +7 # 1.Now assume P k is true:8+10+12+…+(2k+6)=k 2 +7k.Add 2(k+1)+6=2k+8 to both sides:8+10+12+…+(2k+6)+[2(k+1)+6]=k 2 +7k+2k+8=(k 2 +2k+1)+7k+7=(k+1) 2 +7(k+1), so P k+1 is true.Therefore, P n is true for all n Ú 1.3. P n : 6+10+14+…+(4n+2)=n(2n+4).P 1 is true: 4(1)+2=1(2+4).=21k + 1231k + 12 + 1 = 2k + 23k + 4Copyright © 2011 Pearson Education, Inc. Publishing as Addison-Wesley.
ch09_p355_387.qxd 3/17/10 12:44 PM Page 369Section 9.6 Mathematical Induction 369Now assume P k is true:6+10+14+…+(4k+2)=k(2k+4).Add 4(k+1)+2=4k+6 to both sides:6+10+14+…+(4k+2)+[4(k+1)+2]=k(2k+4)+4k+6=2k 2 +8k+6=(k+1)(2k+6)=(k+1)[2(k+1)+4], so P k+1is true. Therefore, P n is true for all n Ú 1.4. P n : 14+18+22+…+(4n+10)=2n(n+6).P 1 is true: 4(1)+10=2 # 1(1+6). Now assume P k istrue:14+18+22+…+(4k+10)=2k(k+6). Add4(k+1)+10=4k+14 to both sides:14+18+22+…+(4k+10)+[4(k+1)+10]=2k(k+6)+(4k+14)=2(k 2 +8k+7)=2(k+1)(k+7)=2(k+1)(k+1+6), so P k+1 is true. Therefore, P n istrue for all n Ú 1.5. P n : 5n-2. P 1 is true: a 1 =5 # 1-2=3.Now assume P k is true: a k =5k-2.To get a k+1 , add 5 to a k ; that is,a k+1 =(5k-2)+5=5(k+1)-2 This shows thatP k+1 is true. Therefore, P n is true for all n Ú 1.6. P n : a n =2n+5. P 1 is true: a 1 =2 # 1+5=7.Now assume P k is true: a k =2k+5.To get a k+1 , add 2 to a k ; that is,a k+1 =(2k+5)+2=2(k+1)+5 This shows thatP k+1 is true. Therefore, P n is true for all n Ú 1.7. P n : a n =2 # 3 n-1 .P 1 is true: a 1 =2 # 3 1-1 =2 # 3 0 =2.Now assume P k is true: a k =2 # 3 k-1 .To get a k+1 , multiply a k by 3; that is,a k+1 = 3 # 2 # 3 k - 1 =2 # 3k=2 # 3 (k+1)-1 . This showsthat P k+1 is true. Therefore, P n is true for all n Ú 1.8. P n : a n =3 # 5 n-1 .P 1 is true: a 1 =3 # 5 1-1 =3 # 5 0 =3.Now assume P k is true: a k =3 # 5 k-1 .To get a k+1 , multiply a k by 5; that is,a k+1 =5#3#5 k-1 =3 # 5 k =3 # 5 (k+1)-1 . This showsthat P k+1 is true. Therefore, P n is true for all n Ú 1.111 + 129. P 1 : 1= .2k1k + 12P k : 1 + 2 + … + k =21k + 121k + 22P k+1 : 1+2+…+k+(k+1)=.2112 - 1212 + 1210. P 1 : (2(1)-1) 2 =.3k12k - 1212k + 12P k : 1 2 +3 2 +…+(2k-1) 2 =.3P k+1 : 1 2 +3 2 +…+(2k-1) 2 +(2k+1) 211. P 1 :1k + 1212k + 1212k + 32= .3P k :11 # 2 = 11 + 1 .11 # 2 + 12 # 3 +… + 1= kk1k + 12 k + 1 .1P k+1 :…1+1 # 2 + 12 # 3 + k1k + 121+1k + 121k + 22 = k + 1k + 2 .111 + 1212 + 1213 + 3 - 1212. P 1 : 1 4 =#30P k : 1 4 +2 4 +…+k 4k1k + 1212k + 1213k 2 + 3k - 12=.30P k+1 : 1 4 +2 4 +…+k 4 +(k+1) 41k + 121k + 2212k + 3213k 2 + 9k + 52=.3013. P n : 1+5+9+…+(4n-3)=n(2n-1).P 1 is true: 4(1)-3=1 # (2 # 1-1).Now assume P k is true:1+5+9+…+(4k-3)=k(2k-1).Add 4(k+1)-3=4k+1 to both sides:1+5+9+…+(4k-3)+[4(k+1)-3]=k(2k-1)+4k+1=2k 2 +3k+1=(k+1)(2k+1)=(k+1)[2(k+1)-1],so P k+1 is true.Therefore, P n is true for all n Ú 1.14. P n : 1+2+2 2 +…+2 n-1 =2 n -1.P 1 is true: 2 1-1 =2 1 -1.Now assume P k is true:1+2+2 2 +…+2 k-1 =2 k -1.Add 2 k to both sides,1+2+2 2 +…+2 k-1 +2 k=2 k -1+2 k =2 # 2 k -1=2 k+1 -1, soP k+1 is true.Therefore, P n is true for all n Ú 1.1115. P n : +…+n1n + 12 = n1 # 2 + 12 # 3n + 1 .1P 1 is true:1 # 2 = 11 + 1 .Now assume P k is true:1…1+1 # 2 + 12 # 3 + k1k + 12 .1Addto both sides:1k + 121k + 2211+…+k1k + 12 + 11 # 2 + 12 # 31k + 121k + 22k1k1k + 22 + 1= +=k + 1 1k + 121k + 22 1k + 121k + 221k + 121k + 12=1k + 121k + 22 = k + 1k + 2 = k + 11k + 12 + 1 ,so P k+1 is true.Therefore, P n is true for all n Ú 1.11n16. P n : +…+ =1 # 3 + 13 # 5 12n - 1212n + 12 2n + 1 .1P 1 is true: it says that1 # 3 = 12 # 1 + 1 .Now assume P k is true:Copyright © 2011 Pearson Education, Inc. Publishing as Addison-Wesley.
ch09_p355_387.qxd 3/17/10 12:44 PM Page 370370 <strong>Chapter</strong> 9 Discrete Mathematics11… +12k - 1212k + 12 = k1 # 3 + 13 # 5 + 2k + 1 .a 1 11 - r k 2 + a 1 r k 11 - r2=1 - r1Adda 1 - a 1 r k + 1321k + 12 - 14321k + 12 + 14= a 1 - a 1 r k - a 1 r k + 1=,1 - r1 - r1= to both sides, and we haveso P k+1 is true. Therefore, P n is true for all positive12k + 1212k + 32integers n.11+…+1 # 3 + 13 # n5 12k - 1212k + 1222. P n : S n = [2a 1 +(n-1)d].12+First note that a321k + 12 - 14321k + 12 + 14n =a 1 +(n-1)d. P 1 is true:11k1k12k + 32 + 1S 1 = [2a 1 +(1-1)d]= (2a 1 )=a 1 .= + = …222k + 1 12k + 1212k + 32 12k + 1212k + 32kNow assume P12k + 121k + 12k is true: S k = [2a 1 +(k-1)d].k + 1= =21k + 12 + 1 ,212k + 1212k + 32Add a k+1 =a 1 +kd to both sides, and observe thatso P k+1 is true.S k +a k+1 =S k+1 .Therefore, P n is true for all n Ú 1.Then we have17. P n : 2 n kÚ 2n.SP 1 is true: 2 1 2#k+1 = [2a 1 +(k-1)d]+a 1 +kdÚ 1 (in fact, they are equal). Now assume2P k is true: 2 k Ú 2k.1=kaThen 2 k+1 =2 # 2 k 2#1 + k(k-1)d+a 1 +kdÚ 2k2=2 # (k+k) Ú 2(k+1), so P k+1 is true. Therefore, P1n =(k+1)a 1 + k(k+1)dis true for all n Ú 1.2k + 118. P n : 3 n Ú 3n.= [2a 1 +(k+1-1)d].P 1 is true: 3 1 3#2Ú 1 (in fact, they are equal). Now assumeTherefore, PP k is true: 3 k k+1 is true, so P n is true for all n Ú 1.Ú 3k.Then 3 k+1 =3 3 k n# Ú 3#3kn1n + 12=3 #23. P n :(k+2k) Ú 3(k+1), so P k+1 is true.a k = .k = 1 2Therefore, P n is true for all n Ú 1.119. P n : 3 is a factor of n 3 +2n.P 1 is true: a k = 1 = 1 # 2P 1 is true: 3 is a factor of 1 3 k = 1 2+2 #.1=3.Now assume P k is true: 3 is a factor of k 3 kk1k + 12+2k.Now assume P k is true:Then (k+1) 3 a i = .+2(k+1)i = 1 2=(k 3 +3k 2 +3k+1)+(2k+2)Add (k+1) to both sides, and we have=(k 3 +2k)+3(k 2 +k+1).k + 1k1k + 12Since 3 is a factor of both terms, it is a factor of the sum,a i = + 1k + 12i = 1 2so P k+1 is true. Therefore, P n is true for all n Ú 1.k1k + 12 21k + 12 1k + 121k + 2220. P n : 6 is a factor of 7 n -1.= + =P 1 is true: 6 is a factor of 7 1 222-1=6.1k + 121k + 1 + 12Now assume P k is true, so that 6 is a factor of= , so P k+1 is true.7 k -1=6.2Then 7 k+1 -1=7 # 7 k -1=7(7 k -1)+6. Since 6 is a Therefore, P n is true for all n Ú 1.factor of both terms of this sum, it is a factor of the sum,nso P k+1 is true.24. P n : P 1 is true: 1 3 = 12 2 2a k 3 = n2 1n + 12 2.4 .k = 1 4Therefore, P n is true for all positive integers n.Now assume P k is true so that21. P n : The sum of the first n terms of a geometric sequencek 2 1k + 12a 1 11 - r n 21 3 +2 3 + Á2+k 3 = . Add (k+1) 3 to bothwith first term a 1 and common ratio r Z 1 is .41 - ra 1 11 - r 1 sides and we have2P 1 is true: a 1 = .1 3 +2 3 + Á +k 3 +(k+1) 31 - rk 2 1k + 12 2k 2 1k + 12 2 + 41k + 12 3Now assume P k is true so that= +(k+1) 3 =a 1 11 - r k 442a 1 +a 1 r+ Á =a 1 r k-1 =11 - r2 .1k + 12 2 1k 2 + 4k + 42 1k + 12 2 11k + 12 + 12 2= =Add a 1 r k 44to both sides:a 1 +a 1 r+ Á =a 1 r k-1so P k+1 is true. Therefore, P n is true for all positiveintegers.+a 1 r k = a 111 - r k 2+ a11 - r2 1 r kCopyright © 2011 Pearson Education, Inc. Publishing as Addison-Wesley.
ch09_p355_387.qxd 3/17/10 12:44 PM Page 371Section 9.6 Mathematical Induction 37150025. Use the formula in 23: a k = 1500215012= 125,250k = 1 225026. Use the formula in Example 2: a k 2 = 125021251215012k = 16=5,239,625n n 327. Use the formula in 23: a k = a k - a kk = 4 k = 1 k = 1n1n + 12n 2 + n - 12 1n - 321n + 42= - 3 # 4==2 2 227528. Use the formula in 24: a k 3 = 1752 2176 2 2k = 1 4=8,122,5003529. Use the formula in 14: a 2 k - 1 =2 35 -1≠3.44*10 10k = 11530. Use the formula in 24: a k 3 = 1152 2116 2 2k = 1 4=14,400nn nn31. a (k 2 -3k+4)= a k 2 - a 3k + a 4k = 1k = 1 k = 1 k = 1n1n + 1212n + 12 n1n + 12=- 3 c d + 4n62n1n 2 - 3n + 82=3nnn32. a 12k 2 n+ 5k - 22 = a 2k 2 + a 5k - a 2 = 2k = 1k = 1 k = 1 k = 1n1n + 1212n + 12 n1n + 12= 2c d+5c d - 2n62n1n + 5214n + 12= n14n2 + 21n + 52=66nn33. a 1 = n2 1n + 12 2a 1k 3 n- 12 = a k 3 --nk = 1k = 1 k = 1 4n1n 3 + 2n 2 + n - 42== n1n - 121n2 + 3n + 4244nn34. a 1k 3 n+ 4k - 52 = a k 3 n+ a 4k- a 5k = 1k = 1n 2 1n + 12 2 n1n + 12=+ 4c d - 5n42n1n 3 + 2n 2 + 9n - 122=4= n1n - 121n2 + 3n + 122435. The inductive step does not work for two people. Sendingthem alternately out of the room leaves one person (andone blood type) each time, but we cannot conclude thattheir blood types will match each other.36. The number k is a fixed number for which the statementP k is known to be true. Once the anchor is established, wecan assume that such a number k exists. We cannotassume that P n is true, because n is not fixed.37. False. Mathematical induction is used to show that a statementP n is true for all positive integers.38. True. (1+1) 2 =4=4(1). P n is false, however, for allother values of n.k = 1k = 139. The inductive step assumes that the statement is true forsome positive integer k. The answer is E.40. The anchor step, proving P 1 , comes first. The answer is A.41. Mathematical induction could be used, but the formulafor a finite arithmetic sequence with a 1 =1, d=2 wouldalso work. The answer is B.42. The first two partial sums are 1 and 9. That eliminates allanswers except C. Mathematical induction can be used toshow directly that C is the correct answer.43. P n : 2 is a factor of (n+1)(n+2). P 1 is true because 2 isa factor of (2)(3). Now assume P k is true so that 2 is afactor of (k+1)(k+2). Then[(k+1)+1][(k+1)+2]=(k+2)(k+3)=k 2 +5k+6=k 2 +3k+2+2k+4=(k+1)(k+2)+2(x+2). Since 2 is a factor ofboth terms of this sum, it is a factor of the sum, and so P k+1is true. Therefore, P n is true for all positive integers n.44. P n : 6 is a factor of n(n+1)(n+2). P 1 is true because 6is a factor of (1)(2)(3). Now assume P k is true so that 6 isa factor of k(k+1)(k+2). Then (k+1)[(k+1)+1][(k+1)+2]=k(k+1)(k+2) +3(k+1)(k+2).Since 2 is a factor of (k+1)(k+2), 6 is a factor of bothterms of the sum and thus of the sum itself, and so P k+1 istrue.45. Given any two consecutive integers, one of them must beeven. Therefore, their product is even. Since n+1 andn+2 are consecutive integers, their product is even.Therefore, 2 is a factor of (n+1)(n+2).46. Given any three consecutive integers, one of them mustbe a multiple of 3, and at least one of them must be even.Therefore, their product is a multiple of 6. Since n, n+1and n+2 are three consecutive integers, 6 is a factor ofn(n+1)(n+2).47. P n : F n+2 -1= a F k . P 1 is true sincek = 1F 1+2 =1=F 3 -1=2-1=1, which equals1a F k = 1. Now assume that P k is true:k = 1kF k + 2 - 1 = a F i . Then F (k+1)+2 -1=F k+3 -1=F k+1 +F k+2 -1k=(F k+2 -1)+F k+1 = a a F i b + F k + 1k + 1i = 1n= a F i , so P k+1 is true. Therefore, P n is truei = 1for all n 1.48. P n : a n
ch09_p355_387.qxd 3/17/10 12:44 PM Page 372372 <strong>Chapter</strong> 9 Discrete Mathematics49. P n : a-1 is a factor of a n -1. P 1 is true because a-1 isa factor of a-1. Now assume P k is true so that a-1 isa factor of a k -1. Then a k+1 -1=a # ak-1=a(a k -1)+(a-1). Since a-1 is a factor of bothterms in the sum, it is a factor of the sum, and so P k+1 istrue. Therefore, P n is true for all positive integers n.50. Let P(a)=a n -1. Since P(1)=1 n -1=0, the FactorTheorem for polynomials allows us to conclude thata-1 is a factor of P.51. P n : 3n-4 n for n 2. P 2 is true since3 # 2 - 4 … 2. Now assume that P k is true:3k-4 2. Then 3(k+1)-4=3k+3-4=(3k-4)+3k k+3 k+1, so P k+1 is true.Therefore, P n is true for all n 2.52. P n : 2 n n 2 for n 4. P 4 is true since 2 4 4 2 . Now assumethat P k is true: 2 k k 2 . Then2 k+1 = 2 # 2 k Ú 2 # k 2 Ú 2 # k 2 Ú k 2 + 2k + 1=(k+1) 2 . The inequality 2k 2 k 2 +2k+1, or equivalently,k 2 2k+1, is true for all k 4 becausek 2 = k # k 4k=2k+2k>2k+1.) Thus P k+1 is true,so P n is true for all n 4.53. Use P 3 as the anchor and obtain the inductive step byrepresenting any n-gon as the union of a triangle and an(n-1)-gon.■ Section 9.7 Statistics and Data (Graphical)Exploration 11. We observe that the numbers seem to be centered a bitbelow 13. We would need to take into account thedifferent state populations (not given in the table) inorder to compute the national average exactly; but, justfor the record, it was about 12.5 percent.2. We observe in the stemplot that three states havepercentages above 15.3. We observe in the stemplot that the bottom five states areall below 10%. Returning to the table, we pick these outas Alaska, Colorado, Ge<strong>org</strong>ia, Texas, and Utah.4. The low outlier is Alaska, where older people would be lesswilling or able to cope with the harsh winter conditions.The high outlier is Florida, where the mild weather andabundant retirement communities attract older residents.Quick Review 9.71. ≠15.48%2. ≠20.94%3. ≠14.44%4. ≠27.22%5. ≠17236. 92007. 235 thousand8. 238 million9. 1 million10. 1 billionSection 9.7 Exercises1. 0 5 8 91 3 4 62 3 6 83 3 9456 161 is an outlier.2. Maris Aaron9 8 5 06 4 3 1 0 2 38 6 3 2 0 4 6 7 99 3 3 0 2 4 4 8 9 94 0 0 4 4 4 4 5 751 6Except for Maris’s one record-breaking year, his homerun output falls well short of Aaron’s.3. Males6 0 36 7 8 8 87 1 2 2 3 3 37This stemplot shows the life expectancies of males in thenations of South America are clustered near 70, with twolower values clustered near 60.4. Females6 5 87 1 27 5 6 7 7 9 98 0 0This stemplot shows the life expectancies of females inthe nations of South America are clustered in the high70s and at 80, with four lower values in the high 60s andlow 70s.5. Males Females3 0 68 8 8 7 6 5 83 3 3 2 2 1 7 1 27 5 6 7 7 9 98 0 0This stemplot shows that the life expectancies of thewomen in the nations of South America are about5–6 years higher than that of the men in the nations ofSouth America.6. Female-Male Difference3 54 7 85 2 4 86 3 5 77 7 88 1This stemplot of differences shows that the women in thenations of South America have higher life expectanciesthan the men by about 5–6 years.Copyright © 2011 Pearson Education, Inc. Publishing as Addison-Wesley.
ch09_p355_387.qxd 3/17/10 12:44 PM Page 373Section 9.7 Statistics and Data (Graphical) 3737.Lifeexpectancy(years)Frequency14.60.0–64.9 265.0–69.9 470.0–74.9 6[–1, 20] by [–5, 60]Lifeexpectancy(years)Frequency15.60.0–64.9 165.0–69.9 19.70.0–74.9 275.0–79.9 816.[–1.5, 17] by [–2, 80]10.[50, 80] by [–1, 9]17.[–1, 25] by [–5, 50]11.12.[50, 80] by [–1, 9][0, 60] by [–1, 5][1965, 2008] by [–1000, 11000]The top male’s earnings appear to be growing exponentially,with unusually high earnings in 1999 and 2000.Since the graph shows the earnings of only the top player(as opposed to a mean or median for all players), it canbehave strangely if the top player has a very good year —as Tiger Woods did in 1999 and 2000.18.13.[0, 60] by [–1, 5][1965, 2008] by [–500, 3400]The top female’s earnings appear to be growing fasterthan linearly, but with more fluctuation since 1990. Sincethe graph shows the earnings of only the top player(as opposed to a mean or median for all players), it isstrongly affected by how well the top player does in agiven year. The top player in 2002 and 2003, AnnikaSörenstam, played in fewer tournaments in 2003 andhence won less money.[–1, 25] by [–5, 60]Copyright © 2011 Pearson Education, Inc. Publishing as Addison-Wesley.
ch09_p355_387.qxd 3/17/10 12:44 PM Page 374374 <strong>Chapter</strong> 9 Discrete Mathematics19.[1965, 2008] by [–1000, 11000]After approaching parity in 1985, the top PGA player’searnings have grown much faster than the top LPGA player’searnings, even if the unusually good years for TigerWoods (1999 and 2000) are not considered part of the trend.20. The 1985 male total was unusually low in the context ofthe rest of the time plot. If it had been in line with theearlier and later data, the difference between the PGAand LPGA growth rates would not have been that large.(The unusual PGA point in 1985, oddly enough, belongedto Curtis Strange.)21.(d) Time is not a variable in the data.24. (a) Stem Leaf6 2 3 9 97 7 9 9 98 0 2 3 7 7 8 9 99 0 0 0 1 1 2 3 3 3 4 5 6 710 2 3 3 6 6 6 7 911 0 2 4 612 0 5 9(b)(c)Interval Frequency6.0–6.9 47.0–7.9 48.0–8.9 89.0–9.9 1310.0–10.9 811.0–11.9 412.0–12.9 3[–1, 25] by [–5, 60]The two home run hitters enjoyed similar success, with Maysenjoying a bit of an edge in the earlier and later years of hiscareer, and Mantle enjoying an edge in the middle years.22.[5, 14] by [–2, 15](d) The data are not categorical.25.The two home run hitters were comparable for the firstseven years of their careers.23. (a) Stem Leaf28 229 † 3 73031 6 732 7 833 ∞ 5 5 5 834 2 8 835 3 3 436 3 737 †38 5(b)(c)[–1, 25] by [–5, 80]Interval Frequency25.0–29.9 330.0–34.9 1135.0–39.9 626.[1890, 2010] by [–4, 40]=CA +=NY ■=TX[1890, 2010] by [–4, 40]=PA +=IL ■=FL27. False. The empty branches are important for visualizingthe true distribution of the data values.28. False. They are outliers only if they are significantlyhigher or lower than the other numbers in the data set.29. A time plot uses a continuous line. The answer is C.30. Back-to-back stemplots are designed for comparing datasets. The answer is B.31. The histogram suggests data values clustered near anupper limit — such as the maximum possible score on aneasy test. The answer is A.32. 45° is 1/8, or 12.5%, of 360°. The answer is B.33. Answers will vary. Possible outliers could be the pulserates of long-distance runners and swimmers, which areoften unusually low. Students who have had to run toclass from across campus might have pulse rates thatare unusually high.[20, 45] by [–1, 13]Copyright © 2011 Pearson Education, Inc. Publishing as Addison-Wesley.
ch09_p355_387.qxd 3/17/10 12:44 PM Page 375Section 9.8 Statistics and Data (Algebraic) 37534. Answers will vary. Female heights typically have adistribution that is uniformly lower than male heights,but the difference might not be apparent from a stemplot, especially if the sample is small.35.The maxima of both graphs occur in July (t=7), whilethe minima occur in January (t=1), so we should chooseb=3.5 or 4 (if the latter is halfway between themaximum and the minimum, but the former gives abetter match for some values of t). Thus, for the hightemperature, use f(t)≠17+15 sinc p 1t - 3.52dor617+15 sin c p , and for the low temperature, use6 1t - 42d[0, 13] by [–15, 40]36. Using the form f(t)=k+a sin[b(t-h)], we have b = p>6for both graphs since the period is 12. a is half the differencebetween the extremes, and k is the average of the extremes.Finally, h is the offset to the first time this average occurs.g(t)≠6.5 +15.5 sin6.5+15.5 sinc p 1t - 42d.6c p 1t - 3.52d6or■ Section 9.8 Statistics and Data (Algebraic)Exploration 11. In Figure 9.19 (a) the extreme values will cause the range to be big, but the compact distribution of the rest of the data indicatea small interquartile range. The data in Figure 9.19 (b) exhibit high variability.2. Figure 9.20 (b) has a longer “tail” to the right (skewed right), so the values in the tail pull the mean to the right of the (resistant)median. Figure 9.20 (c) is skewed left, so the values in the tail pull the mean to the left of the median. Figure 9.20 (a) issymmetric about a vertical line, so the median and the mean are close together.Quick Review 9.81.7a x i = x 1 + x 2 + x 3 + x 4 + x 5 + x 6 + x 7i = 152. a 1x i - x2 = 1x 1 - x2 + 1x 2 - x2 + 1x 3 - x2+ 1x 4 - x2 + 1x 5 - x2=x 1 +x 2 +x 3 +x 4 +x 5 -5 x.i = 1Note that, since x = 1 5this simplifies to 0.5 a x i ,i = 1713. (x 1 +x 2 +x 3 +x 4 +x 5 +x 6 +x 7 )7 a i =i = 1x 1 75114. (x 1 +x 2 +x 3 +x 4 +x 5 -5 )= (x 1 +x 2 +x 3 +x 4 +x 5 )- . Note that, since x = 1 5xx5 a x , this5 a i - x2 = ii = 11x 1 55i = 1simplifies to 0.5. The expression at the end of the first line is a simple expansion of the sum (and is a reasonable answer to the given question).By expanding further, we can also arrive at the final expression below, which is somewhat simpler.511=5 31x 1 - x2 2 + 1x 2 - x2 2 + Á + 1x 5 - x2 2 45 a 1x i - x2 2 i = 11=5 31x2 1 - 2x 1 x + x 2 2 + 1x 2 2 - 2x 2 x + x 2 2 + Á + 1x 2 5 - 2x 5 x + x 2 241=5 3x2 1 + x 2 2 + Á + x 2 5 - 2x1x 1 + x 2 + Á + x 5 24 + x 21=1x 2 1 + x 2 2 + Á + x 2 52 - x 25 1x2 1 + x 2 2 + Á + x 2 52 - 2x 2 + x 2 = 1 56. The square root does not allow for further simplication. The final answer is the square root of the expression from Exercise 5: either1B5 31x 1 - x 2 2 + 1x 2 - x 2 2 + Á + 1x 5 - x 2 2 487. a x i f ii = 1108. a 1x i - x2 2i = 1or B15 1x2 1 + x 2 2 + Á + x 2 52 - x 2 .9.50150 a 1x i - x2 2i = 110. B17 a 7i = 11x i - x2 2Copyright © 2011 Pearson Education, Inc. Publishing as Addison-Wesley.
ch09_p355_387.qxd 3/17/10 12:44 PM Page 376376 <strong>Chapter</strong> 9 Discrete MathematicsSection 9.8 Exercises1. (a) Statistic. The number characterizes a set of known data values.(b) Parameter. The number describes an entire population, on the basis of some statistical inference.(c) Statistic. The number is calculated from information about all rats in a small, experimental population.2. (a) Mode. No quantitative measure beyond what is “most common” is implied.(b) Mean. A pitcher’s earned run average (ERA) is total earned runs divided by number of nine-inning blocks pitched.(c) Median. The middle height is considered average.11343. x= (12+23+15+48+36)= = 26.8554. x 157= 14 + 8 + 11 + 6 + 21 + 72 =6 6 = 9.51360.75. x= (32.4+48.1+85.3+67.2+72.4+55.3)= = 3607 L 60.1266 601198.46. x= (27.4+3.1+9.7+32.3+12.8+39.4+73.7)= = 99277 35 L 28.317. x= (1.5+0.5+4.8+7.3+6.3+3.0)=3.9 million, or 3,900,000618. x= (29.8+12.9+11.4+18.0+11.9+17.0)≠16.83 million, or about 16,830,000611379. x= (0+0+1+2+61+33+26+13+1)= L 15.2 satellites991110. x= (30,065,000+13,209,000+44,579,000+…+17,819,000)= 21,171,000 km 277 1148,196,0002 =11. There are 9 data values, which is an odd number, so the median is the middle data value when they are arranged in order. Inorder, the data values are {0, 0, 1, 1, 2, 8, 21, 28, 30}. The median is 2.12. There are 7 data values, which is an odd number, so the median is the middle data value when they are arranged in order. Inorder, the data values are {7687, 9938, 13,209, 17,819, 24,256, 30,065, 44,579} in thousands of km 2 . The median is 17,819,000 km 2 .66053613. Mays: home runs/year. Mantle: 29.8 home runs/year. Mays had the greater production rate.22 = 30 18 L121514. State College: 2.4 houses/day. College Station: houses/day. The State College workers were faster.5 = 7 L 2.1415. No computations needed — Hip-Hop House: 1147 skirts in 4 weeks. What-Next Fashions: 1516 skirts in 4 weeks. What-NextFashions had the greater production rate.311,000,000,00016. India: L $352.882,575,000218,000,000,000Mexico:L $2485. Mexico has the higher PCI.87,715,00017. For {79.5, 82.1, 82.3, 82.9, 84.0, 84.5, 84.8, 85.4, 85.7, 85.7, 86.0, 86.4, 87.0, 87.5, 87.7, 88.0, 88.0, 88.2, 88.2, 88.5, 89.2, 89.2, 89.8, 89.8,87.7 + 88.091.1, 91.3, 91.6, 91.8, 94.1, 94.4}, the median is = 87.85, and there is no mode.218.19.5118,5222 + 4127,1022 + 3131,2862 + 2120,7322 + 1118,8752x =18,522 + 27,102 + 31,286 + 20,732 + 18,875L 3.055198792 + 4151192 + 3161432 + 2126162 + 1130272x =9879 + 5119 + 6143 + 2616 + 3027L 3.6120. (a) Non-weighted: x = 1221(2+5+12+20+…+10+3)= 18.42°C1212 L1221312 + 1521282 + 11221312 + 12021302 + Á + 11021302 + 1321312(b) Weighted: x =31 + 28 + 31 + 30 + Á + 30 + 31(c) The weighted average is the better indicator.= 6748365 L18.49°CCopyright © 2011 Pearson Education, Inc. Publishing as Addison-Wesley.
ch09_p355_387.qxd 3/17/10 12:44 PM Page 377Section 9.8 Statistics and Data (Algebraic) 37721. (a) Non-weighted: x = 177(–9-7-1+7+…-1-7)= 6.42°C1212 L1-921312 + 1-721282 + 1-121312 + 1721302 + Á + 1-121302 + 1-721312(b) Weighted: x =31 + 28 + 31 + 30 + Á + 30 + 31(c) The weighted average is the better indicator.= 2370365 L6.49°C22. The ordered data for Mark McGwire is {3,9,9,22,29, 32,32, 33,39, 39, 42, 49,52, 58, 65, 70} so the median is33 + 39= 36,249 + 5222 + 29Q 3 = = 50.5, Q 1 = = 25.5.22Five-number summary: {3, 25.5, 36, 50.5, 70}Range: 70-3=67IQR: 50.5-25.5=25No outliersThe ordered data for Barry Bonds is:{16, 19, 24, 25,25, 33, 33, 34,34, 37, 37, 40,42, 46, 49, 73} so34 + 34the median is = 34,240 + 4225 + 25Q 3 = = 41, and Q 1 = = 25.22Five-number summary: {16, 25, 34, 41, 73}Range: 73-16=57IQR: 41-25=16Outlier: 7323. The ordered data for Willie Mays is: {4, 6, 8, 13, 18, 20, 22,23, 28, 29, 29,34, 35, 36, 37, 38, 40, 41, 47, 49, 51, 52} so themedian is29 + 34= 31.5, Q 3 =40, and Q 1 =20.2Five-number summary: {4, 20, 31.5, 40, 52}Range: 52-4=48IQR: 40-20=20No outliersThe ordered data for Mickey Mantle is:{13, 15, 18, 19, 21, 22, 23, 23, 27,30, 31, 34, 35, 37, 40, 42,52, 54} so the median is27 + 30= 28.5, Q , and Q 1 = 21.23 = 37Five-number summary: {13, 21, 28.5, 37, 54}Range: 54-13=41IQR: 37-21=16No outliers24. While a stemplot is not needed to answer this question,the sorted stemplot below is more compact than a sortedlist of the 44 numbers. The underlined numbers are theones used for the five-number summary, which isMin Q 1 Median Q 3 Max6.2 8.5 9.25 10.6 12.98.3 + 8.7 9.2 + 9.3 10.6 + 10.6= = =222The range is 12.9-6.2=6.7 andIQR =10.6-8.5=2.1 . There are no outliers, sincenone of the numbers fall below Q 1 -1.5*IQR=5.35or above Q 3 +1.5*IQG=13.75.Stem Leaf6 2 3 9 97 7 9 9 98 0 2 3 7 7 8 9 99 0 0 0 1 1 2 3 3 3 4 5 6 710 2 3 3 6 6 6 7 911 0 2 4 612 0 5 925. The sorted list is {28.2, 29.3, 29.7, 31.6, 31.7, 32.7, 32.8, 33.5,33.5, 33.5, 33.8, 34.2, 34.8, 34.8, 35.3, 35.4, 36.7, 37.3, 38.5}. Sincethere are 19 numbers, the median is the 10th, Q 1 is the5th, and Q 3 is the 15th; no additional computations areneeded. All these numbers are underlined above, and aresummarized below:Min Q 1 Median Q 3 Max28.2 31.7 33.5 35.3 38.5The range is 38.5-28.2=10.3, andIQR =35.3-31.7=3.6. There are no outliers, sincenone of the numbers fall below Q 1 -1.5* IQR=26.3or above Q 3 +1.5* IQR=40.7.Note that some computer software may return31.7 + 32.734.8 + 35.3Q 1 =32.2= and Q 3 =35.05= .22(The TI calculator which computes the five-numbersummary produces the results shown in the table.)26. (a)(b)27. (a)(b)[–4, 80] by [–1, 3][–4, 80] by [–1, 3][–3, 80] by [–1, 2][–3, 80] by [–1, 2]Copyright © 2011 Pearson Education, Inc. Publishing as Addison-Wesley.
ch09_p355_387.qxd 3/17/10 12:44 PM Page 378378 <strong>Chapter</strong> 9 Discrete Mathematics28. (a)(b)[0, 1000] by [1, 3][0, 1000] by [1, 3]1229. 12 of the 44 numbers are more than 10.5:44 = 3114030. All but 4 of the 44 numbers are more than 7.5:44 = 101131. The five-number summaries are:Mays: 4, 20, 31.5, 40, 52 (top of graph)Mantle: 13, 21, 28.5, 37, 54(bottom of graph)(a) Mays’s data set has the greater range.(b) Mays’s data set also has the greater IQR.[0, 60] by [0, 3]32. Although Mays’ best year was not as good as Mantle’sbest year, Mays was in some ways a better home runhitter: His median season was 31.5 home runs, comparedto 28.5 for Mantle. However, Mays had three extremelybad years when he hit fewer than 10 home runs; thesemake him look worse by comparison.For #33–38, the best way to do the computation is with the statisticsfeatures of a calcuator.33. Í≠9.08, Í 2 =82.534. Í≠23.99, Í 2 =575.6435. Í≠120.69; Í 2 ≠14,566.5936. Í≠93.99; Í 2 ≠8834.5037. Í≠1.53, Í 2 ≠2.3438. Í≠2.60, Í 2 ≠6.7739. No. An outlier would need to be less than25.5-1.5(25)=–12 or greater than50.5+1.5(25)=88.40. The standard deviation of a set can never be negative,since it is the (positive) square root of the variance. It ispossible for the standard deviation of a set to be zero, butall the numbers in the set would have to be the same.41. (a) 68%(b) 2.5%(c) A parameter, since it applies to the entire population.42. (a) 16%(b) 32.7(c) No. The mean would have to be weighted according tothe number of people in each state who took the ACT.43. False. The median is a resistant measure. The mean isstrongly affected by outliers.44. True. The box extends from the first quartile, Q 1 , to thethird quartile, Q 3 , and Q 3 -Q 1 is the interquartile range.45. The plot of an ideal normal distribution is a symmetric“bell curve.” The answer is A.10132 + 9132 + 8152 + 7162 + 6142 + 5132 + 4112 + 3102 + 2102 + 110246. x =25The answer is B.= 7.2847. The total number of points from all 30 exams combined is30*81.3=2439. Adding 9 more points and recalculatingproduces a new mean of (2439+9)/30=81.6.Themedian will be unaffected by an adjustment in the topscore. The answer is B.48. In a normal distribution, 95% of the data values lie within2 standard deviations of the mean. The answer is C.49. There are many possible answers; examples are given.(a) {2, 2, 2, 3, 6, 8, 20} — mode=2, median=3, and43x= ≠6.14.7(b) {1, 2, 3, 4, 6, 48, 48} — median=4, x=16, andmode=48.1(c) {–20, 1, 1, 1, 2, 3, 4, 5, 6} — x= , mode=1, and3median=2.50. There are many possible answers; examples are given.(a) {2, 4, 6, 8} — Í≠2.24 and IQR=7-3=4.(b) {1, 5, 5, 6, 6, 9} — IQR=6-5=1, and Í≠2.36.(c) {3, 3, 3, 9, 9, 9} — IQR=9-3=6 andrange=9-3=6.51. No: (x i - x) … (max-min) 2 =(range) 2 , so thatnnÍ 2 11= x) 2 … (range) 2 =(range) 2 . Thenn a 1x i -n a i = 1i = 1Í= 2s 2 … 21range2 2 = range.52. The mode does not have to be a numerical value becauseit is the data value that occurs most often; both the meanand the median are for numerical data only, since theyboth involve calculations of numerical data values.Copyright © 2011 Pearson Education, Inc. Publishing as Addison-Wesley.
ch09_p355_387.qxd 3/17/10 12:44 PM Page 379Section 9.9 Statistical Literacy 37953. There are many possible answers; example data sets aregiven.(a) {1, 1, 2, 6, 7} — median=2 and x=3.4.(c) {1, 1, 2, 6, 7} — range=7-1=6 and2*IQR=2(6-1)=10.[–1, 10] by [–1, 5](b) {1, 6, 6, 6, 6, 10} — 2*IQR=2(6-6)=0 andrange=10-1=9.[–1, 12] by [–1, 5]54. One possible answer: {1, 2, 3, 4, 5, 6, 6, 6, 30}.[–1, 12] by [–1, 5]55. For women living in South American nations, the mean life expectancy is179.72139.12 + 167.9218.72 + Á + 179.2213.42 + 177.32125.02 27,825.56x = = L 75.9 years.39.1 + 8.7 + Á + 3.4 + 25.0366.456. For men living in South American nations, the mean life expectancy is172.02139.12 + 162.5218.72 + Á + 172.7213.42 + 171.02125.02 25,160.64x = = L 68.7 years.39.1 + 8.7 + Á + 3.4 + 25.0366.457. Since Í=0.05 mm, we have 2Í=0.1 mm, so 95% of theball bearings will be acceptable. Therefore, 5% will berejected.58. Use Â=12.08 and Í=0.04.Then Â-2Í=12.00 and Â+2Í=12.16, so 95% of thecans contain 12 to 12.16 oz of cola, 2.5% contain less than12 oz, and 2.5% contain more than 12 oz. Therefore, 2.5%of the cans contain less than the advertised amount.■ Section 9.9 Statistical LiteracyExploration 11. Correlation begins with a scatter plot, which requiresnumerical data from two quantitative variables (likeheight and weight). “Gender” is a categorical variable.2. The sample did not represent the population of all votersvery well, as many subgroups were underrepresented (forexample, office workers). It was also not a good idea forthe mayor to use his own staff to gather this kind of data.3. The doctor’s “experiment” proves nothing about theeffect of vanilla gum on headache pain unless we cancompare these subjects with a similar group that does notuse vanilla gum. Many headaches are gone in two hoursanyway.4. The negative correlation simply showed that studentswith more absences tended to have lower GPAs. Thisdoes not imply that the absences caused the grades to godown. Perhaps getting low grades caused students to skipschool. Or another variable, such as parental involvement,might have influenced both grades and absences.Correlation does not imply causation.Quick Review 9.911.652.3643.52 = 11314.1015.106. a 1 10 ba 110 b = 11007.8.9.a 1 510 b = 0.00001a 9 510 b = 0.590491 - a 910 b 5= 0.4095110. a 1 510 b + 5a 1 410 b a 910 b = 0.00046Section 9.9 Exercises1. Correlation is being used incorrectly. Intelligence might beassociated with some quantitative variable, but beauty iscategorical.2. Correlation is being used incorrectly. The correlation coefficientis usually not the slope of the regression line.Copyright © 2011 Pearson Education, Inc. Publishing as Addison-Wesley.
ch09_p355_387.qxd 3/17/10 12:44 PM Page 380380 <strong>Chapter</strong> 9 Discrete Mathematics3. Correlation is being used incorrectly. The high correlationcoefficient does nothing to support Sean’s crazy theory,because the great blue whale (with a long name and a hugeweight) is an unusual point that lies far away from the otherthree.4. Correlation is being used correctly. Notice that Jenna’s firstobservation does not commit her to a linear model, but hersecond sentence does. Her observation about the model isthen appropriate.5. Correlation is being used incorrectly. Marcus is OK with hisfirst observation, but not with his second. While his linearmodel is a bad fit, he should not conclude that “there is nosignificant mathematical relationship.” In fact, check outthis sinusoidal fit:6. Correlation is being used incorrectly. Correlation does notimply causation! (If you are wondering, though, the rainwatertheory has been confirmed for several sea snakespecies through controlled laboratory experiments.)7. This is a random sample (technically pseudo-random, but itshould suffice).8. This is not a random sample, since the students will come infive chunks of ten, each chunk probably being alphabetical.This could result in an unusual number of students in thesample who are related to each other.9. This is not a random sample of all Reno citizens. All 50selected are likely to be from early in the alphabet.10. This is not a random sample, although there is quite a bit ofchance involved. Notice, for example, that the five chosenstudents will probably not include any pairs of best friends,since they would probably have lined up together.11. This is not a random sample, nor does it apparently try tobe.12. This is not a random sample, because it relies on humanchoice. It might appear to be random, but what would happenif one of the winners turned out (even by chance) to berelated to one of the ushers?13. Voluntary response bias. The students most likely torespond were those who felt strongly about suggestions forimprovement, so the rate of negative responses was probablyhigher than the parameter. He could have gotten a lessbiased response with an in-class census of all his students(ideally in multiple-choice form so that their handwritingwould not betray their identities).14. Response bias. By giving free samples and money to therespondents, the company was influencing them to respondpositively. The support for the new cereal was likely to behigher in this sample than in the population. The companywould have gotten more reliable data with a comparativestudy of the new flavor against an established product.15. Undercoverage bias. The survey systematically excluded thestudents who were not actually eating in the dining hall, sothe sample statistic was bound to be higher than the populationparameter. A better method would have been tochoose a random sample from the student body first, thenseek them out for the survey (perhaps in their homerooms).16. Response bias and undercoverage bias. The question itselfwas clearly designed to elicit a positive response, and thesampling frame (PTA parents) was designed to survey onlypeople who actually had children. The support in the samplewas probably higher than in the population. Getting atruly random sample for polling purposes is notoriously difficult,but even a random sample from the phone bookwould improve on this one.17. Response bias. The question was designed to elicit a negativeresponse, and it never even mentioned stop signs. The97% was much higher than it would have been with a simplequestion like, “Should citizens be allowed to ignore stopsigns?”18. Undercoverage bias. Mrs. Bohackett (who never attemptedto sample randomly) is making a global conclusion on thebasis of her own back yard, which might reflect whitethroatedsparrow migration in unusual ways. The evidencewould be more persuasive if similar observations weremade at many other feeders, but causation could not beestablished without experimentation.19. This is an observational study since no treatment wasimposed.20. This is an observational study since no treatment wasimposed.21. This is an observational study since no treatment wasimposed.22. This is an experiment since a treatment was imposed.23. This is an experiment since a treatment was imposed.24. This is an observational study since no treatment wasimposed.25. Using random numbers, select 12 of the 24 plots to get thenew fertilizer. Use the original fertilizer on the other 12plots. Compare the yields at harvest time.26. The plots can be arranged in pairs of similar productivity,then one member of each pair can be randomly assigned toreceive the new fertilizer.27. This requires three treatments. Split the 24 plots randomlyinto three groups of 8: new fertilizer 1, new fertilizer 2, andoriginal fertilizer.28. This requires four treatments. Split the 24 plots randomlyinto four groups of 6: crop 1 with new fertilizer, crop 1 withoriginal fertilizer, crop 2 with new fertilizer, and crop 2 withoriginal fertilizer.29. Fatigue may be a factor after they have driven 20 golf balls.They could gather the data on different days, or they couldrandomly choose half the golfers to drive the new ball first.30. The tasters should be blinded to the containers. Ideally, thedrinks should be poured into identical cups for presentation(being careful not to lose the fizz).31. The music assignment should be randomized, not left to thechoice of the mother. Otherwise, the mother’s music preference(with possible lifestyle implications) becomes a potentiallysignificant confounding variable.32. Because of the nature of the experiment, it should beblocked for gender to be sure that each group contains anequal mix of male and female listeners.Copyright © 2011 Pearson Education, Inc. Publishing as Addison-Wesley.
ch09_p355_387.qxd 3/17/10 12:44 PM Page 381<strong>Chapter</strong> 9 Review 38133. One possible solution: Use the command “randInt(1, 500,50) ” to choose 50 random numbers from 1 to 500. If thereare any repeat numbers in the list, use “randInt(1, 500)” topick additional numbers until you have a sample of 50.34. One possible solution: Use the command “randInt(1, 400,100)” to choose 100 random numbers from 1 to 400. If thereare any repeat numbers in the list, use “randInt(1, 400)” topick additional numbers until you have a sample of 100.35. One possible solution: Enter the numbers 1 to 32 in list L1using the command “seq(X, X, 1, 32) : L1” and enter 32random numbers in list L2 using the command “rand(32): L2 .” Then sort the random numbers into ascendingorder, bringing L1 along for the ride, using the command“SortA(L2,L1).” The numbers in list L1 are now in randomorder.36. One possible solution: Enter the numbers 1 to 28 in list L1using the command “seq(X, X, 1, 28) : L1” and enter 28random numbers in list L2 using the command “rand(28): L2 .” Then sort the random numbers into ascendingorder, bringing L1 along for the ride, using the command“SortA(L2,L1).” Read off the randomly sorted numbers inL1 in order by pairs.37. One possible solution: Use the command “randInt(1, 5, 20)”to generate 20 random numbers from 1 to 5. Let 1 and 2designate donors with O-positive blood. Do this nine timesand keep track of how many strings have fewer than fournumbers that are 1 or 2.38. One possible solution: Use the command “randInt(1, 8)” togenerate random numbers between 1 and 8.39. One possible solution: Use the command “randInt(1, 6)” togenerate random numbers between 1 and 6. Push ENTERtwice to get a roll of two dice. (Note that you do not want togenerate random totals between 2 and 12. You learned inSection 9.3 that those totals are not equally likely.)40. One possible solution: Number the cards from 1 to 52. Usethe command “randInt(1, 52,5)” to choose 5 random numbersfrom 1 to 52. If there are any repeat numbers in thelist, use “randInt(1, 52)” to pick additional numbers untilyou have a sample “hand” of 5.41. False. Observational studies can show strong associations,but experiments would be required to establish causation.42. False. The underlying relationship might not be linear, inwhich case the magnitude of the r value is not a relevantmeasure.43. Freshman science grades is the only quantitative variableamong the choices. The answer is B.44. The answer is A.45. The answer is C.46. People late in the alphabet are less likely to be chosen. Theanswer is D.47. Answers will vary. Note that you should not expect all thecounts to be exactly the same (that would suggest nonrandomnessin itself), but “randomness” would predict anapproximately equal distribution, especially for a large class.48. Answers will vary, but the count is likely to be low. Forexample, a class of 20 would be looking at 500 two-digitnumbers, about 50 of which should be double digited.49. (a) The correlation coefficient will increase and the slopewill remain about the same.(b) The correlation coefficient will increase and the slopewill increase.(c) The correlation coefficient will decrease and the slopewill decrease.50. The points in (b) and (c) are influential points.51. One possible picture:52. One possible picture:53. (a) The size of the hospital is not affecting the death ratesof the patients. The lurking variable is the patient’s condition.Bigger hospitals tend to get the more criticalcases, and critical cases have a higher death rate.(b) The number of seats is not affecting the speed of the jet.The lurking variable is the size of the aircraft. Largerjets generally have more seats and go faster.(c) The size of the shoe does not affect reading ability. Thelurking variable is the age of the student. In general,older students have larger feet and read at a higherlevel.(d) The extra firemen are not causing more damage. Thelurking variable is the size of the fire. Larger fires causemore damage and require more firefighters.(e) The salary is not generally affected by the player’sweight. The lurking variable is the player’s position onthe team. Linemen weigh more and tend to earn lessmoney than the (usually lighter) players in the so-called“skill” positions (e.g., quarterbacks, running backs,receivers, and defensive backs).54. (a) Prospective(b) Retrospective(c) Retrospective(d) Prospective(e) Retrospective■ <strong>Chapter</strong> 9 Review1.2.3.a 12 5 b = 12!5!112 - 52! = 12!5!7! = 792a 789787 b = 789!787!1789 - 7872! = 789!787!2! = 310,86618C 12 =4. 35 C 28 =18!12!118 - 122! = 18!12!6! = 18,56435!28!135 - 282! = 35!28!7! = 6,724,520Copyright © 2011 Pearson Education, Inc. Publishing as Addison-Wesley.
ch09_p355_387.qxd 3/17/10 12:44 PM Page 382382 <strong>Chapter</strong> 9 Discrete Mathematics5.6.12P 7 =15P 8 =#7. 26 36 4 =43,670,016 code words8. 3+(3 # 4)=15 trips9. 26 P 2# 1 0 P 4 + 10 P 3# 26 P 3 =14,508,000 license plates10. 45 C 3 =14,190 committees11. Choose 10 more cards from the other 49:3C 3# 49 C 10 = 49 C 10 =8,217,822,536 hands12. Choose a king, then 8 more cards from the other 44:4C 4# 4 C 1# 44 C 8 = 4 C 1# 44 C 8 =708,930,508 hands13. 5 C 2 + 5 C 3 + 5 C 4 + 5 C 5 =2 5 - 5 C 0 - 5 C 1 =26 outcomes#12!112 - 72! = 12! = 3,991,6805!15!115 - 82! = 15! = 259,459,2007!14. 21 C 2 14 C 2 =19,110 committees15. 5 P 1 + 5 P 2 + 5 P 3 + 5 P 4 + 5 P 5 =32516. 2 4 =16 (This includes the possibility that he has no coinsin his pocket.)17. (a) There are 7 letters, all different. The number of distinguishablepermutations is 7!=5040. (GERMANYcan be rearranged to spell MEG RYAN.)(b) There are 13 letters, where E, R, and S each appeartwice. The number of distinguishable permutations is13!.2!2!2! = 778,377,600(PRESBYTERIANS can be rearranged to spellBRITNEY SPEARS.)18. (a) There are 7 letters, all different. The number of distinguishablepermutations is 7!=5040.(b) There are 11 letters, where A appears 3 times and L,S, and E each appear 2 times. The number of distinguishablepermutations is11!.3!2!2!2! = 831,60019. (2x+y) 5 =(2x) 5 +5(2x) 4 y+10(2x) 3 y 2 +10(2x) 2 y 3 +5(2x)y 4 +y 5 =32x 5 +80x 4 y+80x 3 y 2 +40x 2 y 3 +10xy 4 +y 520. (4a-3b) 7 =(4a) 7 +7(4a) 6 (–3b)+21(4a) 5 (–3b) 2 +35(4a) 4 (–3b) 3 +35(4a) 3 (–3b) 4 +21(4a) 2 (–3b) 5 +7(4a)(–3b) 6 +(–3b) 7=16,384a 7 -86,016a 6 b+193,536a 5 b 2 -241,920a 4 b 3 +181,440a 3 b 4 -81,648a 2 b 5 +20,412ab 6 -2187b 721. (3x 2 +y 3 ) 5 =(3x 2 ) 5 +5(3x 2 ) 4 (y 3 )+10(3x 2 ) 3 (y 3 ) 2 +10(3x 2 ) 2 (y 3 ) 3 +5(3x 2 )(y 3 ) 4 +(y 3 ) 5=243x 10 +405x 8 y 3 +270x 6 y 6 +90x 4 y 9 +15x 2 y 12 +y 1522. a1 + 1 6=1+6(x –1 )+15(x –1 ) 2 +20(x –1 ) 3 +15(x –1 ) 4 +6(x –1 ) 5 +(x –1 ) 6x b =1+6x –1 +15x –2 +20x –3 +15x –4 +6x –5 +x –623. (2a 3 -b 2 ) 9 =(2a 3 ) 9 +9(2a 3 ) 8 (–b 2 )+36(2a 3 ) 7 (–b 2 ) 2 +84(2a 3 ) 6 (–b 2 ) 3 +126(2a 3 ) 5 (–b 2 ) 4 +126(2a 3 ) 4 (–b 2 ) 5 +84(2a 3 ) 3 (–b 2 ) 6+36(2a 3 ) 2 (–b 2 ) 7 +9(2a 3 )(–b 2 ) 8 +(–b 2 ) 9 =512a 27 -2304a 24 b 2 +4608a 21 b 4 -5376a 18 b 6 +4032a 15 b 8-2016a 12 b 10 +672a 9 b 12 -144a 6 b 14 +18a 3 b 16 -b 1824. (x –2 +y –1 ) 4 =(x –2 ) 4 +4(x –2 ) 3 (y –1 )+6(x –2 ) 2 (y –1 ) 2 +4(x –2 )(y –1 ) 3 +(y –1 ) 4 =x –8 +4x –6 y –1 +6x –4 y –2 +4x –2 y –3 +y –425.26.a 11 8 b1128 1-22 3 = - 11!8 #8!3! = -11 10 # 9 # 83 # 2 #= -13201a 8 2 b1222 112 6 = 8!42!6! = 8 # 7 # 42 #= 112127. {1, 2, 3, 4, 5, 6}28. {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), ... ,(6, 6)}29. {13, 16, 31, 36, 61, 63}30. {Defective, Nondefective}31. {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}32. {HHT, HTH, THH, TTH, THT, HTT}33. {HHH, TTT}34. P(at least one head)=1-P(no heads)=1-a 1 32 b7=835. P(HHTHTT)= a 1 62 b = 1 2 = 16 6436. P(2 H and 3 T)= 5C 2# a 1 52 b = 102 = 55 1637. P(1 H and 3 T)= 4 C 1# a 1 42 b = 4 2 4 = 1 438. P(10 nondefectives in a row)=(0.997) 10 ≠0.9739. P(3 successes and 1 failure)= 4C 1# a 1 42 b = 4 2 4 = 1 4 = 0.2540. We can redefine the words “success” and “failure” to meanthe opposite of what they meant before. In this sense, theexperiment is symmetrical, because P(S)=P(F).41. P(SF)=(0.4)(0.6)=0.2442. P(SFS)=(0.4)(0.6)(0.4)=0.09643. P(at least 1 success)=1-P(no successes)=1-(0.6) 2 =0.6444. Successes are less likely than failures, so the two are notinterchangeable.45. (a) P(brand A)=0.5(b) P(cashews from brand A)=(0.5)(0.3)=0.15(c) P(cashew)=(0.5)(0.3)+(0.5)(0.4)=0.350.15(d) P(brand A/cashew)=0.35 L 0.4346. (a) P(track wet and Mudder Earth wins)=(0.80)(0.70)=0.56(b) P(track dry and Mudder Earth wins)=(0.20)(0.40)=0.08(c) 0.56+0.08=0.64Copyright © 2011 Pearson Education, Inc. Publishing as Addison-Wesley.
ch09_p355_387.qxd 3/17/10 12:44 PM Page 383`61. a n =a 1 r n–1 , so –192=a 1 r 3 and 196,608=a 1 r 8 . Thenr 5 -192=–1024, so r=–4, and a 1 =1-42 3 = 3;a n =3(–4) n–1 .62. a n =a 1 +(n-1)d, so 14=a 1 +2d, and–3.5=a 1 +7d. Then 5d=–17.5, sod=–3.5, and a 1 =14-2(–3.5)=21;a n =21-3.5(n-1)=24.5-3.5n.For #63–66, use one of the formulas S n =na a 1 + a nb or2nS n = [2a 1 +(n-1)d]. In most cases, the first of these is2easier (since the last term a n is given); note thata n - a 1n= + 1.d63. 8 # a64.7 # a65. 27 # a66. 31 # a-11 + 10b = 4 # 1-12 = -4213 - 11b = 722.5 - 75.5b = 1 # 27 # 1-732 = -985.52 2-5 + 55b = 31 # 25 = 7752<strong>Chapter</strong> 9 Review 3830.56(d) P(track wet/Mudder Earth wins)=0.64 = 0.875 a 1 11 - r n 2For #67–70, use the formula S n = . Note that1 - rFor #47–48, substitute n=1, n=2, … , n=6, and n=40.a nln ƒa n >a 1 ƒn=1+log47. 0, 1, 2, 3, 4, 5; 39|r| ` =1+ .a 1 ln ƒrƒ4 1648. –1, , –2, , - 16 ≠2.68 10 103 , 647 ;411 - 1-1>22 6 267.= 213 51 - 1-1>22 8For #49–54, use previously computed values of the sequence-311 - 11>32 5 2to find the next term in the sequence.68.= - 1211 - 11>32 2749. –1, 2, 5, 8, 11, 14; 32211 - 3 10 250. 5, 10, 20, 40, 80, 160; 10,24069.= 59,0481 - 351. –5, –3.5, –2, –0.5, 1, 2.5; 11.5111 - 1-22 14 2152. 3, 1,3 , 1 9 , 1 27 , 181 ; 3 - 10 170.= -5461=1 - 1-2259,049153. –3, 1, –2, –1, –3, –4; –7671. Geometric with r= :354. –3, 2, –1, 1, 0, 1; 13218711 - 11>32 10 2S 10 == 29,524 = 3280.4For #55–62, check for common difference or ratios between1 - 11>32 9successive terms.72. Arithmetic with d=–3:55. Arithmetic with d=–2.5;10Sa n =12+(–2.5)(n-1)=14.5-2.5n10 = [2(94)+9(–3)]=5 # 161=805256. Arithmetic with d=4;73.a n =–5+4(n-1)=4n-957. Geometric with r=1.2;a n =10 # (1.2) n158. Geometric with r=–2;1a n = # 1-22 n - 1 = - 18 16 1-22n[0, 15] by [0, 2]59. Arithmetic with d=4.5;74.a n =–11+4.5(n-1)=4.5n-15.5160. Geometric with r= ;b n = 7 # a 1 4n4 b = 28 # a 1 n4 b[0, 16] by [–10, 460]75. With a 1 =$150, r=1+0.08/12, and n=120, the sumbecomes$150 31 - 11 + 0.08>122 120 4= $27,441.91.1 - 11 + 0.08>12276. The payment amount P must be such thatPa1 + 0.08 012 b + Pa1 + 0.08 112 b + Á+ Pa1 + 0.08 119.12 b Ú $30,000Using the formula for the sum of a finite geometric series,P 31 - 11 + 0.08>122 120 4Ú $30,0001 - 11 + 0.08>122-0.08>12or P Ú $30,0001 - 11 + 0.08>122 120L $163.983L $163.99 rounded up.3 377. Converges: geometric with a 1 = and r= , so2 43>2S= .1 - 13>42 = 3>21>4 = 6Copyright © 2011 Pearson Education, Inc. Publishing as Addison-Wesley.
ch09_p355_387.qxd 3/17/10 12:44 PM Page 384384 <strong>Chapter</strong> 9 Discrete Mathematics78. Converges: geometric with a 1 = - 2 and r= - 1 , so3 3-2>3S= .1 - 1-1>32 = -2>34>3 = -1 279. Diverges: geometric with r= - 4 3680. Diverges: geometric with r=581. Converges: geometric with a 1 =1.5 and r=0.5, so1.5S= .1 - 0.5 = 1.50.5 = 382. Diverges; geometric with r=1.2212183. a 3-8 + 51k - 124 = a 15k - 132k = 1k = 11084. a 41-22 k - 1 10= a 1-22 k + 1k = 1k = 1qq85. a 12k + 12 2 or a 12k - 12 2k = 0k = 1q86. a a 1 k qk = 0 2 b or a a 1 k - 1k = 1 2 bnn n87. a 13k + 12 = 3 a k + a 1k = 1k = 1 k = 1n1n + 12= 3 # + n = 3n2 + 5n n13n + 52=222n88. a 3k 2 n n1n + 1212n + 12= 3 a k 2 = 3 #k = 1 k = 16n1n + 1212n + 12=289. ak = 1+ 4 # 25 = 465090.25175ak = 11k 2 - 3k + 42 = 25 # 26 # 5113k 2 - 5k + 12 = 3 # 175 # 176 # 351- 5 # 175 # 17626+ 175 = 5,328,575n1n + 12 n1n + 121n + 2291. P n : 1+3+6+…+ = .26111 + 12 111 + 1211 + 22P 1 is true: =.26k1k + 12Now assume P k is true: 1+3+6+…+2k1k + 121k + 22 1k + 121k + 22= . Add to both sides:62k1k + 12 1k + 121k + 221+3+6+…+ +22k1k + 121k + 22 1k + 121k + 22=+62=(k+1)(k+2) a k 6 + 1 2 b=(k+1)(k+2) a k + 3 b61k + 1211k + 12 + 1211k + 12 + 22= ,6so P k+1 is true. Therefore, P n is true for all n Ú 1.6- 3 # 25 # 262###92. P n : 1 2+2 3+3 4+…+n(n+1)n1n + 121n + 22= . P 1 is true:3111 + 1211 + 221(1+1)=.3Now assume P k is true: 1 # 2+2 # 3+3 # 4+…+k(k+1)k1k + 121k + 22=.3Add (k+1)(k+2) to both sides:1 # 2+2 # 3+… +k(k+1)+(k+1)(k+2)k1k + 121k + 22= +(k+1)(k+2)3=(k+1)(k+2) a k 3 + 1b=(k+1)(k+2) a k + 3 b31k + 1211k + 12 + 1211k + 12 + 22=3;so P k is true. Therefore, P n is true for all n Ú 1.93. P n : 2 n–1 … n!. P 1 is true: it says that 2 1–1 … 1! (they areequal). Now assume P k is true: 2 k–1 … k!. Then2 k1–1 =2 # 2 k–1 … 2 # k! (k+1)k!=(k+1)!, so P k1 is true. Therefore, P n istrue for all n Ú 1.94. P n : n 3 +2n is divisible by 3. P 1 is true because1 3 +2 # 1=3 is divisible by 3. Now assume P k is true:k 3 +2k is divisible by 3. Then note that(k+1) 3 +2(k+1)=(k 3 +3k 2 +3k+1)+(2k+2)=(k 3 +2k)+3(k 2 +k+1).Since both terms are divisible by 3, so is the sum, so P k1is true. Therefore, P n is true for all n Ú 1.95. (a) 9 1 210 6 711 4 5 5 7 712 0 2 4 6 7 713 5 614 1 6 7 7 815 4 816 1 417 0 618192021 92223 4(b) Price Frequency90,000–99,999 2100,000–109,999 2110,000–119,999 5120,000–129,999 6130,000–139,999 2140,000–149,999 5150,000–159,999 2160,000–169,999 2170,000–179,999 2210,000–219,999 1230,000–239,999 1Copyright © 2011 Pearson Education, Inc. Publishing as Addison-Wesley.
ch09_p355_387.qxd 3/17/10 12:44 PM Page 385<strong>Chapter</strong> 9 Review 385(c)(c)[8, 24] by [–1, 7]96. (a) 1 1 1 1 1 1 2 2 3 3 4 41 5 9 9 92 0 0 02 6 8 93 34455 86 167 1(b) Number of Visitors Frequency10,000,000–14,999,999 1115,000,000–19,999,999 420,000,000–24,999,999 325,000,000–29,999,999 330,000,000–34,999,999 135,000,000–39,999,999 040,000,000–44,999,999 045,000,000–49,999,999 050,000,000–54,999,999 055,000,000–59,999,999 160,000,000–64,999,999 165,000,000–69,999,999 070,000,000–74,999,999 1(c)[0.5, 8] by [–1, 12]97. (a) 12 0 0 4 413 1 1 2 6 7 914 0 3 4 815 616 317 7 918 019 0 1 720 2212223 0(b) Length (in seconds) Frequency120—129 4130—139 6140—149 4150—159 1160—169 1170—179 2180—189 1190—199 3200—209 1210—219 0220—229 0230—239 1[120, 240] by [0, 7]98. (a) For the stemplot (but not for the other frequencytable or histogram), round to nearest hundred yardsfirst:Stem Leaf1 12 3 4 5 7 83 1 3 5 5 64 0 2 3 7 75 0 6(b) Yardage Frequency1000–1999 12000–2999 53000–3999 64000–4999 45000–5999 2(c)[0, 6500] by [–1, 6]99. The ordered data is {9.1, 9.2, 10.6, 10.7, 11.4, 11.5, 11.5,11.7, 11.7, 12, 12.2, 12.4, 12.6, 12.7, 12.7, 13.5, 13.6, 14.1,14.6, 14.7, 14.7, 14.8, 15.4, 15.8, 16.1, 16.4, 17.0, 17.6, 21.9,23.4} so the12.7 + 13.5median is = 13.1, Q 3 =15.4, and Q 1 =11.7.2Five-number summary: {9.1, 11.7, 13.1, 15.4, 23.4}Range: 23.4-9.1=14.3 ($143,000)IQR: 15.4-11.7=3.7 ($37,000)Í≠3.19, Í 2 ≠10.14Outliers: 21.9 and 23.4 are greater than15.4+1.5(3.7)=20.95.100. The ordered data is {1.1, 1.1, 1.1, 1.1, 1.1, 1.2, 1.2, 1.3, 1.3,1.4, 1.4, 1.5, 1.9, 1.9, 1.9, 2,0, 2.0, 2.0, 2.6, 2.8, 2.9, 3.3, 5.8,6.1, 7.1} so the median is 1.9,2.6 + 2.8Q 3 = = 2.7 and Q 1 =1.2.2Five-number summary: {1.1, 1.2, 1.9, 2.7, 7.1}Range: 7.1-1.1=6.0 (6 million)IQR: 2.7-1.2=1.5 (1.5 million)Í≠1.63, Í 2 ≠2.64Outliers: 5.8, 6.1, and 7.1 are greater than2.7+1.5(1.5)=4.95.101. The ordered data is {120, 120, 124, 124, 131, 131, 132, 136,137, 139, 140, 143, 144, 148, 156, 163, 177, 179, 180, 190,191, 197, 202, 230} so the median is143 + 144179 + 180= 143.5, Q 3 == 179.5, and22131 + 132Q 1 = = 131.5.2Copyright © 2011 Pearson Education, Inc. Publishing as Addison-Wesley.
ch09_p355_387.qxd 3/17/10 12:44 PM Page 386386 <strong>Chapter</strong> 9 Discrete MathematicsFive-number summary: {120, 131.5, 143.5, 179.5, 230}Range: 230-120=110IQR: 179.5-131.5=48Í=29.9, Í 2 =891.4No Outliers.102. The ordered data is: {1112, 2327, 2382, 2521, 2709, 2806,3127, 3338, 3485, 3489, 3631, 3959, 4228, 4264, 4689, 4690,5000, 5648},3485 + 3489so the median is = 3487,2Q 3 =4264 and Q 1 =2709.Five-number summary: {1112, 2709, 3487, 4264, 5648}Range: 5648-1112=4536IQR: 4264-2709=1555≠1095, 2 ≠1,199,223No Outliers.103. (a)(b)104 (a)[8, 24] by [–1, 1][8, 4] by [–1, 1]106.(a)(b)[0, 6000] by [–1, 1][0, 6000] by [–1, 1]107. 4 0 0 12 49 2 1 13 1 6 78 4 3 0 1415 63 167 17 918 019 0 1 720 2212223 0The songs released in the earlier years tended to beshorter.108. Earlier years are in the upper box plot. The range andinterquartile range are both greater in the lower graph,which shows the times for later years.(b)[0, 8] by [–1, 1]109.[100, 250] by [–5, 10]105. (a)[0, 8] by [–1, 1][–1, 25] by [100, 250]Again, the data demonstrates that songs appearing latertended to be longer in length.110.(b)[100, 250] by [–1, 1][100, 250] by [–1, 1][0, 7] by [50, 250]The average times are {130.5, 132.75, 157, 142.5,168.75, 202}. The trend is clearly increasing overall, withless fluctuation than the time plot for Exercise 105.111. 1 9 36 84 126 126 84 36 9 1Copyright © 2011 Pearson Education, Inc. Publishing as Addison-Wesley.
ch09_p355_387.qxd 3/17/10 12:44 PM Page 387<strong>Chapter</strong> 9 Project 387n! 1n - k2!112. n P k n–k P j =1n - k2! 31n - k2 - j4!n!=1n - k - j2!n!=3n - 1k + j24! = nP k + j113. (a) P(no defective bats)=(0.98) 4 ≠0.922(b) P(one defective bat)= 4 C 1# (0.98) 3 (0.02)=4(0.98) 3 (0.02)≠0.075114. (a) P(no defective light bulbs)=(0.9996) 10 ≠0.996(b) P(two defective light bulbs)= 10 C 2#(0.9996) 8 (0.0004) 2 ≠7.18 * 10 –6115. Incorrect. Intelligence might be associated with somequantitative variable, but strength is categorical.116. The student is incorrect. The correlation coefficient isusually not the slope of the regression line.117. Yes, using a random number generator will result in arandom sample.118. No, this is not a random sample of all Atlanta citizens.All people in the sample are likely to be those whosenames come early in the alphabet.119. This sampling method will have voluntary response bias.120. This is an observational study since no treatment wasimposed.<strong>Chapter</strong> 9 ProjectAnswers are based on the sample data shown in the table.1. Stem Leaf55 96 1 1 2 3 3 3 4 4 4 46 5 6 6 6 7 8 8 9 9 97 0 0 1 1 1 2 2 37 5The average is about 66 or 67 inches.2. A large number of students are between 63 and 64 inchesand also between 69 and 72 inches.3.Height Frequency59–60 161–62 363–64 765–66 467–68 369–70 571–72 573–74 175–76 1speculate about average heights for males and femalesseparately. Possibly the two peaks within the distributionrepresent an average height of 63–64 inches for femalesand 70–71 inches for males.4. Mean=66.9 in; median=66.5 in; mode=64 in.The mean and median both appear to be good measuresof the average, but the mode is too low. Still, the modemight well be similar in other classes.5. The data set is well distributed and probably does nothave outliers.6. The stem and leaf plot puts the data in order.Minimum value: 59Maximum value: 7566 + 67Median: = 66.52Q 1 :64Q 3 :70The five-number summary is {59, 64, 66.5, 70, 75}.7.[56, 78] by [–1, 7]The box plot visually represents the five-number summary.The whisker-to-whisker size of the box plot representsthe range of the data, while the width of the box representsthe interquartile range.8. Mean=67.5; median=67; The new five-number summaryis {59, 64, 67, 71, 86}.[56, 88] by [–1, 7]The minimum and first quartile are unaffected, but themedian, third quartile, and maximum are shifted to varyingdegrees.9. The new student’s height, 86 inches, lies 15 inches awayfrom Q 3 , and that is more than 1.5(Q 3 -Q 1 )=10.5.The height of the new student should probably be tossedout during prediction calculations.10. Mean=68.9; median=68. The new five-numbersummary is {59, 64, 68, 71, 86}.All three of the new students are outliers by the1.5*IQR test. Their heights should be left out duringprediction making.[59, 78] by [–1, 7]Again, the average appears to be about 66 or 67 inches.Since the data are not broken out by gender, one can onlyCopyright © 2011 Pearson Education, Inc. Publishing as Addison-Wesley.
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