Selected Answers for

Core Connections Algebra

2 Core Connections Algebra

Lesson 11.1.1 11-4. a: 2x !1, shift up 2 units b: 4x ! 3 , twice as steep

c: 2x +1 , shift left 2 units d: 4x ! 6 , twice as steep, y-intercept shifts down 3 units, x-intercept does not change 11-5. a: x = 2 b: k ≈ 3.76 or 1.24 c: –2 < x < 10 d: x = 5 11-6. 1, 400, 000 ! 50x >1, 200, 000 , less than 4000 square miles per year 11-7. a: x-intercepts (–2, 0) and (0, 0); y-intercept (0, 0) b: x-intercepts (–3, 0) and (5, 0); y-intercept (0, 3) c: x-intercepts (–1, 0) and (1, 0), y-intercept (0, –1) 11-8. (3, 3) and (–2, –7) 11-9. s = a + 150, 3s + 5a = 4730; 685 students 11-10. a: f (x)+ 2 = x2 !1 ; Shifted up 2 units.

b: 2 f (x) = 2x2 ! 6 ; Twice as steep and the vertex shifts down 3 units. c: f (x + 2) = (x + 2)2 ! 3 ; Shifted left 2 units.

d: f (2x) = 4x2 ! 3 ; Four times as steep. 11-11. a: 1

3 b: !1± !3

2 is not a real number because the square root of a

negative number is an imaginary number. 11-12. a: 3 b: 2 c: does not exist d: 0 e: 1 f: 1 11-13. a: y = x2 + 3x ! 28 b: y = x2 +10x + 24 c: y = x2 + 6x + 9 11-14. A 11-15. S - $1575 ! $24, 000 ; $22, 425 ! S ! $25575

Selected Answers 3

Lesson 11.1.2 11-20. a: f !1(x) = x!3

2 b: g!1(x) = 4x + 5

11-21. a: –1 ≤ x ≤ 3 b: x = !2 ± 7 " 0.65 or ! 4.65 c: all real numbers d: x = –2.5 or 5.5 11-22. a: (x ! 3)2 !1 , shift down 1 unit b: !(x ! 3)2 , reflected over the x-axis

c: (x ! 4)2 , shift right 1 unit d: (!x ! 3)2 , reflected over the y-axis 11-23. a: one b: none c: two d: one 11-24. m ! 31 " 5 ; 26 ! m ! 36 11-25. The team president is using the mean, and the fans are using the median. A few large “outliers,” such as super star players, have very high salaries.

Lesson 11.2.1 (Day 1) 11-28. (46.4, 48.5, 50.4, 52.5, 55.9) 11-29. a: f !1(x) = 3(x + 2)

b: g!1(x) = (x ! 5)12

= 2(x ! 5)

11-30. a: x = 1 b: x > 3 or x < –3 c: 0 ! x ! 4

3

d: x = 3

4 e: !2 " x " 3 f: x = 3

11-31. 45 miles 11-32. y > 2x !1 11-33. a: x = 1

2 or –2

b: no real solution

4 Core Connections Algebra

Lesson 11.2.1 (Day 2) 11-34. a: (1.58, 2.50, 2.91, 3.49, 4.29)

b: See solution graph at right. c: The median (center) is at 2.91 points.

The shape is symmetric. The IQR (spread) is Q3!Q1= 3.49 ! 2.50 = 0.99 points. There are no apparent outliers.

11-35. The graph starts at (3, 1); D: x ! 3 , R: y !1 . See graph at right. 11-36. While there are multiple ways to write the equation, one possible way is y = (x + 2)(x + 3)+1 . However, all equations should be equivalent to y = x2 + 5x + 7 . 11-37. 1980 + 30m + 2590 + 20m = 5000 , m = 8.6; It will be full after 8 months; there will not be

enough room for songs in the 9th month. 11-38. a: !2 < x < 2 b: x ! 2.5 c: x = 1

4

d: no solution e: x = –12 f: !5 " x " 3 11-39. a: The vertex is at (2, 6). The coefficient of –2 means the graph

is pointing downward, so the vertex ix a maximum.

b: The y- intercept is at !2(!2)2 + 6 = !2 . Along with the vertex, and knowing the parabola is pointing downward, there is enough information to make a sketch of the graph. See graph at right.

1 2 3 4 GPA

0

5

10

# of

Stu

dent

s

x

y

x

y

Selected Answers 5

Lesson 11.2.2 (Day 1) 11-46. a: See solution graph at right.

b: The median is 257 rpm. The graph is single-peaked and skewed. The IQR is Q3!Q1= 263! 253!10 rpm. 291 rpm is apparently an outlier.

c: The median. Because the data is not symmetrical and has an outlier the mean is not an appropriate measure of center.

11-47. a: t(n) = !3n +10 or t(n) = 7 ! 3(n !1) b: t(n) = 2000

3!3n"1 or an = 2000

9(3)n

11-48. a: x2 ! 5 , shift down 2 units

b: !2x2 + 6 , reflected over x-axis, stretched

c: (x ! 2)2 ! 3 , shift right 2 units

d: (!2x)2 ! 3 or 4x2 ! 3 , stretched, and reflected over y-axis onto itself 11-49. a: The vertex is (–1, –5) and the point is a minimum.

b: –5 11-50. They will be the same after 20 years, when both are $1800. 11-51. a: 25a!22b36 b: 5 !3"1x"9y5 11-52. 72 represents room temperature and a horizontal asymptote of the graph, 0.7 represents a

temperature loss of 30% per unit of time.

245 265 285 300 RPM

0

5

Freq

uenc

y

6 Core Connections Algebra

Lesson 11.2.2 (Day 2) 11-53. a: See graphs at right. The median of the two groups are

virtually identical. Both groups have uniform distributions of ages. Neither group has any outliers. However, the ages in Group 7B are much more widely distributed—have much more variability—than the ages in Group 7A. The IQR for 7A is only 70 ! 53= 17 years, while the IQR for 7B is more than twice as wide at 77 ! 39.5 " 38 years. The minimum for 7A is 20 years older than the minimum for 7B, and the maximum for 7A is 22 years younger than the maximum for 7B. Note that the small number of data points does not allow for bin widths on the histogram much narrower than 20 years; it is not appropriate to create bin widths of 10 years.

b: Either. Since the data distributions are symmetric and there are no outliers, either measure of center is appropriate.

11-54. a: f !1(x) = (x+2)

7 b: Yes

11-55. a: x !12 b: !10 < x <10 c: x < 0 d: x < !5 , x >1 11-56. y = 2(x +1)2 + 4 11-57. a: x = ± 4 b: (–5, –17)

c: x = 4 or –2 d: x = !1± 57

4"1.64 or –2.14

11-58. See graph at right. 11-59. Based on direction and vertex of the parabola compared with the slope and y-intercept of

the line, there are two points of intersection.

20 40 80 200 Age (years)

0

4

8 7B

60

20 40 80 200 Age (years)

0

5

10 7A

60

Selected Answers 7

Lesson 11.2.3 11-67. See graph at right. The distribution of weights is symmetric

with no outliers (as determined by the modified boxplot). The mean is 40 kg with a standard deviation of 16 kg. The weights are rounded to the nearest whole number.

11-68. a: See boxplots at right below. Unequivocally, the farmer

should plant in shade. The median crop is about 7 bushels higher in shade. The minimum, maximum, first quartile, and third quartile are all higher in shade. Both distributions are skewed in the same direction. The spread in data (IQR) is almost the same for both type of tree—the middle box is the same size for both boxplots. The maximum of 127 bushels from one of the shady trees is almost certainly an outlier.

b: No. Neither of the boxplots are symmetrical; the distributions are skewed. The maximum on the shady plot may be an outlier.

11-69. a: y = x2 ! 4x b: y = x2 ! 4 c: y = 6x2 ! x ! 2 11-70. (–2, –10) 11-71. a: all real numbers b: –5 < x < 4

c: no solution d: x = 1

3

11-72. a–iv, b–ii, c–v, d–i, e–iii. b is the only histogram with a narrow range, so it matches to ii.

The two skewed histograms are straightforward to match. c has a uniform distribution, so the quartiles on the boxplot must be of even length, as in v. d has a lot of data at the two edges, and the data in the middle is more spread out, so the “whiskers” of the boxplot must be narrow, and the box must be wide, as in i.

11-73.

10 30 50 70

0 30 60 100 130

sunny

shady

8 Core Connections Algebra

Lesson 11.3.1 11-75. Edison is correct because 3(2)+ 2(!3) = 2 and 5(2)!12 = !2 . 11-76. 16 11-77. See solution graph at right. D: x ! 3 , R: y ! 0 11-78. a: not possible b: !27 c: 8 d: 0 e: –1 11-79. See graph at right. The distribution is symmetric with no

outliers. The mean is 50.7 cm and the standard deviation is 2.6 cm. The lengths were measured to the nearest tenth of a centimeter.

11-80. (1, 12) and (–5, 42) 11-81. If d = number of dimes and q = number of quarters, then q = 2d ! 6 and d + q = 147 .

Then d = 51 and q = 96 , so Jessica has 51(0.10)+ 96(0.25) = $29.10 . 11-82. f (x) = !(x ! 3)(x +1) = !(x !1)2 + 4 = !x

2+ 2x + 3

11-83. a: x = 1 b: x < –1 or x > 7 c: x = 7 11-84. See graph at right. The vertex is at (3, –1); x-intercepts (4, 0) and

(2, 0); y-intercept (0, 2) 11-85. a: Team 2 works, on average, a little faster—the median number of widgets per team

member is slightly higher. The distributions for both teams are similarly symmetric. However, the members of Team 1 are much more consistent than Team 2. The variability (IQR) of Team 1 is almost half that of Team 2, and Team 1’s range is less too. Neither team had outliers.

b: Since both distributions are nearly symmetric with no outliers, it is appropriate to compare standard deviations. Since Team 1 had both IQR and range smaller than Team 2, we would expect that Team 1 has a smaller standard deviation.

46 48 50 52 54 56

x

y

Selected Answers 9

Lesson 11.3.2 11-88. a: There is a strong positive linear association between the depth of a water well and the

cost to install it. There are no apparent outliers. b: On average every foot deeper you drill the well the cost increases by $14.65. c: The coefficient of correlation is 0.929, R-squared = 0.864. 86% of the variation in the

cost of drilling a water well can be explained by a linear association with its depth. d: $1395 represents the cost of a well that has no depth. It would be roughly the cost of

the pump e: 1395 + 14.65(80) = $2567, 1395 + 14.65(150) = $3593,

1395 + 14.65(200) = $4325 f: From part (e), the predicted cost is $2567. Actual – $2567 = $363;

actual cost was $2930. g: A linear model looks the most appropriate because there is no pattern in the residual

plot. 11-89. y = 1

2(x ! 2)2 + 3

11-90. a: 3 b: 1 c: 2 11-91. 0.50c + 0.75b !100, c ! 0, b ! 0 11-92. The parabola has vertex (1, –3) and points down. The line has y-intercept at

(0, –5) and decreases. There are two points of intersection. 11-93. w !10 > 0.13 ; w < 9.87 or w >10.13

10 Core Connections Algebra

Lesson 11.3.3 11-98. No 11-99. a: yes b: no; most inputs have two outputs c: no; x = !1 has two outputs 11-100. a: x = 3± 21 b: x = 2 ± 7 11-101. a: none b: two c: two d: one 11-102. a: The slope of the line of best fit is –75.907. Jeremiah has been giving coins away at a

rate of about 76 coins a year. b: In 2010 he had 1295 coins. If c is the number of coins, and y is the number of years

since 2010, then c = 1295 ! 76y . When c = 0 coins, y ≈ 17 years from now. In 2027 he will have only 3 coins left.

11-103. a: The IQR for W is more than for Z because the middle of the boxplot for W is wider.

The standard deviation for Z is greater because overall, including the outliers, the data for Z is spread out more than for W. Since mean is impacted by outliers more than median, the standard deviation (which is based on mean) is impacted by more by the outliers in Chip Z. Mean and standard deviation are not appropriate for Chip Z because the shape is skewed and there are outliers.

b: Chip Z appears to be the more energy efficient. It has a lower median use of current. Also, for most of the data sets tested, chip Z uses the same or slightly less current than chip W. Chip Z has smaller IQR: it is more consistent in current usage. However, all these benefits may be offset by the two high outliers belonging to Chip Z which might indicate a reliability problem.

Selected Answers 11

Lesson 11.3.4 11-110. (!2.5, ! 36.75) ; Students can complete the square or they can use the fact that due to

the parabola’s symmetry, the vertex must have an x-coordinate that is halfway between the x-intercepts.

11-111. a: y = 2!or !!2 b: x = 8 or ! 3 11-112. (6, 8) and (5, 3) 11-113. D: x < 2 and x > 2 , R: y < 0 and y > 0 Solution graph shown at right. 11-114. 8x +10(6) = 8.5(x + 6) , 18 pounds 11-115. See solution graph at right. The shape is double-peaked

and symmetric. There are no outliers. The mean speed is 79.5 mph with a standard deviation of 6.8 mph.

11-116. (–2, 5) and (6, 21) 11-117. See solution at right. 11-118. a: none b: two c: one d: two 11-119. a: x = –12 b: 4 ! 12 < b < 4 + 12

c: !33" x " 27 d: n = 1

5 or 2

11-120. x-intercepts: (! 0.2, 0) and (! 3.1, 0) , vertex (!1.7, ! "6.3) ;

Solution graph is shown at right. 11-121. a: See table and graph at far right.

b: y = (x ! 2)(x ! 5) = x2 ! 7x +10 c: See graph at right. d: 4 seconds, 256 feet

70 74 86 90 Speed (mph)

0

5

10

78 82

x y –2 5 –1 0 0 –3 1 –4 2 –3 3 0 4 5

x

y

(–1, 0) (–7, 0)

y

(–4, –9)

12 Core Connections Algebra

Lesson 11.3.5 11-126. Based on direction and vertex of the parabola compared with the intercepts of the line,

there are two points of intersection. 11-127. a: x ! 4 b: x > 20.5 c: !5 " x "1 d: x >19 or x < !12 11-128. x = –9 or –5 11-129. See graph at right. 11-130. 25 homes 11-131. See graph at right.

x-intercept: (–2, 0), y-intercept: (0, –2); there is no value for g(1) , which creates a break in the graph.

11-132. B 11-133. y = !

3

2x !1

11-134. Answers vary, but likely answers are 6(m ! 2) , 2(3m ! 6) , 3(2m ! 4) , and 1(6m !12) . 11-135. a: –1 b: 2 c: –2 d: –1 11-136. a: 5m2 + 9m ! 2 b: !x2 + 4x +12

c:25x2 !10xy + y2 d: 6x2 !15xy +12x 11-137. x = !3 or –11

x

y