Answer Key Chapter 4 - College Algebra | OpenStax (2024)

4.1 Linear Functions

4.2 Modeling with Linear Functions

4.3 Fitting Linear Models to Data

1.

Terry starts at an elevation of 3000 feet and descends 70 feet per second.

3.

$d\left(t\right)=100-10t$

5.

The point of intersection is $\left(a,a\right).$ This is because for the horizontal line, all of the $y$ coordinates are $a$ and for the vertical line, all of the $x$ coordinates are $a.$ The point of intersection is on both lines and therefore will have these two characteristics.

7.

Yes

9.

Yes

11.

No

13.

Yes

15.

Increasing

17.

Decreasing

19.

Decreasing

21.

Increasing

23.

Decreasing

25.

2

27.

–2

29.

$y=\frac{3}{5}x-1$

31.

$y=3x-2$

33.

$y=-\frac{1}{3}x+\frac{11}{3}$

35.

$y=-1.5x-3$

37.

perpendicular

39.

parallel

41.

$\begin{array}{l}f(0)=-(0)+2\\ f(0)=2\\ y-\mathrm{int}:(0,2)\\ 0=-x+2\\ x-\mathrm{int}:(2,0)\end{array}$

43.

$\begin{array}{l}h(0)=3(0)-5\\ h(0)=-5\\ y-\mathrm{int}:(0,-5)\\ 0=3x-5\\ x-\mathrm{int}:\left(\frac{5}{3},0\right)\end{array}$

45.

$\begin{array}{l}-2x+5y=20\\ -2(0)+5y=20\\ 5y=20\\ y=4\\ y-\mathrm{int}:(0,4)\\ -2x+5(0)=20\\ x=-10\\ x-\mathrm{int}:(-10,0)\end{array}$

47.

Line 1: m = –10 Line 2: m = –10 Parallel

49.

Line 1: m = –2 Line 2: m = 1 Neither

51.

$\text{Line1}:m=–2\text{Line2}:m=–2\text{Parallel}$

53.

$y=3x-3$

55.

$y=-\frac{1}{3}t+2$

57.

59.

$y=-\frac{5}{4}x+5$

61.

$y=3x-1$

63.

$y=-2.5$

65.

F

67.

C

69.

A

71.

73.

75.

77.

79.

81.

83.

85.

$y=\text{3}$

87.

$x=-3$

89.

Linear, $g(x)=-3x+5$

91.

Linear, $f(x)=5x-5$

93.

Linear, $g(x)=-\frac{25}{2}x+6$

95.

Linear, $f(x)=10x-24$

97.

$f(x)=-58x+17.3$

99.

101.

103.

105.

$y=-\frac{16}{3}$

107.

$x=a$

109.

$y=\frac{d}{c–a}x–\frac{ad}{c–a}$

111.

$y=100x–98$

113.

$x<\frac{1999}{201}\text{,}x>\frac{1999}{201}$

115.

$45 per training session.

117.

The rate of change is 0.1. For every additional minute talked, the monthly charge increases by $0.1 or 10 cents. The initial value is 24. When there are no minutes talked, initially the charge is $24.

119.

The slope is –400. this means for every year between 1960 and 1989, the population dropped by 400 per year in the city.

121.

C

1.

Determine the independent variable. This is the variable upon which the output depends.

3.

To determine the initial value, find the output when the input is equal to zero.

5.

6 square units

7.

20.01 square units

9.

2,300

11.

64,170

13.

$P\left(t\right)=75,000+2500t$

15.

(–30, 0) Thirty years before the start of this model, the town had no citizens. (0, 75,000) Initially, the town had a population of 75,000.

17.

Ten years after the model began

19.

$W(t)=0.5t+7.5$

21.

$\left(-15,0\right)$ : The x-intercept is not a plausible set of data for this model because it means the baby weighed 0 pounds 15 months prior to birth. $\left(0,\text{7}.\text{5}\right)$ : The baby weighed 7.5 pounds at birth.

23.

At age 5.8 months

25.

$C\left(t\right)=12,025-205t$

27.

$\left(\text{58}.\text{7},0\right):$ In roughly 59 years, the number of people inflicted with the common cold would be 0. $\left(0,\text{12},0\text{25}\right)$ Initially there were 12,025 people afflicted by the common cold.

29.

2063

31.

$y=-2t\text{+180}$

33.

In 2070, the company’s profit will be zero.

35.

$y=\text{3}0t-\text{3}00$

37.

(10, 0) In the year 1990, the company’s profits were zero

39.

Hawaii

41.

During the year 1933

43.

$105,620

45.

47.

49.

P(t) = 190t + 4,360

51.

53.

More than 133 minutes

55.

More than $42,857.14 worth of jewelry

57.

More than $66,666.67 in sales

1.

When our model no longer applies, after some value in the domain, the model itself doesn’t hold.

3.

We predict a value outside the domain and range of the data.

5.

The closer the number is to 1, the less scattered the data, the closer the number is to 0, the more scattered the data.

7.

61.966 years

9.

11.

13.

15.

$\text{Thisvalueofrindicatesastrongnegativecorrelationorslope,soC}$

17.

$\text{Thisvalueofrindicatesaweaknegativecorrelation,soB}$

19.

21.

23.

Yes, trend appears linear because $r=0.\text{985}$ and will exceed 12,000 near midyear, 2016, 24.6 years since 1992.

25.

$y=\text{1}.\text{64}0x+\text{13}.\text{8}00,$ $r=0.\text{987}$

27.

$y=-0.962x+26.86,r=-0.965$

29.

$y=-\text{1}.\text{981}x+\text{6}0.\text{197;}$ $r=-0.\text{998}$

31.

$y=0.\text{121}x-38.841,r=0.998$

33.

$(\mathrm{-2},\mathrm{-6}),(1,\text{−12}),(5,\mathrm{-20}),(6,\text{−22}),(9,\text{−28});$ Yes, the function is a good fit.

35.

$(\text{189}.8,0)$ If 18,980 units are sold, the company will have a profit of zero dollars.

37.

$y=0.00587x+\text{1985}.4\text{1}$

39.

$y=\text{2}0.\text{25}x-\text{671}.\text{5}$

41.

$y=-\text{1}0.\text{75}x+\text{742}.\text{5}0$

1.

Yes

3.

Increasing

5.

$y=-\text{3}x+\text{26}$

7.

3

9.

$y=\text{2}x-\text{2}$

11.

Not linear.

13.

parallel

15.

$(\mathrm{–9},0);(0,\mathrm{–7})$

17.

Line 1: $m=-2;$ Line 2: $m=-2;$ Parallel

19.

$y=-0.2x+21$

21.

23.

More than 250

25.

118,000

27.

$y=-\text{3}00x+\text{11},\text{5}00$

29.

31.

18,500

33.

$91,625

35.

37.

39.

2023

41.

$y=-1.294x+49.412;r=-0.974$

43.

2027

45.

7,660

1.

Yes

3.

Increasing

5.

y = −1.5x − 6

7.

y = −2x − 1

9.

No

11.

Perpendicular

13.

(−7, 0); (0, −2)

15.

y = −0.25x + 12

17.

19.

150

21.

165,000

23.

y = 875x + 10,625

25.

27.

29.

In early 2018

31.

y = 0.00455x + 1979.5

33.

r = 0.999