Question

1. which situation could be modeled by a linear function?

a. the value of a car depreciates by 7% annually.
b. a gym charges a $50 initial fee and then $30 monthly.
c. the number of bacteria in a lab doubles weekly.
d. the amount of money in a bank account increases by 0.1% monthly.

1. the table below shows the value of a particular car over time.

time (years)	value (dollars)
5	10550
10	5570
15	2940
20	1550

determine whether a linear or exponential function is more appropriate for modeling this data. explain your choice.

1. joey recorded his heart - rate, in beats per minute (bpm), after doing different numbers of jumping jacks. his results are shown in the table below.

number of jumping jacks x	heart rate (bpm) y
10	84
15	104
20	100
30	120

state the linear regression equation that estimates the heart rate per number of jumping jacks.
state the correlation coefficient of the linear regression, rounded to the nearest hundredth.
explain what the correlation coefficient suggests in the context of this problem.

1. student scores on a recent test are shown in the table below.

85

96

92

87

Explanation:

Step1: Analyze question 8

A linear - function has a constant rate of change.

• Option A: The value of a car depreciates by 7% annually. This is an exponential decay (formula $A = P(1 - r)^t$), not linear.
• Option B: A gym charges a $50 initial fee and then $30 monthly. The cost $C$ as a function of the number of months $m$ is $C(m)=30m + 50$, which is a linear function ($y = mx + b$ form).
• Option C: The number of bacteria in a lab doubles weekly. This is an exponential growth (formula $A = P\cdot2^{\frac{t}{w}}$ where $w$ is the time - interval for doubling), not linear.
• Option D: The amount of money in a bank account increases by 0.1% monthly. This is an exponential growth (formula $A = P(1 + r)^t$), not linear. So the answer to question 8 is B.

Step2: Analyze question 9

To determine if a linear or exponential function is more appropriate for the car - value data:
Let $t$ be the time in years and $V$ be the value of the car.

• Calculate the differences in value for equal time - intervals.

From $t = 0$ to $t = 5$: $\Delta V=20000 - 10550 = 9450$
From $t = 5$ to $t = 10$: $\Delta V=10550 - 5570 = 4980$
From $t = 10$ to $t = 15$: $\Delta V=5570 - 2940 = 2630$
From $t = 15$ to $t = 20$: $\Delta V=2940 - 1550 = 1390$
The differences are not constant, so it is not a linear function.
Let's check for an exponential function. Calculate the ratios of consecutive values:
$\frac{V(5)}{V(0)}=\frac{10550}{20000}=0.5275$
$\frac{V(10)}{V(5)}=\frac{5570}{10550}\approx0.528$
$\frac{V(15)}{V(10)}=\frac{2940}{5570}\approx0.528$
$\frac{V(20)}{V(15)}=\frac{1550}{2940}\approx0.527$
The ratios are approximately constant, so an exponential function is more appropriate for modeling this data.

Step3: Analyze question 10

For the heart - rate data:
We have data points $(x,y)$ where $x$ is the number of jumping jacks and $y$ is the heart - rate.
The linear regression equation is of the form $y = mx + b$. Using a calculator or statistical software (or the following formulas for $n$ data points $(x_i,y_i)$:
$n = 6$
$\sum_{i = 1}^{n}x_i=0 + 10+15 + 20+30=75$
$\sum_{i = 1}^{n}y_i=68 + 84+104+100+120 = 476$
$\sum_{i = 1}^{n}x_i^2=0^2+10^2 + 15^2+20^2+30^2=0 + 100+225+400+900 = 1625$
$\sum_{i = 1}^{n}x_iy_i=0\times68+10\times84 + 15\times104+20\times100+30\times120=0 + 840+1560+2000+3600 = 8000$
$m=\frac{n\sum_{i = 1}^{n}x_iy_i-\sum_{i = 1}^{n}x_i\sum_{i = 1}^{n}y_i}{n\sum_{i = 1}^{n}x_i^2-(\sum_{i = 1}^{n}x_i)^2}$
$m=\frac{6\times8000 - 75\times476}{6\times1625-75^2}=\frac{48000 - 35700}{9750 - 5625}=\frac{12300}{4125}\approx2.98$
$\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}=\frac{75}{6}=12.5$
$\bar{y}=\frac{\sum_{i = 1}^{n}y_i}{n}=\frac{476}{6}\approx79.33$
$b=\bar{y}-m\bar{x}=79.33-2.98\times12.5=79.33 - 37.25 = 42.08$
The linear regression equation is $y = 2.98x+42.08$
The correlation coefficient $r$ is given by:
$r=\frac{n\sum_{i = 1}^{n}x_iy_i-\sum_{i = 1}^{n}x_i\sum_{i = 1}^{n}y_i}{\sqrt{[n\sum_{i = 1}^{n}x_i^2-(\sum_{i = 1}^{n}x_i)^2][n\sum_{i = 1}^{n}y_i^2-(\sum_{i = 1}^{n}y_i)^2]}}$
First, calculate $\sum_{i = 1}^{n}y_i^2=68^2+84^2+104^2+100^2+120^2=4624+7056+10816+10000+14400 = 46906$
$r=\frac{6\times8000 - 75\times476}{\sqrt{(6\times1625 - 75^2)(6\times46906-476^2)}}$
$r=\frac{12300}{\sqrt{4125\times(281436 - 226576)}}=\frac{12300}{\sqrt{4125\times54860}}\approx0.89$
A correlation coefficient $r\approx0.89$ (close to 1) suggests a strong positive linear relationship between the number of jumping jacks and the heart - rate. That is, as the number of jumping jacks increases, the heart - rate tends to increase in a fairly linear fashion.

A correlation coefficient of approximately 0.89 suggests a strong positive linear relationship between the number of jumping jacks and the heart - rate.

Answer:

1. B. A gym charges a $50 initial fee and then $30 monthly.
2. An exponential function is more appropriate for modeling the car - value data because the ratios of consecutive values are approximately constant while the differences in value for equal time - intervals are not constant.
3. Linear regression equation: $y = 2.98x + 42.08$

Correlation coefficient: $r\approx0.89$