Question

1. the table below shows the value of a particular car over time. time (years) value (dollars) 0 20,000 5 10,550 10 5570 15 2940 20 1550 determine whether a linear or exponential function is more appropriate for modeling this data. explain your choice. 10. joey recorded his heart - rate, in beats per minute (bpm), doing different numbers of jumping jacks. his results are in the table below. number of jumping jacks x heart rate (bpm) y 0 68 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 about this data.

Explanation:

9.

Step1: Calculate differences for linear - like behavior

For the time - value data, find the differences in time ($\Delta t$) and values ($\Delta V$).
$\Delta t_1=5 - 0=5$, $\Delta V_1 = 10550 - 20000=- 9450$; $\Delta t_2=10 - 5 = 5$, $\Delta V_2=5570 - 10550=-4980$; $\Delta t_3=15 - 10 = 5$, $\Delta V_3=2940 - 5570=-2630$; $\Delta t_4=20 - 15 = 5$, $\Delta V_4=1550 - 2940=-1390$. The differences in values are not constant, so it is not a linear function.

Step2: Calculate ratios for exponential - like behavior

Find the ratios of consecutive values. $\frac{V_1}{V_0}=\frac{10550}{20000}=0.5275$, $\frac{V_2}{V_1}=\frac{5570}{10550}\approx0.528$, $\frac{V_3}{V_2}=\frac{2940}{5570}\approx0.528$, $\frac{V_4}{V_3}=\frac{1550}{2940}\approx0.527$. The ratios of consecutive values are approximately constant.

Step1: Calculate the necessary sums

Let $n = 6$.
$\sum_{i = 1}^{n}x_i=0 + 10+15 + 20+30=75$, $\sum_{i = 1}^{n}y_i=68 + 84+104+100+120 = 476$, $\sum_{i = 1}^{n}x_i^2=0^2+10^2 + 15^2+20^2+30^2=0 + 100+225+400+900 = 1625$, $\sum_{i = 1}^{n}y_i^2=68^2+84^2+104^2+100^2+120^2=4624+7056+10816+10000+14400 = 46906$, $\sum_{i = 1}^{n}x_iy_i=0\times68+10\times84 + 15\times104+20\times100+30\times120=0 + 840+1560+2000+3600 = 8000$.

Step2: Calculate the slope ($m$) and y - intercept ($b$) of the linear regression line

The slope $m=\frac{n\sum_{i = 1}^{n}x_iy_i-\sum_{i = 1}^{n}x_i\sum_{i = 1}^{n}y_i}{n\sum_{i = 1}^{n}x_i^2-(\sum_{i = 1}^{n}x_i)^2}$
$m=\frac{5\times8000 - 75\times476}{5\times1625-75^2}=\frac{40000 - 35700}{8125 - 5625}=\frac{4300}{2500}=1.72$
The y - intercept $b=\frac{\sum_{i = 1}^{n}y_i-m\sum_{i = 1}^{n}x_i}{n}$
$b=\frac{476-1.72\times75}{5}=\frac{476 - 129}{5}=\frac{347}{5}=69.4$
The linear regression equation is $y = 1.72x+69.4$

Step3: Calculate the correlation coefficient ($r$)

$r=\frac{n\sum_{i = 1}^{n}x_iy_i-\sum_{i = 1}^{n}x_i\sum_{i = 1}^{n}y_i}{\sqrt{(n\sum_{i = 1}^{n}x_i^2-(\sum_{i = 1}^{n}x_i)^2)(n\sum_{i = 1}^{n}y_i^2-(\sum_{i = 1}^{n}y_i)^2)}}$
$r=\frac{5\times8000 - 75\times476}{\sqrt{(5\times1625 - 75^2)(5\times46906-476^2)}}$
$r=\frac{40000 - 35700}{\sqrt{(8125 - 5625)(234530 - 226576)}}$
$r=\frac{4300}{\sqrt{2500\times7954}}$
$r=\frac{4300}{\sqrt{19885000}}\approx\frac{4300}{4459.26}\approx0.96$

Step4: Interpret the correlation coefficient

A correlation coefficient of $r\approx0.96$ (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 significantly in a linear fashion.

Answer:

An exponential function is more appropriate for modeling this data because the ratios of consecutive values are approximately constant while the differences in values are not constant.

10.