The difference between the gas prices in 2013 and 2001 is 3.00
<h3>Graphing tool</h3>
See attachment for the graph of the line of best fit
<h3>The equation</h3>
Start by calculating the slope of the line of best fit using:
![b = \frac{y_2 -y_1}{x_2 -x_1}](https://tex.z-dn.net/?f=b%20%3D%20%5Cfrac%7By_2%20-y_1%7D%7Bx_2%20-x_1%7D)
From the line of best fit, we have:
![b = \frac{3 -2}{8 -4}](https://tex.z-dn.net/?f=b%20%3D%20%5Cfrac%7B3%20-2%7D%7B8%20-4%7D)
The regression equation is then represented as:
![\^y = b(\^x - x_1) + y_1](https://tex.z-dn.net/?f=%5C%5Ey%20%3D%20b%28%5C%5Ex%20-%20x_1%29%20%2B%20y_1)
So, we have:
![\^y = \frac 14 (\^x - 4) + 2](https://tex.z-dn.net/?f=%5C%5Ey%20%3D%20%5Cfrac%2014%20%28%5C%5Ex%20-%204%29%20%2B%202)
![\^y = \frac 14\^x - 1 + 2](https://tex.z-dn.net/?f=%5C%5Ey%20%3D%20%5Cfrac%2014%5C%5Ex%20-%201%20%2B%202)
![\^y = \frac 14\^x + 1](https://tex.z-dn.net/?f=%5C%5Ey%20%3D%20%5Cfrac%2014%5C%5Ex%20%2B%201)
So, the linear regression equation is:
![\^y = \frac 14\^x + 1](https://tex.z-dn.net/?f=%5C%5Ey%20%3D%20%5Cfrac%2014%5C%5Ex%20%2B%201)
In 2001, the value of x is 1.
So, we have:
![\^y_1 = \frac 14\times (1) + 1](https://tex.z-dn.net/?f=%5C%5Ey_1%20%3D%20%5Cfrac%2014%5Ctimes%20%281%29%20%2B%201)
![\^y_1 = 1.25 \\](https://tex.z-dn.net/?f=%5C%5Ey_1%20%3D%201.25%20%5C%5C)
In 2013, the value of x is 13.
So, we have:
![\^y_2 = \frac 14\times (13) + 1](https://tex.z-dn.net/?f=%5C%5Ey_2%20%3D%20%5Cfrac%2014%5Ctimes%20%2813%29%20%2B%201)
![\^y_2 = 4.25](https://tex.z-dn.net/?f=%5C%5Ey_2%20%3D%204.25)
So, the difference (d) between the prices is:
![d = 4.25 - 1.25](https://tex.z-dn.net/?f=d%20%3D%204.25%20-%201.25)
![d = 3.00](https://tex.z-dn.net/?f=d%20%3D%203.00)
Hence, the difference between the gas prices in 2013 and 2001 is 3.00
Read more about linear regression at:
brainly.com/question/17844286