Answer:
The best fit is <em>A. Linear model</em>
<em></em>
Step-by-step explanation:
Given:
Monthly Rate = $20, Number of customers = 5000
If there is a decrease of $1 in the monthly rate, the number of customers increase by 500.
To find:
The type of model that best fits the given situation?
Solution:
Monthly Rate = $20, Number of customers = 5000
Let us decrease the monthly rate by $1.
Monthly Rate = $20 - $1 = $19, Number of customers = 5000 + 500 = 5500
Let us decrease the monthly rate by $1 more.
Monthly Rate = $19 - $1 = $18, Number of customers = 5500 + 500 = 6000
Here, we can see that there is a <em>linear change </em> in the number of customers whenever there is decrease in the monthly rate.
We have 2 pair of values here,
x = 20, y = 5000
x = 19, y = 5500
Let us write the equation in slope intercept form:
![y =mx+c](https://tex.z-dn.net/?f=y%20%3Dmx%2Bc)
Slope of a function:
![m=\dfrac{y_2-y_1}{x_2-x_1}](https://tex.z-dn.net/?f=m%3D%5Cdfrac%7By_2-y_1%7D%7Bx_2-x_1%7D)
![m=\dfrac{5500-5000}{19-20}\\\Rightarrow -500](https://tex.z-dn.net/?f=m%3D%5Cdfrac%7B5500-5000%7D%7B19-20%7D%5C%5C%5CRightarrow%20-500)
So, the equation is:
![y =-500x+c](https://tex.z-dn.net/?f=y%20%3D-500x%2Bc)
Putting x = 20, y = 5000:
![5000 =-500\times 20+c\\\Rightarrow c = 5000 +10000 = 15000](https://tex.z-dn.net/?f=5000%20%3D-500%5Ctimes%2020%2Bc%5C%5C%5CRightarrow%20c%20%3D%205000%20%2B10000%20%3D%2015000)
![\Rightarrow \bold{y =-500x+15000}](https://tex.z-dn.net/?f=%5CRightarrow%20%5Cbold%7By%20%3D-500x%2B15000%7D)
Let us check whether (18, 6000) satisfies it.
Putting x = 18:
so, it is true.
So, the answer is:
The best fit is <em>A. Linear model</em>