Answer:
(a) As time increases, the amount of water in the pool increases.
11 gallons per minute
(b) 65 gallons
Step-by-step explanation:
From inspection of the table, we can see that <u>as time increases, the amount of water in the pool increases</u>.
We are told that Ann adds water at a constant rate. Therefore, this can be modeled as a linear function.
The rate at which the water is increasing is the <em>rate of change</em> (which is also the <em>slope </em>of a linear function).
Choose 2 ordered pairs from the table:
![\textsf{let}\:(x_1,y_1)=(8, 153)](https://tex.z-dn.net/?f=%5Ctextsf%7Blet%7D%5C%3A%28x_1%2Cy_1%29%3D%288%2C%20153%29)
![\textsf{let}\:(x_2,y_2)=(12,197)](https://tex.z-dn.net/?f=%5Ctextsf%7Blet%7D%5C%3A%28x_2%2Cy_2%29%3D%2812%2C197%29)
Input these into the slope formula:
![\textsf{slope}\:(m)=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{197-153}{12-8}=\dfrac{44}{4}=11](https://tex.z-dn.net/?f=%5Ctextsf%7Bslope%7D%5C%3A%28m%29%3D%5Cdfrac%7By_2-y_1%7D%7Bx_2-x_1%7D%3D%5Cdfrac%7B197-153%7D%7B12-8%7D%3D%5Cdfrac%7B44%7D%7B4%7D%3D11)
Therefore, the rate at which the water in the pool is increasing is:
<u>11 gallons per minute</u>
To find the amount of water that was already in the pool when Ann started adding water, we need to create a linear equation using the found slope and one of the ordered pairs with the point-slope formula:
![y-y_1=m(x-x_1)](https://tex.z-dn.net/?f=y-y_1%3Dm%28x-x_1%29)
![\implies y-153=11(x-8)](https://tex.z-dn.net/?f=%5Cimplies%20y-153%3D11%28x-8%29)
![\implies y-153=11x-88](https://tex.z-dn.net/?f=%5Cimplies%20y-153%3D11x-88)
![\implies y=11x-88+153](https://tex.z-dn.net/?f=%5Cimplies%20y%3D11x-88%2B153)
![\implies y=11x+65](https://tex.z-dn.net/?f=%5Cimplies%20y%3D11x%2B65)
When Ann had added no water, x = 0. Therefore,
![y=11(0)+65](https://tex.z-dn.net/?f=y%3D11%280%29%2B65)
![y=65](https://tex.z-dn.net/?f=y%3D65)
So there was <u>65 gallons</u> of water in the pool before Ann starting adding water.