Question

Ann is adding water to a swimming pool at a constant rate. The table below shows the amount of water in the pool after different amounts of time.
Time (minutes) 8, 12, 16, 20
Water (gallons) 153,197,241,285

Answer the following questions.


Choose the statement that best describes how the time and the amount of water in the pool are related. Then give the value requested.

As time increases, the amount of water in the pool decreases.

gallons per minute

As time increases, the amount of water in the pool increases.

gallons per minute

(b)How much water was already in the pool when Ann started adding water?
gallons

Answer 1

#a

Take 2 points

• (12,197)
• (16,241)

Slope=Rate of water flow

• m=241-197/16-12
• m=44/4
• m=11

The water increases with respect to time as slope is positive.

It's

• 11gallons/min

#2

We need amount of water when time is 0

So

Let that be y and point is (0,y) and comaparing with (8,153)

• 153-y/8=11
• 153-y=88
• y=153-88
• y=65gallons

Answer 2

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 as time increases, the amount of water in the pool increases.

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 rate of change (which is also the slope of a linear function).

Choose 2 ordered pairs from the table:

$\textsf{let}\:(x_1,y_1)=(8, 153)$

$\textsf{let}\:(x_2,y_2)=(12,197)$

Input these into the slope formula:

$\textsf{slope}\:(m)=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{197-153}{12-8}=\dfrac{44}{4}=11$

Therefore, the rate at which the water in the pool is increasing is:

11 gallons per minute

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:

$y-y_1=m(x-x_1)$

$\implies y-153=11(x-8)$

$\implies y-153=11x-88$

$\implies y=11x-88+153$

$\implies y=11x+65$

When Ann had added no water, x = 0.  Therefore,

$y=11(0)+65$

$y=65$

So there was 65 gallons of water in the pool before Ann starting adding water.