Question

1. an inflatable pool is leaking at a constant rate of 25 fluid ounces every 4 seconds. which table shows the relationship between the time in seconds, x, and the water, y?

a.

time, x	water, y
12	1.92
16	2.56
20	3.2

b.

time, x	water, y
12	300
16	400
20	500

c.

time, x	water, y
12	50
16	75
20	100

d.

time, x	water, y
12	75
16	100
20	125

Explanation:

Step1: Find the rate of change

The pool leaks 25 fluid ounces every 4 seconds. So the rate of change (slope) is $\frac{25}{4} = 6.25$ fluid ounces per second? Wait, no—wait, actually, when time increases, the water should decrease? Wait, no, the problem says "leaking", so as time \( x \) (seconds) increases, the water \( y \) (fluid ounces) should decrease? Wait, but the tables given: let's check the rate. Wait, maybe I misread. Wait, the problem says "the relationship between the time in seconds, \( x \), and the water, \( y \)". Wait, maybe the rate is 25 ounces every 4 seconds, so the amount of water lost per second is \( \frac{25}{4} = 6.25 \) ounces per second. But let's check the tables. Wait, maybe the tables are showing the amount of water remaining? Wait, no, let's check the options. Wait, maybe I made a mistake. Wait, let's calculate the rate for each table.

For a linear relationship, the rate (slope) should be constant. Let's check the change in \( y \) over change in \( x \).

First, let's find the correct rate. The leak rate is 25 ounces every 4 seconds, so the rate of change of \( y \) with respect to \( x \) is \( \frac{25}{4} = 6.25 \) ounces per second? Wait, no—if it's leaking, then as \( x \) increases by 4, \( y \) decreases by 25? Wait, but the tables have \( y \) increasing. Wait, maybe the tables are showing the amount of water leaked, not remaining? Wait, the problem says "the water, \( y \)"—maybe \( y \) is the amount leaked? So as time increases, the amount leaked increases. So the rate is 25 ounces every 4 seconds, so the slope is \( \frac{25}{4} = 6.25 \) ounces per second. Let's check each table:

Table A:
\( x \) from 8 to 12: change in \( x = 4 \), change in \( y = 1.92 - 1.28 = 0.64 \). Rate: \( 0.64 / 4 = 0.16 \). Not 6.25.

Table B:
\( x \) from 8 to 12: change in \( x = 4 \), change in \( y = 300 - 200 = 100 \). Rate: \( 100 / 4 = 25 \). Wait, 25 per 4 seconds? Wait, 100 per 4 seconds? No, 300 - 200 = 100 when \( x \) increases by 4 (from 8 to 12). So rate is 100 / 4 = 25 per second? No, that's too fast. Wait, no, 25 ounces every 4 seconds, so per second it's 6.25. Wait, maybe I messed up.

Wait, let's calculate the slope for each table:

Table D:
\( x \): 8, 12, 16, 20.
\( y \): 50, 75, 100, 125.

Change in \( x \) from 8 to 12: 4 seconds. Change in \( y \): 75 - 50 = 25. So rate: 25 / 4 = 6.25? Wait, 25 over 4 seconds. Wait, 25 ounces every 4 seconds. So when \( x \) increases by 4, \( y \) increases by 25. That matches the leak rate. Let's check:

From 8 to 12: \( x \) +4, \( y \) +25 (50 to 75: 25). From 12 to 16: \( x \) +4, \( y \) +25 (75 to 100: 25). From 16 to 20: \( x \) +4, \( y \) +25 (100 to 125: 25). Perfect! So the slope is 25/4 = 6.25, which is consistent. Let's check other tables:

Table B:
\( x \) 8 to 12: change in \( x = 4 \), change in \( y = 300 - 200 = 100 \). 100/4 = 25, which is too high (since the rate is 25 every 4 seconds, not 100 every 4 seconds).

Table C:
\( x \) 8 to 12: change in \( x = 4 \), change in \( y = 50 - 25 = 25 \)? Wait, no, \( x \) 8: \( y =25 \); \( x=12 \): \( y=50 \). Change in \( x =4 \), change in \( y=25 \). Then \( x=16 \): \( y=75 \) (change 25), \( x=20 \): \( y=100 \) (change 25). Wait, that's also 25 every 4 seconds. Wait, but Table D also has 25 every 4 seconds. Wait, what's the difference?

Wait, Table C: \( x=8 \), \( y=25 \); Table D: \( x=8 \), \( y=50 \). Let's calculate the initial amount? Wait, no, the problem says "the relationship"—maybe the rate is 25 ounces every 4 seconds, so the equation is \( y = \frac{25}{4}x \). Let's check for \( x=8 \): \( y = \frac{25}{4} \times 8 = 50 \). Ah! So that's the key. So the equation is \( y = \frac{25}{4}x \). Let's check:

For \( x=8 \): \( y = \frac{25}{4} \times 8 = 50 \). Which matches Table D (x=8, y=50).

For \( x=12 \): \( y = \frac{25}{4} \times 12 = 75 \). Which matches Table D (x=12, y=75).

For \( x=16 \): \( y = \frac{25}{4} \times 16 = 100 \). Which matches Table D (x=16, y=100).

For \( x=20 \): \( y = \frac{25}{4} \times 20 = 125 \). Which matches Table D (x=20, y=125).

Now check Table C: \( x=8 \), \( y=25 \). \( \frac{25}{4} \times 8 = 50 eq 25 \). So Table C is wrong.

Table B: \( x=8 \), \( y=200 \). \( \frac{25}{4} \times 8 = 50 eq 200 \). Wrong.

Table A: \( x=8 \), \( y=1.28 \). \( \frac{25}{4} \times 8 = 50 eq 1.28 \). Wrong.

So the correct table is D.

Answer:

D.

$$\begin{tabular}{|c|c|} \hline Time, \( x \) & Water, \( y \) \\ \hline 8 & 50 \\ \hline 12 & 75 \\ \hline 16 & 100 \\ \hline 20 & 125 \\ \hline \end{tabular}$$