Question

kevin is preparing his swimming pool for the winter. he connects a pump to the pool, and the water in the pool slowly drains out. this situation can be modeled as a linear relationship. chart: x-axis time draining (minutes) from 0 to 100, y-axis water in the pool (gallons) from 0 to 10,000, line starting at (0,8000) and decreasing

Explanation:

Step1: Identify two points on the line

From the graph, when \( x = 0 \) (time = 0 minutes), \( y = 8000 \) gallons. When \( x = 30 \) minutes, \( y = 7000 \) gallons. So the two points are \((0, 8000)\) and \((30, 7000)\).

Step2: Calculate the slope (\( m \))

The slope formula is \( m=\frac{y_2 - y_1}{x_2 - x_1} \). Substituting the points: \( m=\frac{7000 - 8000}{30 - 0}=\frac{- 1000}{30}=-\frac{100}{3}\approx - 33.33 \) gallons per minute.

Step3: Determine the equation of the line

Using the slope - intercept form \( y=mx + b \), where \( b \) is the y - intercept. From the point \((0, 8000)\), we know \( b = 8000 \). So the equation is \( y=-\frac{100}{3}x + 8000 \).

(If we want to find when the pool is empty, set \( y = 0 \):
\( 0=-\frac{100}{3}x + 8000 \)
\(\frac{100}{3}x=8000\)
\(x = 8000\times\frac{3}{100}=240\) minutes. But since the problem is about modeling the linear relationship, the key steps are to find the slope and the equation of the line as above.)

Answer:

The linear equation modeling the water in the pool over time is \( y =-\frac{100}{3}x + 8000 \) (or if finding the time to empty, the time is 240 minutes).