Titus works at a hotel. Part of his job is to keep the complimentary pitcher of water at least half full and always with ice. When he starts his shift, the water level shows 8 gallons, or 128 cups of water. As the shift progresses, he records the level of the water every 10 minutes. After 2 hours, he uses a regression calculator to compute an equation for the decrease in water. His equation is [tex] W \approx -0.414 t + 129.549 [/tex], where [tex] t [/tex] is the number of minutes and [tex] W [/tex] is the level of water.

According to the equation, after about how many minutes would the water level be less than or equal to 64 cups?

A. 150 minutes
B. 160 minutes
C. 170 minutes
D. 180 minutes



Answer :

To determine after how many minutes the water level would be less than or equal to 64 cups, we start with the given equation that models the decrease in water level over time:

[tex]\[ W \approx -0.414 t + 129.549 \][/tex]

Here, [tex]\( W \)[/tex] is the water level in cups, and [tex]\( t \)[/tex] is the number of minutes that have passed. We need to find the value of [tex]\( t \)[/tex] when [tex]\( W \leq 64 \)[/tex] cups.

Let's set [tex]\( W \)[/tex] to 64 and solve for [tex]\( t \)[/tex]:

[tex]\[ 64 = -0.414 t + 129.549 \][/tex]

To isolate [tex]\( t \)[/tex] on one side, we rearrange the equation as follows:

1. Subtract 64 from both sides:

[tex]\[ 129.549 - 64 = -0.414 t \][/tex]

2. Simplify the left side:

[tex]\[ 65.549 = -0.414 t \][/tex]

3. Divide both sides by -0.414 to solve for [tex]\( t \)[/tex]:

[tex]\[ t = \frac{65.549}{0.414} \][/tex]

By performing the division, we find:

[tex]\[ t \approx 158.33 \][/tex]

Now, we compare the calculated [tex]\( t \)[/tex] to the provided options: 150, 160, 170, and 180 minutes. The calculated time of approximately 158.33 minutes is closest to 160 minutes.

Thus, according to the equation, the water level would be less than or equal to 64 cups after about 160 minutes.