As Ryan drained his pool this fall he collected the following data. Time is the independent variable.TimeHeight of water(hours) (cm)0.7 7211.5 6492.0 6313.5 4964.1 4695.0 3616.7 2627.2 2447.9 1638.7 100(b) What is the slope of the line?_____cm/hours *(c) What is the height intercept?(d) At what time would you expect the pool to be empty?(e) What would you expect the height of the water to be after 5.5 hours?

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JoanaQ281407go JoanaQ281407go · Answer 1

ANSWER:

(b) -77.6 cm/hours

(c) 775 cms

(d) 10 hours

(e) 348 cms

STEP-BY-STEP EXPLANATION:

(b)

We calculate the slope of the line using the following formula:

[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]

We use the two extreme points, like this:

[tex]m=\frac{100-721}{8.7-0.7}=\frac{-621}{8}=-77.6\text{ cm/hours}[/tex]

(c)

We calculate the y-intercept, which would be the height intercept with the slope and the highest point known and closest to time 0, like this:

[tex]\begin{gathered} 721=-77.6\cdot0.7+b \\ \\ b=721+54.32 \\ \\ b\approx775 \end{gathered}[/tex]

(d)

The expected time that the pool is empty is when the height is equal to 0, therefore:

[tex]\begin{gathered} 0=-77.6\cdot \:x+775 \\ \\ x=\frac{-775}{-77.6} \\ \\ x\approx10\text{ hours} \end{gathered}[/tex]

(e)

Now, to determine the height at 5.5 hours, we perform the following equation obtained:

[tex]\begin{gathered} h=-77.6\cdot(5.5)+775 \\ \\ h=-426.8+775 \\ \\ h\approx348\text{ cms} \end{gathered}[/tex]