Question

The level of water in a draining pool can be modeled with the function d(t)= 12 (.72) ^ t, where d is the depth of water, measured in feet, and t is the amount of time it’s been draining, measure in days.

Calculate the average rate of change for 0 greater or equal to t and 4 is greater or equal to t.

Answer 1

`\approx` -2.195

Explanation:

Given the function:

`d(t)= 12 (.72) ^ t`

`d` is the depth of water in feet and

`t` is the amount of time

To find:

The average rate of change for
`t` in the interval [0, 4].

Solution:

The required rate of change in the time interval [0, 4] can be represented as:

Change in the function over the interval [0, 4] and the change in the interval.

OR

`\Rightarrow (d(4)-d(0))/(4-0)\\\Rightarrow (12 (.72) ^ 4-12 (.72) ^ 0)/(4-0)\\\Rightarrow (3.22-12 * 1)/(4)\\\Rightarrow (3.22-12)/(4)\\\Rightarrow (-8.78)/(4)\\\Rightarrow \bold{-2.195}`