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

Answer:

$\approx$ -2.195

Step-by-step 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

$\Rightarrow \dfrac{d(4)-d(0)}{4-0}\\\Rightarrow \dfrac{12 (.72) ^ 4-12 (.72) ^ 0}{4-0}\\\Rightarrow \dfrac{3.22-12 \times 1}{4}\\\Rightarrow \dfrac{3.22-12}{4}\\\Rightarrow \dfrac{-8.78}{4}\\\Rightarrow \bold{-2.195}$