The average rate of change of a function $C(x)$ over an interval $[x_1, x_2]$ is given by the formula:

$\text{Average rate of change} = \frac{C(x_2) - C(x_1)}{x_2 - x_1}$

In this problem, the cost function is $C(x) = 20 + 50\sqrt{x}$, and the interval of students enrolled is from $x_1 = 940$ to $x_2 = 1080$.

Step 1: Calculate $C(x_1)$ and $C(x_2)$

First, we calculate $C(940)$ and $C(1080)$.

$C(940) = 20 + 50\sqrt{940}$ $C(1080) = 20 + 50\sqrt{1080}$

Step 2: Compute the average rate of change

Substitute these values into the formula for the average rate of change:

$\text{Average rate of change} = \frac{C(1080) - C(940)}{1080 - 940}$

Let me compute these values for you.### Step 1: Calculate $C(940)$ and $C(1080)$ $C(940) = 1552.97 \, \text{dollars}$ $C(1080) = 1663.17 \, \text{dollars}$

Step 2: Compute the average rate of change

The average rate of change is: $\text{Average rate of change} = \frac{1663.17 - 1552.97}{1080 - 940} = 0.79 \, \text{dollars per student}$

So, the average rate of change of the cost with respect to the number of students enrolled is approximately 0.79 dollars per student.

Would you like further details on this, or do you have any questions?

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Here are 5 related questions:

1. How do you interpret the average rate of change in this context?
2. What is the instantaneous rate of change of the cost function at a specific enrollment value?
3. How would the cost change if the number of students increased by 100?
4. How does the square root in the cost function influence the rate of change?
5. How would you model a similar cost function with a different fixed fee?

Tip: The average rate of change is similar to the slope of a secant line connecting two points on a curve!