The function that models the spread of measles is given by:

$P(t) = \frac{200}{1 + 199(1.2)^{-t}}$

where $P(t)$ represents the number of students infected at day $t$, and $t$ is the number of days since the outbreak.

To find the average rate of infection over the first 20 days, we use the following formula for the average rate of change of a function over an interval $[a, b]$:

$\text{Average rate of infection} = \frac{P(b) - P(a)}{b - a}$

In this case:

• $a = 0$ (starting at day 0),
• $b = 20$ (ending at day 20),
• $P(0)$ and $P(20)$ are the values of the function at these points.

Step 1: Find $P(0)$

$P(0) = \frac{200}{1 + 199(1.2)^{0}} = \frac{200}{1 + 199} = \frac{200}{200} = 1$

Step 2: Find $P(20)$

$P(20) = \frac{200}{1 + 199(1.2)^{-20}}$

I'll calculate $P(20)$ now.After calculating, we find that:

$P(20) \approx 32.31$

Step 3: Calculate the average rate of infection

Now, using the formula for the average rate of change:

$\text{Average rate of infection} = \frac{P(20) - P(0)}{20 - 0} = \frac{32.31 - 1}{20} = \frac{31.31}{20} \approx 1.57$

Thus, the average rate of infection over the first 20 days is approximately 1.57 students per day.

Would you like more details or explanations on any of the steps?

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Here are 5 related questions to deepen your understanding:

1. How would the rate of infection change if the spread of measles is faster (e.g., if 1.2 is replaced by a larger number)?
2. What is the total number of students that can be infected as $t$ approaches infinity?
3. How would you find the instantaneous rate of infection at a particular day, like day 10?
4. How does the function $P(t)$ behave as $t \to 0$ and as $t \to \infty$?
5. Can we determine at which day the infection rate is the fastest based on this model?

Tip: The average rate of change gives a general trend over an interval, but the instantaneous rate of change (using derivatives) provides more specific information at any particular point in time.