Questions: The total number of people, P, who have been infected with a contagious virus t weeks after the epidemic began is given by the following formula. P = 2660 / (1 + 6(0.5)^t) Complete parts a through d below. b. Complete the table. t P 0 380 1 665 2 1064 3 1520 6 2432 c. In the long run, how many total people will be infected? 2660 people d. Represent your answer to part c using limit notation by dragging the appropriate choices in the template. lim 2432 P(t) lim 1 -> 2660 lim infinity 2660 Drag the given expressions or values into the appropriate locations below to set up a limit which represents your answer to part c.

The total number of people, P, who have been infected with a contagious virus t weeks after the epidemic began is given by the following formula.

P = 2660 / (1 + 6(0.5)^t)

Complete parts a through d below.

b. Complete the table.

t P
0 380
1 665
2 1064
3 1520
6 2432

c. In the long run, how many total people will be infected?

2660 people

d. Represent your answer to part c using limit notation by dragging the appropriate choices in the template.

lim
2432
P(t)
lim 1 -> 2660
lim
infinity
2660

Drag the given expressions or values into the appropriate locations below to set up a limit which represents your answer to part c.

Transcript text: The total number of people, $P$, who have been infected with a contagious virus $t$ weeks after the epidemic began is given by the following formula. \[ P=\frac{2660}{1+6(0.5)^{t}} \] Complete parts a through d below. b. Complete the table. \begin{tabular}{|c|c|} \hline $\mathbf{t}$ & $\boldsymbol{P}$ \\ \hline 0 & 380 \\ \hline 1 & 665 \\ \hline 2 & 1064 \\ \hline 3 & 1520 \\ \hline 6 & 2432 \\ \hline \end{tabular} c. In the long run, how many total people will be infected? 2660 people d. Represent your answer to part c using limit notation by dragging the appropriate choices in the template. $\lim$ 2432 $\mathrm{P}(\mathrm{t})$ $\lim _{1 \rightarrow 2660}$ $\lim$ $\infty$ 2660 Drag the given expressions or values into the appropriate locations below to set up a limit which represents your answer to part c. $\square$ $\square$ $\square$

Solution

Solution Steps

To solve part b, we need to calculate the number of people infected, $ P $, at different weeks $ t $ using the given formula. For part c, we need to determine the long-term behavior of the function as $ t $ approaches infinity.

Solution Approach

For part b, use the given formula to calculate $ P $ for each $ t $ value in the table.
For part c, analyze the formula to determine the limit of $ P $ as $ t $ approaches infinity.

Step 1: Calculate $ P $ for Given $ t $ Values

Using the formula

\[ P = \frac{2660}{1 + 6(0.5)^{t}} \]

we calculate $ P $ for the specified values of $ t $:

For $ t = 0 $: \[ P(0) = \frac{2660}{1 + 6(0.5)^{0}} = \frac{2660}{1 + 6 \cdot 1} = \frac{2660}{7} \approx 380.00 \]
For $ t = 1 $: \[ P(1) = \frac{2660}{1 + 6(0.5)^{1}} = \frac{2660}{1 + 6 \cdot 0.5} = \frac{2660}{4} \approx 665.00 \]
For $ t = 2 $: \[ P(2) = \frac{2660}{1 + 6(0.5)^{2}} = \frac{2660}{1 + 6 \cdot 0.25} = \frac{2660}{2.5} \approx 1064.00 \]
For $ t = 3 $: \[ P(3) = \frac{2660}{1 + 6(0.5)^{3}} = \frac{2660}{1 + 6 \cdot 0.125} = \frac{2660}{1.75} \approx 1520.00 \]
For $ t = 6 $: \[ P(6) = \frac{2660}{1 + 6(0.5)^{6}} = \frac{2660}{1 + 6 \cdot 0.015625} = \frac{2660}{1.09375} \approx 2432.00 \]

Step 2: Determine the Long-Term Behavior

To find the long-term number of people infected as $ t $ approaches infinity, we analyze the limit:

\[ \lim_{t \to \infty} P(t) = \frac{2660}{1 + 6(0.5)^{t}} \]

As $ t $ approaches infinity, $ (0.5)^{t} $ approaches $ 0 $:

\[ \lim_{t \to \infty} P(t) = \frac{2660}{1 + 0} = 2660 \]