Question

the function $n(t)=\frac{70,000}{1 + 30e^{-1.5t}}$ describes the number of people, $n(t)$, who become ill with a virus $t$ weeks after its initial outbreak in a town with 70,000 inhabitants. the horizontal asymptote in the graph indicates that there is a limit to the epidemics growth. complete parts (a) through (c) below.
a. how many people became ill with the virus when the epidemic began? (when the epidemic began, $t = 0$.)
when the epidemic began, approximately 2258 people were ill with the virus. (round to the nearest person as needed.)
b. how many people were ill by the end of the second week?
by the end of the second week, approximately $\square$ people were ill with the virus. (round to the nearest person as needed.)

Explanation:

Step1: Substitute \(t = 0\) into the function

When \(t = 0\), \(N(0)=\frac{70000}{1 + 30e^{-1\times0}}\).
Since \(e^{0}=1\), then \(N(0)=\frac{70000}{1 + 30\times1}=\frac{70000}{31}\approx2258\).

Step2: Substitute \(t = 2\) into the function

When \(t = 2\), \(N(2)=\frac{70000}{1+30e^{-1\times2}}\).
First, calculate \(e^{- 2}\approx0.1353\).
Then \(30e^{-2}\approx30\times0.1353 = 4.059\).
So \(N(2)=\frac{70000}{1 + 4.059}=\frac{70000}{5.059}\approx13837\).

Answer:

a. \(2258\) people
b. \(13837\) people