Question

the function $n(t)=\frac{50000}{1 + 20e^{-2.5t}}$ describes the number of people, $n(t)$, who become ill with a virus $t$ weeks after its initial outbreak in a town with 50000 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 2381 people were ill with the virus.
(round to the nearest person as needed.)
b. how many people were ill by the end of the fourth week?
by the end of the fourth week, approximately $square$ people were ill with the virus.
(round to the nearest person as needed.)

Explanation:

Step1: Substitute t = 4 into the function

We have $N(t)=\frac{50000}{1 + 20e^{-2.5t}}$, substituting $t = 4$ gives $N(4)=\frac{50000}{1+20e^{-2.5\times4}}$.

Step2: Calculate the value of the exponent

First, calculate $-2.5\times4=- 10$. Then $e^{-10}=\frac{1}{e^{10}}$. Since $e^{10}\approx22026.47$, $e^{-10}\approx\frac{1}{22026.47}\approx0.0000454$.

Step3: Calculate the denominator

Next, calculate $1 + 20e^{-10}=1+20\times0.0000454=1 + 0.000908 = 1.000908$.

Step4: Calculate N(4)

Finally, $N(4)=\frac{50000}{1.000908}\approx49955$.

Answer:

49955