In a city with a population of 1,200,000, there are currently 1000 cases of infection with the HIV virus. If the spread of the disease is projected by the formula $$P = \frac { 1,200,000 } { 1 + ( 1200 - 1 ) e ^ { - 0.4 t } }$$
how many people will be infected in 8 years?
A) $\quad 855,800$ people
B) $\quad 24,600$ people
C) $\quad 24,100$ people
D) $\quad 9,100$ people
E) 211,100 people

Question

Quizplus · Accepted Answer

We can use the given formula to calculate the projected number of HIV cases after 8 years.

First, we need to find the value of t when there are currently 1000 cases.

1000 = (1,200,000)/(1 + (1200 - 1)e^(-0.4t))

Simplifying this equation, we get:

1 + (1199)e^(-0.4t) = 1200

(1199)e^(-0.4t) = 1199

e^(-0.4t) = 1

-0.4t = 0

t = 0

So, we can plug in t = 8 (8 years from now) in the formula and solve for P:

P = (1,200,000)/(1 + (1200 - 1)e^(-0.4(8)))

P = (1,200,000)/(1 + 1199e^(-3.2))

P ≈ 24,097

Therefore, the projected number of HIV cases in 8 years is 24,097, which corresponds to choice C.

In a City with a Population of 1,200,000, There Are P=1,200,0001+(1200−1)e−0.4tP = \frac { 1,200,000 } { 1 + ( 1200 - 1 ) e ^ { - 0.4 t } }P=1+(1200−1)e−0.4t1,200,000​

In a City with a Population of 1,200,000, There Are $P = \frac { 1,200,000 } { 1 + ( 1200 - 1 ) e ^ { - 0.4 t } }$