a.
To find the initial number of people infected, let's calculate the value of f(t) for t = 0:
[tex]f(0)=\frac{102000}{1+5200e^0}=\frac{102000}{1+5200}=\frac{102000}{5201}=19.61[/tex]
Rounding to the nearest whole number, the initial value is 20 people.
b.
Using t = 4, let's calculate the value of f(t):
[tex]f(4)=\frac{102000}{1+5200e^{-4}}=\frac{102000}{1+5200\cdot0.0183156}=\frac{102000}{96.24112}=1.059.84[/tex]
Therefore the number of infected people is 1060.
c.
When t tends to infinity, the value of 5200e^-t will tend to zero, therefore we have:
[tex]\lim_{t\to\infty}f(t)=\frac{102000}{1+0}=102000[/tex]
Therefore the limiting size is 102000 people.
