Answer :
Answer:
1560
Step-by-step explanation:
The rate of Increase of the population (P) of the bacteria is given as:
[tex]\frac{dP}{dt} =4.058e^{1.6t}[/tex]
[tex]dP =4.058e^{1.6t}}dt\\Taking\: Integrals\\\int dP =4.058 \int e^{1.6t}dt\\P(t)=\frac{4.058}{1.6} (e^{1.6t} +K)\\P(t)=2.53625 (e^{1.6t} +K)[/tex]
Where k is a constant of Integration.
At t=0, P(t)=36
[tex]36=2.53625 (e^{1.6*0} +K)\\36=2.53625 (e^{0} +K)\\36=2.53625 (1 +K)\\36=2.53625 +2.53625K\\K=13.19[/tex]
Therefore:
[tex]P(t)=2.53625 (e^{1.6t} +13.19)\\At \: t=4\\P(4)=2.53625 (e^{1.6*4} +13.19)\\=1559.88\\P(4)=1560[/tex]