Answer:
3271
Step-by-step explanation:
In this problem, we have:
[tex]p_0 = 1000[/tex] is the population of bacteria at the beginning, at time t = 0
After 2 hours, we have a number of bacteria equal to
[tex]p(2)=1200[/tex]
This means that the growth of the population for every 2 hours is:
[tex]\frac{p(2)}{p_0}=\frac{1200}{1000}=1.2[/tex]
This means that we can write an expression for the population after n pairs of hours as follows:
[tex]p(n)=p_0 1.2^n[/tex]
where
n is the number of "pairs of hours" passed from t = 0
In this problem, we want to find the population of bacteria after 13 hours, which contains a number of pairs of hours equal to:
[tex]n=\frac{13}{2}=6.5[/tex]
Therefore, substituting 6.5 in the expression, we find the population after 13 hours:
[tex]p_{13}=(1000)1.2^{6.5}=3271[/tex]
