Answer:
3271
Step-by-step explanation:
In this problem, we have:
is the population of bacteria at the beginning, at time t = 0
After 2 hours, we have a number of bacteria equal to
![p(2)=1200](https://tex.z-dn.net/?f=p%282%29%3D1200)
This means that the growth of the population for every 2 hours is:
![\frac{p(2)}{p_0}=\frac{1200}{1000}=1.2](https://tex.z-dn.net/?f=%5Cfrac%7Bp%282%29%7D%7Bp_0%7D%3D%5Cfrac%7B1200%7D%7B1000%7D%3D1.2)
This means that we can write an expression for the population after n pairs of hours as follows:
![p(n)=p_0 1.2^n](https://tex.z-dn.net/?f=p%28n%29%3Dp_0%201.2%5En)
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:
![n=\frac{13}{2}=6.5](https://tex.z-dn.net/?f=n%3D%5Cfrac%7B13%7D%7B2%7D%3D6.5)
Therefore, substituting 6.5 in the expression, we find the population after 13 hours:
![p_{13}=(1000)1.2^{6.5}=3271](https://tex.z-dn.net/?f=p_%7B13%7D%3D%281000%291.2%5E%7B6.5%7D%3D3271)