Answer:
3271
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://img.qammunity.org/2021/formulas/mathematics/high-school/d6doln7pvb0bftbciu16v2b7m7ccpcifk5.png)
This means that the growth of the population for every 2 hours is:
![(p(2))/(p_0)=(1200)/(1000)=1.2](https://img.qammunity.org/2021/formulas/mathematics/high-school/b0qz7lxq6t8geyraola3jptpemopb7w0ua.png)
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://img.qammunity.org/2021/formulas/mathematics/high-school/7y9ibpvpimwjad97sbk0f5uclmh3jzh3sd.png)
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=(13)/(2)=6.5](https://img.qammunity.org/2021/formulas/mathematics/high-school/pyzodbdeieo4uqij08n8z3mu1129lbjakn.png)
Therefore, substituting 6.5 in the expression, we find the population after 13 hours:
![p_(13)=(1000)1.2^(6.5)=3271](https://img.qammunity.org/2021/formulas/mathematics/high-school/u0wwkn2g4zkfvp73ovk0px3iknf0r8djv5.png)