Answer
$40.71\%$
Work Step by Step
The population of the bacteria after one hour is:
$$p_{1}=p(0)\cdot 1.05$$ where $p(0)$ is the initial population.
The population of the bacteria after $2$ hours is:
$$p_{2}=p(1)\cdot 1.05=p(0)\cdot 1.05 \cdot 1.05$$
Following the pattern, the population of the bacteria after $t$ hour is:
$$p_{t}=p(0)\cdot (1.05)^{t}$$
The percent of increase is:
$$\frac{p(7)-p(0)}{p(0)}\cdot 100\%=\frac{p(0)\cdot (1.05)^{7}-p(0)}{p(0)}\cdot 100\%=((1.05)^{7}-1)\cdot 100\%=40.71\%$$
