Check below, please.
1) Since the function is modeled like this:
[tex]N(t)=905e^{0.35t}[/tex]a) Let's plug into that the given data and find the relative growth
[tex]\begin{gathered} N(1)=1284.2 \\ N(2)=1822.4 \\ \frac{1822.4}{1284.2}=1.4190-1=0.4190\Rightarrow41.9\% \end{gathered}[/tex]Plugging into that t=1 and t=2 we could find two values, and dividing we found that there was a growth of approximately 41.9% in the Bacterium population.
b) The initial population of the culture is found when we plug t=0
[tex]N(0)=905\left(e\right)^{0.35t}\Rightarrow N(0)=905[/tex]c) Let's plug into t=5 and find the number:
[tex]\begin{gathered} N(5)=905e^{0.35(5)} \\ N(5)=5207.91542\approx5208 \end{gathered}[/tex]