The given functions are
[tex]\begin{gathered} f(t)=20.52e^{0.0195t}\rightarrow(1) \\ g(t)=11.35e^{0.0111t}\rightarrow(2) \end{gathered}[/tex]Where t is the time in minutes
f(t) and g(t) are the numbers of bacterias
We need to find the difference in 5 hours
Change the time to minutes by multiplying 5 hours by 60, then substitute it in the functions to find the values, then subtract them
[tex]\begin{gathered} t=5\times60 \\ t=300 \end{gathered}[/tex][tex]\begin{gathered} f(300)=20.52e^{0.0195(300)} \\ f(300)=7125.25(2d\mathrm{}p) \end{gathered}[/tex][tex]\begin{gathered} g(300)=11.35e^{0.0111t} \\ g(300)=317.10(2d\mathrm{}p) \end{gathered}[/tex]Now, we will subtract them to find the difference
[tex]\begin{gathered} D=7125.25-317.10 \\ D=6808.15 \end{gathered}[/tex]Then the difference is about 6808