Answer;
[tex]\begin{gathered} a)\text{ 195 bacteria} \\ b)\text{ 3,291,055,916 bacteria} \end{gathered}[/tex]Explanation;
a) We want to get the initial population of the bacteria
We start by writing a formula that links the initial bacteria population to a later bacteria population after time t
[tex]A(t)=I(1+r)^t[/tex]where A(t) is the bacteria population at time t
I is the initial bacteria population
r is the rate of increase in population
t is time
Now, let us find r
At t = 10; we know that A(t) = 2I
Thus, we have it that;
[tex]\begin{gathered} 2I=I(1+r)^{10} \\ (1+r)^{10}\text{ = 2} \\ 1+r\text{ = 1.0718} \\ r\text{ = 1.0718-1} \\ r\text{ = 0.0718} \end{gathered}[/tex]Now, let us find I, since we have r. But we have to make use of t= 80 and A(t) = 50,000
Thus, we have;
[tex]\begin{gathered} 50,000=I(1+0.0718)^{80} \\ I\text{ = }\frac{50,000}{(1+0.0718)^{80}} \\ I\text{ = 195} \end{gathered}[/tex]The initial population is 195 bacteria
b) For after 4 hours, we have to convert to minutes
We know that there are 60 minutes in an hour
So, in 4 hours, we have 4 * 60 = 240 minutes
Now, we proceed to use the formula above with I = 195 and t = 240
We have that as;
[tex]\begin{gathered} A(240)=195(1+0.0718)^{240} \\ A(240)\text{ = 3,291,055,916 bacteria} \end{gathered}[/tex]