Step1 \(P(t) = \frac{14,250}{1+29e^(-0*62t)}\)

\(As t \Rightarrow \infty, e^{-0*62t} \Rightarrow 0\)

\(\therefore P(t) \Rightarrow 14,250\).

This is the carrying capacity of the bacten`al culture.

To determine the time it take to reach 75% of thr carlying capacity, we write \(P(t) = \frac{3}{4}(14,250).\)

Step 2

\( \Rightarrow \not{14,250}{1+29e^{(-0*62t)}} = \frac{3}{4}(\not 14,250)\)

\(\Rightarrow 1+29e^{-0*62t} = \frac{3}{4}\)

\(\Rightarrow e^{(-0*62t)} = \frac{1}{87}\)

\(\Rightarrow t=\frac{1}{0*62} In(87) = 7\cdot2\) Hense it takes the bacterial Popylation \(7\cdot2\) days to reach 75% of the calling capacity.

Step 3 The growth of any population is curtailed by the amount of resources available. This is better modeled by a logistic S curve rather than an exponential growth model which only works in ideal situations and not for modeling real life situations.