$\begin{gathered} P(t)=\frac{K}{1+Ae^{-kt}};A=\frac{K-P_{0_{}}}{P_0}_{} \\ \text{From the question} \\ P_0=\text{ Initial Plants=27} \\ K=\text{Carrying capacity =140} \end{gathered}$

Workings

Step 1: We would need to get the value of A using the carrying capacity and initial plants that started growing in the yard.

This gives;

$\begin{gathered} A=\frac{140-27}{27} \\ A=\frac{113}{27} \end{gathered}$

Step 2: Substitute the value of A into the formula.

$P(t)=\frac{140}{1+\frac{113}{27}e^{-kt}}$

Step 3: Find the value of the constant k

Kindly recall that we are told that the plants increase by 80% after each month. Therefore, after one month we would have;

$\begin{gathered} P(1)=27+(\frac{80}{100}\times27) \\ P(1)=\frac{243}{5} \end{gathered}$

We can then have that after t= 1month

$\begin{gathered} \frac{140}{1+\frac{113}{27}e^{-k\times1}}=\frac{243}{5} \\ Flip\text{ the equation} \\ \frac{1+\frac{113}{27}e^{-k}}{140}=\frac{5}{243} \\ 243(1+\frac{113}{27}e^{-k})=700 \\ 243+1017e^{-k}=700 \\ 1017e^{-k}=700-243 \\ 1017e^{-k}=457 \\ e^{-k}=\frac{457}{1017} \\ -k=\ln (\frac{457}{1017}) \end{gathered}$

Step 4: Substitute -k back into the initial formula.

$\begin{gathered} P(t)=\frac{140}{1+\frac{113}{27}e^{\ln (\frac{457}{1017})t}} \\ =\frac{140}{1+\frac{113}{27}(e^{\ln (\frac{457}{1017})})^t} \\ P(t)=\frac{140}{1+\frac{113}{27}(\frac{457}{1017}^{})^t} \\ \end{gathered}$

The above model is can be used to find the population at any time in the future.

Therefore after 5 months, we can estimate the model to be;

$\begin{gathered} P(5)=\frac{140}{1+\frac{113}{27}(\frac{457}{1017}^{})^5} \\ P(5)=\frac{140}{1.07668} \\ P(5)=130.029\approx130 \end{gathered}$

Answer: The estimated number of plants after 5 months is 130 plants.

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