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3 years ago

100 POINTS. HELP ASAP !!! PLEASE PLEASE

Mathematics

2 answers:

defon3 years ago

6 0

Answer:

I dont know sorry

Step-by-step explanation:

natali 33 [55]3 years ago

3 0

Answer:

task 1 answer is 14

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I need help I don't understand and whats the answers for all of them

Furkat [3]

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6 0

3 years ago

Find the surface area of this cube. Be sure to include the correct unit in your answer.

anygoal [31]

Answer:

294ft

Step-by-step explanation:

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6 0

3 years ago

2. Suppose 27 blackberry plants started growing in a yard. Absent constraint, the blackberry plants will spread by 80% a month.

Marta_Voda [28]

Explanation

The question indicates we should use a logistic model to estimate the number of plants after 5 months.

This can be done using the formula below;

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

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;