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3 years ago

HELP PLEASE

Mathematics

2 answers:

kifflom [539]3 years ago

5 0

Answer:

2nd choice

Step-by-step explanation:

to find the inital peanuts, we multiply 32 ounces by 3/10 or 0.3. then we add x which are peanuts as well. that makes 0.3 x 32 + x. this is our total peanuts. to find the percantage, we find the whole mixture which is now 32 + x, divide peanuts by the whole mixture and multiply by 100.

Ronch [10]3 years ago

3 0

Answer:

b on egde

Step-by-step explanation:

i took the test

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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;

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

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

8 0

1 year ago

Each morning there is a probability of 3/4 that Julie will catch her bus. Over 2 consecutive days, the probability that she will

krek1111 [17]

The probability is 3/16.

The probability that she catches the bus on one of the days is 3/4; the probability she does not on the other day is 1/4. Together we have
3/4(1/4) = 3/16

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jek_recluse [69]

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3 years ago

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What is 0.15% of 250? in the simplest form

grin007 [14]

Answer:

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