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4 years ago

Identify the dependent variable in the statement: the quality of a public school teacher and the amount of student learning.

2 answers:

6 0

Dependent variables are something that depends on other factors. therefore, if the quality of the public school teacher is poor, it will affect the student's learning badly, and if the teacher's teaching is strong, it affects the student's learning greatly. The student's learning depends on how well the teacher teaches.

Alexxandr [17]4 years ago

4 0

The dependent variable is the amount of student learning.

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3y = -x-3 <br> 2y-14 = 4x<br> 4x-3-y = 0<br> x-12 = -3

geniusboy [140]

x=-3 y=0 probably wrong but it’s the root of at least one of these

5 0

4 years ago

#1: 36 in.=__________ft?

KatRina [158]

1. 3 foot
2. 4 yard
3. 1 yard
4. 4 foot
5. 6 foot
6. 108 inch
7. 7.5 foot
8. 66 inch
9. 216 inch
10. 36 foot
11. 3.75 foot
12. 30 yard

3 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;

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

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

8 0

1 year ago

5. The data below shows the high temperature and coffee sales for the day.

mr Goodwill [35]

The answer would be a

7 0

3 years ago

Solve the inequality for u<br>-5u+27 [less than or equal to] 2

baherus [9]

Answer:

5≤u

Step-by-step explanation:

Put the words and numbers into an equation:

-5u+27≤2

Put all like terms on one side:

27-2≤5u ---> Do you see what I did there? Now you don't have to work with negatives. :)

Simplify:

25≤5u

Solve:

5≤u

6 0

3 years ago