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Ilia_Sergeevich [38]
1 year ago
12

Drag each situation to show whether it can be modeled using a linear or an exponential function. If the situation cannot be mode

led with either function type, drag it into the container labeled Neither.

Mathematics
1 answer:
lawyer [7]1 year ago
3 0

Answer:

Step-by-step explanation:

A). An investment account earns 2.8% simple interest.

    Since investment amount increases every year linearly, therefore, the modeled situation represents a LINEAR FUNCTION.

B). The price of a stock varies by 2.8% each week.

    Since price of the stock may increase or decrease every week, therefore, this situation can't be modeled by any function.

Therefore, the answer is NEITHER.

C). An investment account earns 2.8% compound interest, compounded monthly.

Formula to get the value of the final amount in the account is,

Final value = Initial value × (1+\frac{.028}{100})^t

Here 't' = Duration of investment

It's an EXPONENTIAL FUNCTION.

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A random sample of 20 individuals who graduated from college five years ago were asked to report the total amount of debt (in $)
Gekata [30.6K]

Answer:

a. As college debt increases current investment decreases.

b. Y= 68778.2406 - 1.9112X

Every time the college debt increases one dollar, the estimated mean of the current investments decreases 1.9112 dollars.

c. There is a significant linear relationship between college debt and current investment because the P-value is less than 0.1.

d. Y= $59222.2406

e. R²= 0.9818

Step-by-step explanation:

Hello!

You have the information on a random sample of 20 individuals who graduated from college five years ago. The variables of interest are:

Y: Current investment of an individual that graduated from college 5 years ago.

X: Total debt of an individual when he graduated from college 5 years ago.

a)

To see the relationship between the information about the debt and the investment is it best to make a scatterplot with the sample information.

As you can see in the scatterplot (attachment) there is a negative relationship between the current investment and the debt after college, this means that the greater the debt these individuals had, the less they are currently investing.

The statement that best describes it is: As college debt increases current investment decreases.

b)

The population regression equation is Y= α + βX +Ei

To develope the regression equation you have to estimate alpha and beta:

a= Y[bar] -bX[bar]

a= 44248.55 - (-1.91)*12829.70

a= 68778.2406

b= \frac{sumXY-\frac{(sumX)(sumY)}{n} }{sumX^2-\frac{(sumX)^2}{n} }

b=\frac{9014653088-\frac{(256594)(884971)}{20} }{4515520748-\frac{(256594)^2}{20} }

b= -1.9112

∑X= 256594

∑X²= 4515520748

∑Y= 884971

∑Y²= 43710429303

∑XY= 9014653088

n= 20

Means:

Y[bar]= ∑Y/n= 884971/20= 44248.55

X[bar]= ∑X/n= 256594/20= 12829.70

The estimated regression equation is:

Y= 68778.2406 - 1.9112X

Every time the college debt increases one dollar, the estimated mean of the current investments decreases 1.9112 dollars.

c)

The hypotheses to test if there is a linear regression between the two variables are two tailed:

H₀: β = 0

H₁: β ≠ 0

α: 0.01

To make this test you can use either a Student t or the Snedecor's F (ANOVA)

Using t=<u>  b - β  </u>=<u>  -1.91 - 0  </u>= -31.83

                 Sb         0.06

The critical region and the p-value for this test are two tailed.

The p-value is: 0.0001

The p-value is less than the level of signification, the decision is to reject the null hypothesis.

Using the

F= \frac{MSTr}{MSEr}= \frac{4472537017.96}{4400485.72} =1016.37

The rejection region using the ANOVA is one-tailed to the right, and so is the p-value.

The p-value is: 0.0001

Using this approach, the decision is also to reject the null hypothesis.

The conclusion is that at a 1% significance level, there is a linear regression between the current investment and the college debt.

The correct statement is:

There is a significant linear relationship between college debt and current investment because the P-value is less than 0.1.

d)

To predict what value will take Y to a given value of X you have to replace it in the estimated regression equation.

Y/X=$5000

Y= 68778.2406 - 1.9112*5000

Y= $59222.2406

The current investment of an individual that had a $5000 college debt is $59222.2406.

e)

To estimate the proportion of variation of the dependent variable that is explained/ given by the independent variable you have to calculate the coefficient of determination R².

R^2= \frac{b^2[sumX^2-\frac{(sumX)^2}{n} ]}{sumY^2-\frac{(sumY)^2}{n} }

R^2= \frac{-1.9112^2[4515520748-\frac{(256594)^2}{20} ]}{43710429303-\frac{(884971)^2}{20} }

R²= 0.9818

This means that 98.18% of the variability of the current investments are explained by the college debt at graduation under the estimated regression model: Y= 68778.2406 - 1.9112X

I hope it helps!

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Answer:

A  AND D

Step-by-step explanation:

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If you think about how it works when cups are stacked. you will begin with the height of the cup plus the height of the lip you would then add just the heigt of the lips until you get to the height needed. so your answer should be 19 cups
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Paul bakes loaves of bread and bread rolls in the ratio of 2:5. If he bakes 750 bread rolls, how many loaves will he bake? A rec
Zanzabum
Paul bakes 300 bread loaves. I don’t know the question you’re asking for the second question.
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Read 2 more answers
It takes one week for a crew of workers to pave 3/5 kilometer of a road. At that rate, how long will it take to pave 1 kilometer
kondaur [170]

1 kilometer =5/3 week

Step-by-step explanation:

Given that the crew paves 3/5 kilometer of a road in 1 week then this can be written as;

3/5 k = 1 week

For 1 kilometer of a road,

3/5k = 1 week

1k=?

perform cross product as;/multiplication equation

1*1 / 3/5

1*5/3

5/3, 1 2/3 weeks

A division equation will be'

3/5 kilometers = 1 week

Divide both sides by 3/5

1 kilometer =5/3 week

Learn More

Multiplication /Division :brainly.com/question/11892889

Keywords: kilometers, equation, multiplication.division

#LearnwithBrainly

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2 years ago
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