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Pavel
2 months ago
8

Jenn will use 18 connecting cubes to make a model of a park. The model will be in the shape of a rectangle and will have a heigh

t of one cube. In how many different ways can Jenn make the model of the park?
Mathematics
1 answer:
Zina [12.3K]2 months ago
3 0
This question asks how many distinct pairs of integers multiply to give 18.

(Put another way: how many rectangular dimensions can Jenn make using 18 blocks with height 1.)

The factor pairs of 18 are:

1 x 18
2 x 9
3 x 6

Therefore, Jenn can construct the park model in 3 different ways.
You might be interested in
A random sample of 20 individuals who graduated from college five years ago were asked to report the total amount of debt (in $)
AnnZ [12381]

Response:

a. As student debt rises, current investment diminishes.

b. Y= 68778.2406 - 1.9112X

For each dollar increase in college debt, the average current investments decrease by 1.9112 dollars.

c. A substantial linear correlation exists between college debt and current investment as the P-value falls below 0.1.

d. Y= $59222.2406

e. R²= 0.9818

Step-by-step breakdown:

Hello!

Data has been gathered on a random sample of 20 individuals who completed their college education five years ago. The variables under consideration are:

Y: Current investment by an individual who graduated from college five years prior.

X: Total debt of an individual upon graduating five years ago.

a)

To explore the relationship between debt and investment, creating a scatterplot with the sample data is ideal.

The scatterplot demonstrates a negative correlation, indicating that as these individuals' debt increases, their current investments decrease.

Therefore, the statement that accurately describes this is: As college debt rises, current investment decreases.

b)

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

To develop this equation, estimates for alpha and beta are required:

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

Averages:

Y[bar]= ∑Y/n= 884971/20= 44248.55

X[bar]= ∑X/n= 256594/20= 12829.70

The estimated regression equation becomes:

Y= 68778.2406 - 1.9112X

For every dollar increase in college debt, the average current investments drop by 1.9112 dollars.

c)

To evaluate if there's a linear regression between these variables, the following null hypotheses are formulated:

H₀: β = 0

H₁: β ≠ 0

α: 0.01

Testing can be performed utilizing either a Student t-test or Snedecor's F (ANOVA)

Using t=  b - β  =  -1.91 - 0  = -31.83

                 Sb         0.06

The critical area and P-value for this test is two-tailed. The P-value equals: 0.0001

Since this P-value is underneath the significance level, we reject the null hypothesis.

In the case of ANOVA, the rejection area is also one-tailed to the right, corresponding to the P-value.

The P-value remains: 0.0001

Using this method, we similarly reject the null hypothesis.F= \frac{MSTr}{MSEr}= \frac{4472537017.96}{4400485.72} =1016.37

In conclusion, at a significance level of 1%, there exists a linear relationship linking current investment to college debt.

The accurate statement is:

There exists a significant linear association between college debt and current investment since the P-value is less than 0.1.

d)

To forecast the value of Y when X is set, it is essential to substitute X in the estimated regression equation.

Y/$5000

Y= 68778.2406 - 1.9112*5000

Y= $59222.2406

The anticipated investment for someone with a college debt of $5000 is $59222.2406.

e)

To determine the proportion of variation in the dependent variable that the independent variable accounts for, the coefficient of determination R² must be calculated.

R²= 0.9818

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} }

This indicates that 98.18% of the variability in current investments relates to college graduation debt within the projected regression model: Y= 68778.2406 - 1.9112X

I trust this is beneficial!

5 0
2 months ago
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