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sesenic
1 month ago
12

Disney held a breakfast for parents and their children to eat with Mickey and Minnie Mouse (and the rest of the gang, too!) Adul

t tickets cost $17.95 and children’s tickets cost $12.95. Disney made $7355 from ticket sales from a total of 500 people that attended. How many adults were at the breakfast?
Mathematics
1 answer:
Svet_ta [12.7K]1 month ago
3 0
I trust this information is useful to you.



Let a represent the number of adults.



Then, the number of children will be 500-a.



From here, we set up the equation: 17.95×a + 12.95×(500-a) = 7355.



This simplifies to 17.95a + 6475 - 12.95a = 7355.


Which results in 5a = 880.


Thus, a = 176.
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V = x³ - 6x²y + 12xy² - 8y³

V = (x - 2y)³

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   = (x² - 4xy + 4y²)(x - 2y) ( multiply the terms from the first group with those in the second )

   = x³ - 4x²y + 4xy² - 2x²y + 8xy² - 8y³ ( combine similar terms )

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18 days ago
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Use the Financial database from "Excel Databases.xls" on Blackboard. Use Total Revenues, Total Assets, Return on Equity, Earning
zzz [12365]

Answer:

Refer to the explanation provided

Detailed explanation:

Variable Name R2 value Adj R2 value

Earnings Per Share 0.086535 0.077214

Earnings Per Share,

Dividends Per Share 0.103976 0.085501

Earnings Per Share,

Dividends Per Share,

Average Yield 0.180303 0.154687

Earnings Per Share,

Dividends Per Share,

Average Yield,

Return on Equity 0.183215 0.148824

Earnings Per Share,

Dividends Per Share,

Average Yield,

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Total Assets 0.184669 0.141301

Earnings Per Share,

Dividends Per Share,

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Total Revenues 0.187297 0.134865

4 0
1 month ago
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!

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lawyer [12517]

Answer:

80.7 because you multiply

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16 days ago
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