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kirill
6 days ago
5

Convert 9/20 into a decimal using long division

Mathematics
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Zayed is helping his classmates get ready for their math test by making them identical packages of pencils and calculators. He h
Svet_ta [12734]
To find the maximum number of identical packs we see we have 72 pencils and 24 calculators.

This involves discovering the largest number that divides both 72 and 24 evenly,
which is known as the GCM or greatest common multiplier.

To determine the GCM, factor 72 into primes and group them:
72=2 times 2 times 2 times 3 times 3
24=2 times 2 times 2 times 3
Thus, the common grouping is 2 times 2 times 2 times 3, equating to 24.
Therefore, the maximum number of packs is 24.

For pencils:
72 divided by 24=3
Resulting in 3 pencils per pack.

For calculators:
24 divided by 24=1
So, 1 calculator per pack.

The outcome is 3 pencils and 1 calculator in each pack.
6 0
3 months ago
Read 2 more answers
Suppose that surface σ is parameterized by r(u,v)=⟨ucos(3v),usin(3v),v⟩, 0≤u≤7 and 0≤v≤2π3 and f(x,y,z)=x2+y2+z2. Set up the sur
tester [12383]
It appears you've gathered most of the relevant information, but a critical element is missing. The surface represented is parameterized by... and has a normal vector corresponding to this representation with a specified norm. Thus, the integral of... is...
6 0
2 months ago
Will works 7 hours a day, 6 days a week.
PIT_PIT [12445]

For one day:
= £9.20 × 7
= £64.40

For six days:
£64.40 × 6 = £386.40

After sharing with his mom:
£386.40/7 × 5
= £55.20 × 5
= £276

To purchase a car worth £1932:
£1932/£276 = 7

Thus, he needs 7 weeks to save enough for the car priced at £1932.

6 0
3 months 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!

5 0
1 month ago
If Brian biked at an average speed of 12.2 miles per hour for 1.45 hours, you can find the total distance by multiplying Explain
Inessa [12570]

Answer:

The distance he covered is 17.69.

Step-by-step explanation:

Simply multiply 12.2 by 1.45. This calculation yields a total distance of 17.69.

6 0
3 months ago
Read 2 more answers
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