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katrin2010 [14]
3 years ago
14

Write your answer to the following question on the whiteboard. Show all your work

Mathematics
1 answer:
KatRina [158]3 years ago
8 0
Multiply 48$ by .15 that will give you your answer
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Emma borrowed $300 to repair her car. the finance (interest) charge on the loan was $20, and the term on the loan was 14 days. w
Murrr4er [49]
Given:
loan amount = 300
finance charge = 20
term = 14 days.

To solve for APR.
<span>1. Divide the finance charge by the loan amount.
20/300 = 0.0667

2. Multiply the result by 365. 
0.0667 x 365 = 24.35

3. Divide the result by the term of the loan.
24.35/14 = 1.74 (APR in decimal format)
<span>
4. Multiply the result by 100.
1.74 x 100 = 174% APR</span></span>
7 0
3 years ago
Jensen Tire &amp; Auto is in the process of deciding whether to purchase a maintenance contract for its new computer wheel align
sasho [114]

Answer:

Step-by-step explanation:

Hello!

The given data corresponds to the variables

Y:  Annual Maintenance  Expense ($100s)

X: Weekly Usage  (hours)

n= 10

∑X= 253; ∑X²= 7347; \frac{}{X}= ∑X/n= 253/10= 25.3 Hours

∑Y= 346.50; ∑Y²= 13010.75; \frac{}{Y}= ∑Y/n= 346.50/10= 34.65 $100s

∑XY= 9668.5

a)

To estimate the slope and y-intercept you have to apply the following formulas:

b= \frac{sumXY-\frac{(sumX)(sumY)}{n} }{sumX^2-\frac{(sumX)^2}{n} } = \frac{9668.5-\frac{253*346.5}{10} }{7347-\frac{(253)^2}{10} }= 0.95

a= \frac{}{Y} -b\frac{}{X} = 34.65-0.95*25.3= 10.53

^Y= a + bX

^Y= 10.53 + 0.95X

b)

H₀: β = 0

H₁: β ≠ 0

α:0.05

F= \frac{MS_{Reg}}{MS_{Error}} ~~F_{Df_{Reg}; Df_{Error}}

F= 47.62

p-value: 0.0001

To decide using the p-value you have to compare it against the level of significance:

If p-value ≤ α, reject the null hypothesis.

If p-value > α, do not reject the null hypothesis.

The decision is to reject the null hypothesis.

At a 5% significance level you can conclude that the average annual maintenance expense of the computer wheel alignment and balancing machine is modified when the weekly usage increases one hour.

b= 0.95 $100s/hours is the variation of the estimated average annual maintenance expense of the computer wheel alignment and balancing machine is modified when the weekly usage increases one hour.

a= 10.53 $ 100s is the value of the average annual maintenance expense of the computer wheel alignment and balancing machine when the weekly usage is zero.

c)

The value that determines the % of the variability of the dependent variable that is explained by the response variable is the coefficient of determination. You can calculate it manually using the formula:

R^2 = \frac{b^2[sumX^2-\frac{(sumX)^2}{n} ]}{[sumY^2-\frac{(sumY)^2}{n} ]} = \frac{0.95^2[7347-\frac{(253)^2}{10} ]}{[13010.75-\frac{(346.50)^2}{10} ]} = 0.86

This means that 86% of the variability of the annual maintenance expense of the computer wheel alignment and balancing machine is explained by the weekly usage under the estimated model ^Y= 10.53 + 0.95X

d)

Without usage, you'd expect the annual maintenance expense to be $1053

If used 100 hours weekly the expected maintenance expense will be 10.53+0.95*100= 105.53 $100s⇒ $10553

If used 1000 hours weekly the expected maintenance expense will be $96053

It is recommendable to purchase the contract only if the weekly usage of the computer is greater than 100 hours weekly.

4 0
2 years ago
2x+y=7 for y neeedd help
masya89 [10]

Answer:

  • y = -2x + 7

Step-by-step explanation:

<u>Solve for y:</u>

  • 2x + y = 7

Add - 2x to both sides

  • 2x + y - 2x = 7 - 2x
  • y = 7 - 2x

or

  • y = -2x + 7
8 0
3 years ago
A scale for a scale drawing is 10 cm: 1 mm which is larger the actual object or the scale drawing
Yanka [14]
Since 10 cm is bigger than 1 mm, the scale drawing is bigger

3 0
3 years ago
Yianni typed a 36-word paragraph in 2/3 of a minute<br> what is her typing speed in words per minute
elena-s [515]

I think it would be 54 words per minute

36÷2=18 which is 1/3 of a minute so we multiply 18 by 3 to get how many words in one minute, 18×3=54

4 0
3 years ago
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