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3 years ago

HELPPP PLEASEE I will give good review

Mathematics

1 answer:

givi [52]3 years ago

3 0

ANSWER: slope = $\frac{4}{3}$ and y-intercept = -2

EXPLANATION:

1. The formula for the slope is:

$m=\frac{y_2-y_1}{x_2 - x_1}$

2. Now you just need to pick two points on the graph that the line goes through. E.g. (-3, -6) and (6, 6). You then add it into to formula.

$m = \frac{6-(-6)}{6-(-3)} =\frac{6+6}{6+3}=\frac{12}{9}$

3. Simplify by dividing the numerator and denominator by 3.

$m=\frac{4}{3}$

4. The y-intercept is just where the line hits the y-axis (vertical line). By looking at the graph, we can see the line hits -2.

y-intercept = -2

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Here, first we need to calculate the slope of the line,
Here, Coordinates = (-6, 2) (-1, -4)
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Now, Take first coordinate: y - 2 = -6/5 (x + 6)
Second coordinate: y + 4 = -6/5 (x + 1)

In short, Your Answers would be Option C & D

Hope this helps!

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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

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3 years ago

Jensen Tire & 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

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

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.

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

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.

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Bad White [126]

$\huge \bf༆ Answer ༄$