What are the integer solutions to the inequality below?<br> 9 <_ 2x

The distance between two points knowing theirs coordinates:

AB =√[(x₂-x₁)² +(y₂-y₁)²]; ===>A(5,-4) & B(-3,-1) Given
A(x₁,y₁) & B(x₂,y₂)

AB =√[(-3-5))²+(-1-(-4)²] =√(73) = 8.381 ≈ 5.44 units

3 0

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

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

3 years ago

What is the common denominator of 3/4 and 7/9

Veseljchak [2.6K]

The least common denominator is 36

5 0

3 years ago

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What are the integer solutions to the inequality below? 9 <_ 2x – 1< x +7