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asambeis [7]
3 years ago
13

An apple orchard sold 155 apples in September in November it sold 17 times as many apples as it did in September how many apples

did the apple orchard sell in November?
Mathematics
1 answer:
borishaifa [10]3 years ago
8 0

9514 1404 393

Answer:

  2635

Step-by-step explanation:

17 times the September number is ...

  17×155 = 2635 . . . . apples sold in November

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Which fraction is equivalent to 3/4? 12/12, 9/16,12/16,5/3
Vikki [24]
The answer is 12/16... divide the fraction by 4 and you get 3/4
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What is the sum of two numbers a and b decreased by their product?
belka [17]

The sum of two numbers a and b decreased by their product is (a+b)-ab

Step-by-step explanation:

Given the two numbers as a and b

Sum of these two numbers is a+b

The product of the two numbers is a*b, ab

The sum of the two numbers decreased by their product will be;

(a+b)-ab

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4 years ago
Which is greater 20 cups or 10 pints
Marina CMI [18]
The answer to your question is 10 pints because pints in general are bigger than cups even If it's 20 cups
7 0
3 years ago
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If 2x = 3y + 11 and 2 to the power of x = 2 to the power of 4(y+1), determine the value of x + y
evablogger [386]

Z = -2

Y = 5

is your answer

7 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
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