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vichka [17]
2 years ago
5

PLZ HELP ASAP The numbers of motor vehicle fatalities in the United States for the years 2010 and 2011 are listed in the table b

elow. Which conclusion is true based on the mean fatalities for the two years?
Month 2010 2011
January 2,290 2,365
February 2,016 2,912
March 2,423 3,011
April 2,777 2,638
May 2,934 2,459
June 2,795 2,154
July 3,095 2,268
August 3,083 2,750
September3,024 2,545
October 3,056 2,323
November 2,795 2,741
December 2,597 2,912

A.The variability in the monthly fatalities in 2011 is generally higher than in 2010.
B.The variability in the monthly fatalities in 2010 is equal to the variability in 2011.
C.The monthly fatalities in 2011 are generally higher than in 2010.
D.The monthly fatalities in 2010 are generally higher than in 2011.
E.The monthly fatalities in 2010 are approximately the same as in 2011.

Mathematics
1 answer:
Nonamiya [84]2 years ago
5 0
To find which measure of variability is greater and which average number of monthly fatalities is higher, you will need to calculate the mean and the mean absolute deviation for both years

The mean will tell us which is generally higher, and the mean absolute deviation will tell us which has a greater variability.

The correct answer is D.

Please see the attached picture for the work.

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<h3>Answer:   15x^(7/3) - 8x^(7/4) + x + 9000</h3>

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Explanation:

If you know the cost function C(x), to find the marginal cost, we apply the derivative.

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Since we're given the marginal cost, we'll apply the antiderivative (aka integral) to figure out what C(x) is. This reverses the process described above.

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D represents a fixed constant. I would have used C as the constant of integration, but it's already taken by the cost function C(x).

To determine the value of D, we plug in x = 0 and C(x) = 9000. This is because we're told the fixed costs are $9000. This means that when x = 0 units are made, you still have $9000 in costs to pay. This is the initial value. You'll find that all of this leads to D = 9000 because everything else zeros out.

Therefore, we go from this

C(x) = 15x^{7/3} - 8x^{7/4} + x + D\\\\

to this

C(x) = 15x^{7/3} - 8x^{7/4} + x + 9000\\\\

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