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x, y
14, 4.2
1, 5.8
4.9, 4.7
17, 3.7
4.3, 4.8

"x is the average annual replacement market volume, and y is the cumulative dollars spent per tire.

What does this model predict the cumulative dollars spent per tire will be when the average replacement market is 10 million units?

What does this model predict the average replacement market will be if there cumulative dollars spent per tire is 5.0?"

i'm having trouble with this problem because i dont know what "average annual replacement market volume" is, and i dont know how to enter the info into my calculator. any help would be greatly appreciated!! thanks!

2006-09-19 14:32:00 · 3 answers · asked by Anonymous in Science & Mathematics Mathematics

3 answers

Average annual market volume is X. The best way to solve this problem is to graph the X and Y and see if there is a trend.

MS Excel will let you do this and add a trend line, or a linear regression line. By doing this we discover that X = -.8.1558Y +46.083. This has an R^2 value of .862 meaning that 86.2% of the variance in X is explained by Y.

The negative number before the Y is expected since X (volume) should decrease as Y (price) increases.

To solve the problem place 5 in for Y. The answer is 5.304. This is higher than expected due to the skewedness of the data set.

2006-09-19 14:54:15 · answer #1 · answered by MagicalMke 4 · 0 0

It does not matter if you do not know what those terms mean. If you read the question carefully, and what x and y represent, you will see that it is asking you to find x, when you are given y = 5
.
So in your calculator, you enter the y values as the independent variable and the x values as the dependent variable.Then you can estimate x from any value of y.

The answer I get is 5.0

2006-09-19 22:15:12 · answer #2 · answered by Anonymous · 0 0

Simple question. Lets assume that there is no time period required to source from. All you have to do is "average" the numbers in the y column. 23.2 over 5 is 4.64. Hope I helped.

2006-09-19 21:49:33 · answer #3 · answered by Ricky J. 6 · 0 0

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