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borishaifa [10]
3 years ago
15

Four students are competing in a race. In how many different combinations could the four students place in the race?

Mathematics
1 answer:
VMariaS [17]3 years ago
4 0

"Combinations" isn't really the right word, because a combination doesn't take order of elements into account, and the outcome of a race certainly does depend on order. "Permutations" would be the correct term.

If 4 students are competing, then the race has

4! = 4 • 3 • 2 • 1 = 24

possible outcomes.

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Answer:

The 99% confidence interval for the average length of time all car owners plan to keep their cars is between 3.85 years and 10.55 years.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.99}{2} = 0.005

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.005 = 0.995, so z = 2.575

Now, find the margin of error M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

M = 2.575*\frac{6.5}{\sqrt{25}} = 3.35

The lower end of the interval is the sample mean subtracted by M. So it is 7.2 - 3.35 = 3.85 years

The upper end of the interval is the sample mean added to M. So it is 7.2 + 3.35 = 10.55 years

The 99% confidence interval for the average length of time all car owners plan to keep their cars is between 3.85 years and 10.55 years.

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Answer:

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Step-by-step explanation:

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