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lorasvet [3.4K]
3 years ago
11

an object travels along a horizontal straight path at a constant rate. the object travels 1/20 of the length of the path in 3/4

second. at that rate how many seconds does it take the object to travel the entire length of the path?
Mathematics
1 answer:
neonofarm [45]3 years ago
3 0

Answer:

Step-by-step explanation:

1/20 is a tiny portion of the length that needs to be traveled.  20/20 is the whole length.  If 1/20 gets traveled in 3/4 of a second, multiply 3/4 by 20 to see how long it takes to travel 20(1/20) which is equal to 1, the whole path.

20(3/4) = 15 seconds

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ion: The mean annual tuition and fees in the 2013-2014 academic year for a sample of 13 private colleges in California was $38,0
pogonyaev

Answer:

The pvalue of the test is of 0.0979 > 0.01, which means that we cannot conclude that the mean tuition and fees for private institutions in California is greater than $35,000.

Step-by-step explanation:

Test if the mean tuition and fees for private institutions in California is greater than $35,000.

This means that at the null hypothesis we test if the fee is of $35,000 or less, that is:

H_0: \mu \leq 35000

And at the alternate hypothesis, we test if it is more than this value, that is:

H_a: \mu > 35000

The test statistic is:

As we have the standard deviation of the sample, the t-distribution is used.

t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, s is the standard deviation and n is the size of the sample.

35000 is tested at the null hypothesis:

This means that \mu = 35000

The mean annual tuition and fees in the 2013-2014 academic year for a sample of 13 private colleges in California was $38,000 with a standard deviation of $7900.

This means that n = 13, X = 38000, s = 7900

Value of the test-statistic:

t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}

t = \frac{38000 - 35000}{\frac{7900}{\sqrt{13}}}

t = 1.37

Pvalue of the test and decision:

The pvalue of the test is the probability of finding a mean above 38000, which is the pvalue of t = 1.37, with 13 - 1 = 12 degrees of freedom, using a one-tailed test.

With the help of a calculator, this pvalue is of 0.0979.

The pvalue of the test is of 0.0979 > 0.01, which means that we cannot conclude that the mean tuition and fees for private institutions in California is greater than $35,000.

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