A student ran the 50 yard dash in 5.8 seconds at what speed did the student run in miles per hour round to the nearest 10th
2 answers:
Answer:
The speed of student is 17.634 miles per hour.
Step-by-step explanation:
We need to find the speed of student in miles per hour.
Distance covered by student = 50 yards
We know that there are 0.000568182 miles in 1 yard.
Thus, the student covered
Time taken by the student = 5.8 second
We need to convert second into hours. There are 3600 seconds in 1 hour.
Time taken by student
Formula:
Thus, the speed of student is 17.634 miles per hour.
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Answer:
The p-value of the test is 0.0007 < 0.05, indicating that the the snowfall for the 1993-1994 winters was higher than the previous 20-year average.
Step-by-step explanation:
20-year mean snowfall in the Denver/Boulder region is 28.76 inches. Test if the snowfall for the 1993-1994 winters has higher than the previous 20-year average.
At the null hypothesis, we test if the average was the same, that is, of 28.76 inches. So
At the alternate hypothesis, we test if the average incresaed, that is, it was higher than 28.76 inches. So
The test statistic is:
In which X is the sample mean, is the value tested at the null hypothesis,
is the standard deviation and n is the size of the sample.
28.76 is tested at the null hypothesis:
This means that
Standard deviation of 7.5 inches. However, for the winter of 1993-1994, the average snowfall for a sample of 32 different locations was 33 inches.
This means that .
Value of the test statistic:
P-value of the test and decision:
The p-value of the test is the probability of finding a sample mean above 33, which is 1 subtracted by the p-value of z = 3.2. In this question, we consider the standard level .
Looking at the z-table, z = 3.2 has a p-value of 0.9993.
1 - 0.9993 = 0.0007
The p-value of the test is 0.0007 < 0.05, indicating that the the snowfall for the 1993-1994 winters was higher than the previous 20-year average.