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vaieri [72.5K]
3 years ago
15

Mr. Trager has $500.00 to spend at a bicycle store. All prices listed below include tax.

Mathematics
1 answer:
ICE Princess25 [194]3 years ago
8 0

Answer:

He can buy 2 outfits

Step-by-step explanation:

​How to arrive at the answer is: Since it says that Mr. Trager bought a new bike for $273.98. Subtract $ 273.98 from $ 500.00. The answer is $ 226.02. Then it says he bought 3 bike reflectors for $ 7.23.To find how much the bike reflectors cost, multiply 7.23 by 3. The answer is $21.69. Then subtract $ 21.69 from $ 226.02. The answer is $ 204.33. Then it says he bought a helmet for $ 42.36. Subtract $ 42.36 from  $ 204.33. The answer is $ 161.97. Then it says he plans to buy cycling outfits for $ 78.12 each. To find out how many he can buy, subtract $ 78.12 until you can't anymore. You can subtract $ 78.12 once and the answer is $ 83.85. It looks like we can subtract one more time. Then we subtract $ 78.12 again, and he has $ 5.73 left. We subtracted twice, so he can buy 2 outfits.

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The Rocky Mountain News (January 24, 1994) indicated that the 20-year mean snowfall in the Denver/Boulder region is 28.76 inches
ycow [4]

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

H_0: \mu = 28.76

At the alternate hypothesis, we test if the average incresaed, that is, it was higher than 28.76 inches. So

H_1: \mu > 28.76

The test statistic is:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

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

28.76 is tested at the null hypothesis:

This means that \mu = 28.76

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 \sigma = 7.5, X = 33, n = 32.

Value of the test statistic:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{33 - 28.76}{\frac{7.5}{\sqrt{32}}}

z = 3.2

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 \alpha = 0.05.

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.

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Step-by-step explanation:

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