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Hunter-Best [27]
3 years ago
13

The Smith family is shopping for a new car and they are basing their decision on color and style. Explain how many color choices

and how many style choices they might have if there are 8 possible outcomes. Justify your answer.
Mathematics
1 answer:
Yanka [14]3 years ago
3 0
Sample response: The Smith family has 2 choices to make, color and style. By the fundamental counting principle, the product of the number of choices of color and style must equal 8. So, there could be 1 color and 8 style choices, 2 colors and 4 styles, 4 colors and 2 styles, or 8 colors and 1 style.
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Step-by-step explanation:

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The average undergraduate cost for tuition, fees, and room and board for two-year institutions last year was $13,252. The follow
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Answer:

The p-value of the test is 0.0041 < 0.05, which means that there is sufficient evidence at the 0.05 significance level to conclude that the mean cost has increased.

Step-by-step explanation:

The average undergraduate cost for tuition, fees, and room and board for two-year institutions last year was $13,252. Test if the mean cost has increased.

At the null hypothesis, we test if the mean cost is still the same, that is:

H_0: \mu = 13252

At the alternative hypothesis, we test if the mean cost has increased, that is:

H_1: \mu > 13252

The test statistic is:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question

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.

13252 is tested at the null hypothesis:

This means that \mu = 13252

The following year, a random sample of 20 two-year institutions had a mean of $15,560 and a standard deviation of $3500.

This means that n = 20, X = 15560, s = 3500

Value of the test statistic:

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

t = \frac{15560 - 13252}{\frac{3500}{\sqrt{20}}}

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P-value of the test and decision:

The p-value of the test is found using a t-score calculator, with a right-tailed test, with 20-1 = 19 degrees of freedom and t = 2.95. Thus, the p-value of the test is 0.0041.

The p-value of the test is 0.0041 < 0.05, which means that there is sufficient evidence at the 0.05 significance level to conclude that the mean cost has increased.

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