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OLga [1]
1 year ago
11

An ice cream store sells 28 flavors of ice cream. Determine the number of 4 dip sundaes.

Mathematics
1 answer:
weeeeeb [17]1 year ago
6 0

Answer:

20,475\ ways

Step-by-step explanation:

we know that

<u><em>Combinations</em></u> are a way to calculate the total outcomes of an event where order of the outcomes does not matter.

To calculate combinations, we will use the formula

C(n,r)=\frac{n!}{r!(n-r)!}

where

n represents the total number of items

r represents the number of items being chosen at a time.

In this problem

n=28\\r=4

substitute

C(28,4)=\frac{28!}{4!(28-4)!}\\\\C(28,4)=\frac{28!}{4!(24)!}

simplify

C(28,4)=\frac{(28)(27)(26)(25)(24!)}{4!(24)!}

C(28,4)=\frac{(28)(27)(26)(25)}{(4)(3)(2)(1)}

C(28,4)=20,475\ ways

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A random sample of 160 car purchases are selected and categorized by age. The results are listed below. The age distribution of
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Answer:

The claim that all ages have purchase rates proportional to their driving rates is false.

Step-by-step explanation:

The complete question is:

A random sample of 160 car accidents are selected and categorized by the age of the driver determined to be at fault. The results are listed below. The age distribution of drivers for the given categories is 18% for the under 26 group, 39% for the 26-45 group, 31% for the 45-65 group, and 12% for the group over 65. Calculate the chi-square test statistic used to test the claim that all ages have crash rates proportional to their driving rates.

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Solution:

In this case we need to test whether there is sufficient evidence to warrant rejection of the claim that all ages have crash rates proportional to their driving rates.

A Chi-square test for goodness of fit will be used in this case.

The hypothesis can be defined as:

<em>H₀</em>: The observed frequencies are same as the expected frequencies.

<em>Hₐ</em>: The observed frequencies are not same as the expected frequencies.

The test statistic is given as follows:

\chi^{2}=\sum{\frac{(O-E)^{2}}{E}}

The values are computed in the table.

The test statistic value is 75.10.

The degrees of freedom of the test is:

n - 1 = 4 - 1 = 3

Compute the p-value of the test as follows:

p-value < 0.00001

*Use a Chi-square table.

p-value < 0.00001 < α = 0.05.

So, the null hypothesis will be rejected at any significance level.

Thus, there is sufficient evidence to warrant rejection of the claim that ages have crash rates proportional to their driving rates.

3 0
2 years ago
The approximate line of best fit is given by the equation y=40x-1800. Based on this trend which of the following best predicts t
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Solve for x

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So x = 47.35
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