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salantis [7]
3 years ago
14

Please help! PLSSS! Its for a math test.​

Mathematics
1 answer:
Maslowich3 years ago
8 0

Answer:

330(D)

Step-by-step explanation:

So lets go over what we know:

#1: 44 out of 80 people think they have a average fitness level.

#2: 80 out of the 600 people in the school were surveyed.

Using what we know, lets try and find the answer.

So lets divide the total amount of people, by the amount of people that took the survey:

600/80

=

7.5

Ok, so now we know that we need to take 7.5 surveys to find out what all the students think of their fitness level.

Now lets take teh amount of people in a single survey, with average fitness(44), and multpliy it by how many surveys are needed for the whole school(7.5):

44*7.5

=

330

This is the amount of people in the whole school who think they have an average fitness level.

I hope this helps! :)

<u>Answer:</u>

<u>330</u>

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Find the number of primes less than 190 using the principle of inclusion-exclusion.
Troyanec [42]

The number of primes less than 190 using the principle of inclusion-exclusion are 42

<h3>Principle of inclusion- exclusion</h3>

The principle of inclusion-exclusion is known as a counting technique that computes the number of elements satisfying at least one of several properties and guaranteeing that the numbers are not counted twice.

Prime numbers are numbers only divisible by 1 and itself.

Prime numbers less than 190 are:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181

They are 42 in number

Therefore, the number of primes less than 190 using the principle of inclusion-exclusion are 42

Learn more about prime numbers here:

brainly.com/question/874965

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A person purchased a slot machine and tested it by playing it 1,137 times. There are 10 different categories of outcomes, includ
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Answer:

p_v= P(\chi^2_{9}>11.517)=0.2419

And on this case if we see the significance level given \alpha=0.1 we see that p_v>alpha so we fail to reject the null hypothesis that the observed outcomes agree with the expected frequencies at 10% of significance.

Step-by-step explanation:

A chi-square goodness of fit test determines if a sample data obtained fit to a specified population.

p_v represent the p value for the test

O= obserbed values

E= expected values

The system of hypothesis for this case are:

Null hypothesis: O_i = E_i[/tex[Alternative hypothesis: [tex]O_i \neq E_i

The statistic to check the hypothesis is given by:

\chi^2 =\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}

On this case after calculate the statistic they got: \chi^2 = 11.517

And in order to calculate the p value we need to find first the degrees of freedom given by:

df=n-1=10-1=9, where k represent the number of levels (on this cas we have 10 categories)

And in order to calculate the p value we need to calculate the following probability:

p_v= P(\chi^2_{9}>11.517)=0.2419

And on this case if we see the significance level given \alpha=0.1 we see that p_v>alpha so we fail to reject the null hypothesis that the observed outcomes agree with the expected frequencies at 10% of significance.

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