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lana66690 [7]
3 years ago
12

During an algebra class a teacher asked for examples of non-prime numbers. Jake says 51, Teresa says 41, and let Liu says 21. Wh

ich student(s) are correct in their response
Mathematics
2 answers:
cestrela7 [59]3 years ago
7 0

Answer:

Jake and Liu

Step-by-step explanation:

We are given that an algebra class  teacher asked for example of non prime numbers.

Jake says=51

51 have factors 1,3,17 and 51.

Teresa says=41

41 have factors 1 and 41.

Liu says=21

21 have factors 1,3,7 and 21.

We know that prime number are those numbers which have only two factors 1 and itself.

Composite number:It is that number which have more than two factors 1 and itself.

Therefore, by definition of prime number Teresa is incorrect in their response and Jake and Liu are both correct .

USPshnik [31]3 years ago
3 0
The students that are correct are Jake , and Liu because 51, and 21 are composite.

Hope this helps!
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Test the hypothesis using the P value approach. Be sure the verify the requirements of the test.
Andreas93 [3]

Answer:

p_v =2*P(z  

If we compare the p value and using the significance level given \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion is not significantly different from 0.77.  

Step-by-step explanation:

1) Data given and notation

n=500 represent the random sample taken

X=380 represent the number of people with some characteristic

\hat p=\frac{380}{500}=0.76 estimated proportion of adults that said that it is morally wrong to not report all income on tax returns

p_o=0.76 is the value that we want to test

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is 0.7 .:  

Null hypothesis:p=0.77  

Alternative hypothesis:p \neq 0.77  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

<em>Check for the assumptions that he sample must satisfy in order to apply the test </em>

a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.

b) The sample needs to be large enough

np_o =500*0.77=385>10

n(1-p_o)=384*(1-0.77)=115>10

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.76-0.77}{\sqrt{\frac{0.77(1-0.77)}{500}}}=-0.531  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

p_v =2*P(z  

If we compare the p value and using the significance level given \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion is not significantly different from 0.77.  

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Answer:

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