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Tju [1.3M]
3 years ago
14

Which function does NOT have a range of all real numbers?

Mathematics
1 answer:
lesya692 [45]3 years ago
3 0

Answer:

D

Step-by-step explanation:

D is the correct answer

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Which of the following best describes the algebraic expression x/3 -15
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A. A number divided by three minus fifteen.

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Use the given information to find the p​-value. ​Also, use a 0.05 significance level and state the conclusion about the null hyp
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Answer:

We can assume that the statistic is z_{calc}=0.78

p_v = 2* P(z>0.78) = 0.435

So the p value obtained was a very high 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 true proportion of interest is not different from 3/5

Step-by-step explanation:

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is equal to 3/5 or not.:  

Null hypothesis:p=3/5  

Alternative hypothesis:p \neq 3/5  

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.

Calculate the statistic  

We can assume that the statistic is z_{calc}=0.78

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>0.78) = 0.435

So the p value obtained was a very high 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 true proportion of interest is not different from 3/5

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