According to the Census Bureau, 3.39 people reside in the typical American household. A sample of 26 households in Arizona retirement communities showed the mean number of residents per household was 2.73 residents. The standard deviation of this sample was 1.22 residents. At the .10 significance level, is it reasonable to conclude the mean number of residents in the retirement community household is less than 3.39 persons?

Since the p value is lower than the significance level 0.1 we have enough evidence to reject the null hypothesis, and we can conclude that the true mean is significanlty lower than 3.39 personas at 10% of significance.

Step-by-step explanation:

Data given and notation  

represent the sample mean

represent the sample standard deviation

sample size  

represent the value that we want to test

represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is less than 3.39 persons, the system of hypothesis would be:  

Null hypothesis:  

Alternative hypothesis:  

The statistic is given by:

 (1)  

Calculate the statistic

We can replace in formula (1) the info given like this:  

P-value

The degreed of freedom are given by:

Since is a one sided lower test the p value would be:  

Conclusion  

Since the p value is lower than the significance level 0.1 we have enough evidence to reject the null hypothesis, and we can conclude that the true mean is significanlty lower than 3.39 personas at 10% of significance.

c. prove the lengths are the same.

step-by-step explanation:

you need to prove the segments are congruent. congruent segments have the same length. find the lengths of the segments using the distance formula, and compare them.

answer: c. prove the lengths are the same.

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.