Write the interval notation<br>the domain of the INVERSE<br>of the function shown

CaHeK987 [17]

It's 67/100 because a percent is out of 100 which is already simplified

6 0

3 years ago

Most US adults have social ties with a large number of people, including friends, family, co-workers, and other acquaintances. I

xxTIMURxx [149]

Answer:

$t=\frac{669-634}{\frac{732}{\sqrt{1700}}}=1.971$

$p_v =2*P(t_{(1699)}>1.971)=0.049$

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from 634 at 10% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=669$ represent the sample mean

$s=732$ represent the sample standard deviation

$n=1700$ sample size

$\mu_o =68$ represent the value that we want to test

$\alpha=0.$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ 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 mean is different from 634, the system of hypothesis would be:

Null hypothesis: $\mu = 634$

Alternative hypothesis: $\mu \neq 634$

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

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

$t=\frac{669-634}{\frac{732}{\sqrt{1700}}}=1.971$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=1700-1=1699$

Since is a two sided test the p value would be:

$p_v =2*P(t_{(1699)}>1.971)=0.049$

Conclusion

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from 634 at 10% of signficance.

3 0

3 years ago

What is 25% of success as a fraction and a decimal

Tpy6a [65]

25/100 simplify and get 1/4
.25 as a decimal. 25 hundredths

5 0

4 years ago

Brainliest for whoever gets this right

meriva

Answer

14 degrees

Step-by-step explanation: