- I’ll in box boxes!!!!!

(-2/3) to 2nd Power - 3/4 (2 1/3)<br><br> Will give out 50+25 pts and Brainliest Answer

trasher [3.6K]

I think the answer is -47/36

5 0

4 years ago

The federal government would like to test the hypothesis that the average age of men filing for Social Security is higher than t

BlackZzzverrR [31]

Answer:

C (3²/35+3.5²/39) For more enquiries connect [email protected] AT MY G mail account

4 0

3 years ago

A cylinder has a radius of 3 cm. Sand is poured into the cylinder to form a layer 2 cm deep. What is the volume of sand in the c

netineya [11]

Answer:

Volume=56.54cm³

Step-by-step explanation:

$Solution,\\\\Radius(r)=3cm\\\\Height(h)=2cm\\\\Now,\\\\Volume=\pi r^{2} h\\\\Volume=\pi *(3cm)^{2} *2cm\\\\Volume=18\pi cm^{3} =56.54cm^{3}$

3 0

2 years ago

In a recent poll, 778 adults were asked to identify their favorite seat when they fly, and 492 of them chose a window seat. Use

seropon [69]

Answer:

Null hypothesis: $p \leq 0.5$

Alternative hypothesis: $p > 0.5$

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

$p_v =P(Z>7.36)=9.19x10^{-14}$

Since the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of adults prefer window seats when they fly is significantly higher than 0.5 .

Step-by-step explanation:

1) Data given and notation

n=778 represent the random sample taken

X=492 represent the people that chose a window seat.

$\hat p=\frac{492}{778}=0.632$ estimated proportion of people that chose a window seat.

$p_o=0.5$ is the value that we want to test

$\alpha=0.01$ represent the significance level

Confidence=99% or 0.99

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 majority of adults prefer window seats when they fly:

Null hypothesis: $p \leq 0.5$

Alternative hypothesis: $p > 0.5$

When we conduct a proportion test we need to use the z statisitc, 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 $p_o$ .

3) Calculate the statistic

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

$z=\frac{0.632 -0.5}{\sqrt{\frac{0.5(1-0.5)}{778}}}=7.36$

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 $\alpha=0.01$ . The next step would be calculate the p value for this test.

Since is a right tailed test the p value would be:

$p_v =P(Z>7.36)=9.19x10^{-14}$

5) Conclusion

Since the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of adults prefer window seats when they fly is significantly higher than 0.5 .

6 0

3 years ago

Please Help Me Will Pick the first correct answer brainliest

dimaraw [331]

Answer:

1 and 4 are correct :)

Step-by-step explanation:

8 0

3 years ago