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Xelga [282]
3 years ago
8

Simplify (2u^4/4uv^-4)^-3

Mathematics
1 answer:
aalyn [17]3 years ago
6 0

4v⁶/ u⁵

Simplify

Simplify the expression.

4v6u54v6u5

Tap to view steps...

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A national survey conducted in 2011 among a simple random sample of 1,507 adults shows that 56% of Americans think the Civil War
Drupady [299]

Answer:

a)z=4.658

p_v =P(z>4.65)=1-P(z  

b) Using the significance level assumed \alpha=0.05 we see that p_v so we have enough evidence at this significance level to reject the null hypothesis. And on this case makes sense the claim that the proportion of Americans who thinks that the Civil War is still relevant to American politics and political life is  higher than 50%.

c) The 90% confidence interval would be given (0.527;0.593).

We are confident that about 54% to 59% of all Americans think the Civil War is relevant.

Step-by-step explanation:

I )Part a

1) Data given and notation

n=1507 represent the random sample taken  

X represent the Americans who thinks that the Civil War is still relevant to American politics and political life

\hat p estimated proportion of Americans who thinks that the Civil War is still relevant to American politics and political life in the sample

p_o=0.5 is the value that we want to test since the problem says majority    

\alpha=0.05 represent the significance level (no given, but is assumed)  

z would represent the statistic (variable of interest)  

p_v represent the p value (variable of interest)  

p= population proportion of Americans who thinks that the Civil War is still relevant to American politics and political life

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the proportion of Americans who thinks that the Civil War is still relevant to American politics and political life exceeds 50%(Majority). :  

Null Hypothesis: p \leq 0.5

Alternative Hypothesis: p >0.5

We assume that the proportion follows a normal distribution.  

This is a one tail upper test for the proportion of  union membership.

The One-Sample Proportion Test is "used to assess whether a population proportion \hat p is significantly (different,higher or less) from a hypothesized value p_o".

Check for the assumptions that he sample must satisfy in order to apply the test

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 =1507*0.5=753.5>10

n(1-p_o)=1507*(1-0.5)=753.5>10

3) Calculate the statistic  

The statistic is calculated with the following formula:

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

On this case the value of p_o=0.5 is the value that we are testing and n = 1507.

z=\frac{0.56 -0.5}{\sqrt{\frac{0.5(1-0.5)}{1507}}}=4.658

The p value for the test would be:

p_v =P(z>4.65)=1-P(z  

II) Part b

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.  

Based on the alternative hypothesis the p value would be given by:

p_v =P(z>4.65)=1-P(z  

Using the significance level assumed \alpha=0.05 we see that p_v so we have enough evidence at this significance level to reject the null hypothesis. And on this case makes sense the claim that the proportion of Americans who thinks that the Civil War is still relevant to American politics and political life is  higher than 50%.

III) Part c

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 90% confidence interval the value of \alpha=1-0.9=0.1 and \alpha/2=0.05, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=2.58

And replacing into the confidence interval formula we got:

0.56 - 2.58 \sqrt{\frac{0.56(1-0.56)}{1507}}=0.527

0.56 + 2.58 \sqrt{\frac{0.56(1-0.56)}{1507}}=0.593

And the 90% confidence interval would be given (0.527;0.593).

We are confident that about 54% to 59% of all Americans think the Civil War is relevant.

And this result agrees with the result of part b, since the interval not contains the value of 0.5 we can conclude that the proportion of Americans who thinks that the Civil War is still relevant to American politics and political life it's higher than 0.5 at 90% of confidence.

7 0
3 years ago
What is the area of this irregular shape 28in 7in 20in 30in 25in
Alex777 [14]

Answer:

b1=1b2=1h=7b1=1b2=6h=2b1=6b2=9h=2

7 0
3 years ago
What is 33 1/3 of 51 equals what please show work
Leona [35]
If ur asking for it in percent, it is 17. 33 1/3 percent is the same thing as 1/3 of a number. therefore, u multiply 51 by one third or divide 51 by 3
7 0
3 years ago
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teri and her friends bought a party size sandwich that was 7/9 yard long. they ate 2/3 of a yard. what part of the yard was left
Marysya12 [62]
Teri had 7/9 of a sandwich, you need to find a common denominator for 2/3. Multiply 2/3 by 3 which is 6/9. (6/9 is how much they ate) Then you subtract 7/9 by 6/9 (7/9-6/9=1/9)
Answer:1/9 of a yard

I hope this helped.. not much of a teacher hahah :)
6 0
3 years ago
Simplify the following: Will give brainliest.NEED ANSWER ASAP​
andrey2020 [161]

Answer

the answer is in the picture

Step-by-step explanation:

4 0
3 years ago
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