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Gekata [30.6K]
4 years ago
11

Can someone help me with this

Mathematics
2 answers:
Anvisha [2.4K]4 years ago
7 0

Answer:

D. \frac{{7}^{11} }{ {4}^{11} }

Step-by-step explanation:

( \frac{7}{4}) {}^{11} = \frac{7 {}^{11} }{4 {}^{11} }

Since we are multiplying an exponential number to a fraction the answer is going to be

\frac{7 {}^{11} }{4 {}^{11} }

Hope this helps ❤❤❤ ;)

ira [324]4 years ago
5 0

Answer:

D.  \frac{7^{11} }{4^{11} } \\

Step-by-step explanation:

Hello, I can help you with this

Let's remmber some propiertis of the fraction numbers and the potenciation

Let

a^{m} *  a^{n}=a^{m+n}\\(a^{m})^{n}=a^{m*n}\\(\frac{a}{b})^{m}=\frac{a^{m} }{b^{m}}

Step 1

All you have to do is identify and use the formula

let

(\frac{7}{4})^{11}\\ a=7\\b=4\\m=11\\so\\(\frac{7}{4})^{11}=\frac{7^{11} }{4^{11} } \\

so, the answer is D.

I hope it helps, have a nice day

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Answer:

z=\frac{0.884 -0.72}{\sqrt{\frac{0.72(1-0.72)}{43}}}=2.395  

Now we can find the p value taking incount that we are using a bilateral test

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Since the p value is lower than the significance level of 0.1 we have enough evidence to reject the null hypothesis and we can conclude that the true proportion of Americans that they would prefer an American automobile  is significantly different from 0.72 and then the claim is correct

Step-by-step explanation:

Information given

n=43 represent the random sample taken

X=38 represent the Americans who they would prefer an American automobile

\hat p=\frac{38}{43}=0.884 estimated proportion of Americans that they would prefer an American automobile

p_o=0.72 is the value that we want to check

\alpha=0.1 represent the significance level

z would represent the statistic

p_v represent the p value

Hypothesis to test

We want to verify if the percentage of Americans that they would prefer an American automobile  today differs from 72%, then the system of hypothesis are:

Null hypothesis:p=0.72  

Alternative hypothesis:p \neq 0.72  

The statistic is given by:

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

Replacing the data given we got:

z=\frac{0.884 -0.72}{\sqrt{\frac{0.72(1-0.72)}{43}}}=2.395  

Now we can find the p value taking incount that we are using a bilateral test

p_v =2*P(z>2.395)=0.01662  

Since the p value is lower than the significance level of 0.1 we have enough evidence to reject the null hypothesis and we can conclude that the true proportion of Americans that they would prefer an American automobile  is significantly different from 0.72 and then the claim is correct

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