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Ivanshal [37]
3 years ago
7

Select the correct answer.

Mathematics
1 answer:
Papessa [141]3 years ago
5 0

Answer:

If I had to guess, I'd say its A.

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Alex_Xolod [135]

Answer:

4\times \sqrt{11}\div \(11)

Step-by-step explanation:

\sqrt{16}\div \sqrt{11}

\sqrt{16}\times \sqrt{11}\div \sqrt{11}\times \sqrt{11}

4\times \sqrt{11}\div \(11)  answer

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egoroff_w [7]

Answer:

B. (4x+5)(16x2−20x+25)

Step-by-step explanation:

64x3+125

=(4x+5)(16x2−20x+25)

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2 years ago
There are twenty classes at Northwestern Middle School and twenty classes at Southeastern Middle School. The number of students
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The means and medians are not the same. Or B

Step-by-step explanation:

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2 years ago
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After once again losing a football game to the college'ss arch rival, the alumni association conducted a survey to see if alumni
antoniya [11.8K]

Answer:

a) The 90% confidence interval would be given (0.561;0.719).

b) p_v =P(z>2.8)=1-P(z

c) Using the significance level assumed \alpha=0.01 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 is higher than 0.5 or 50%.    

Step-by-step explanation:

1) Data given and notation  

n=100 represent the random sample taken    

X=64 represent were in favor of firing the coach

\hat p=\frac{64}{100}=0.64 estimated proportion for were in favor of firing the coach

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

\alpha 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 for were in favor of firing the coach

Part a

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}=1.64

And replacing into the confidence interval formula we got:

0.64 - 1.64 \sqrt{\frac{0.64(1-0.64)}{100}}=0.561

0.64 + 1.64 \sqrt{\frac{0.64(1-0.64)}{100}}=0.719

And the 90% confidence interval would be given (0.561;0.719).

Part b

We need to conduct a hypothesis in order to test the claim that the proportion 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 =100*0.64=64>10  

n(1-p_o)=100*(1-0.64)=36>10  

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 = 100.  

z=\frac{0.64 -0.5}{\sqrt{\frac{0.5(1-0.5)}{100}}}=2.8

The p value for the test would be:  

p_v =P(z>2.8)=1-P(z

Part c

Using the significance level assumed \alpha=0.01 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 is higher than 0.5 or 50%.    

6 0
3 years ago
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