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2 years ago

Simplify the cubed root of 56

1 answer:

6 0

The cubed root of 56 is

 $\sqrt[3]{56}$ 

 $\sqrt[3]{8 * 7}$ 

 $\sqrt[3]{8} \sqrt[3]{7}$ 

 $2 \sqrt[3]{7}$ 

The simplified cubed root of 56 is $2 \sqrt[3]{7}$

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3 years ago

36. The price of a new home. According to the U.S. Census

zmey [24]

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2 years ago

After a college football team once again lost a game to their archrival, the alumni association conducted a survey to see if alu

valentina_108 [34]

Answer:

P-value for this hypothesis test is 0.00175.

Step-by-step explanation:

We are given that the alumni association conducted a survey to see if alumni were in favor of firing the coach.

A simple random sample of 100 alumni from the population of all living alumni was taken. Sixty-four of the alumni in the sample were in favor of firing the coach.

Let p = proportion of all living alumni who favored firing the coach

SO, Null Hypothesis, $H_0$ : p = 0.50 {means that the majority of alumni are not in favor of firing the coach}

Alternate Hypothesis, $H_A$ : p > 0.50 {means that the majority of alumni are in favor of firing the coach}

The test statistics that will be used here is One-sample z proportion statistics;

T.S. = $\frac{\hat p-p}{{\sqrt{\frac{\hat p(1-\hat p)}{n} } } } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of the alumni in the sample who were in favor of firing the coach = $\frac{64}{100}$ = 0.64

n = sample of alumni = 100

So, test statistics = $\frac{0.64-0.50}{{\sqrt{\frac{0.64(1-0.64)}{100} } } } }$

= 2.92

Now, P-value of the hypothesis test is given by ;

P-value = P(Z > 2.92) = 1 - P(Z $\leq$ 2.92)

= 1 - 0.99825 = 0.00175

Therefore, the P-value for this hypothesis test is 0.00175.

4 0

3 years ago

What is the answer or solution 9s/2=45

Drupady [299]

S would be equal to 10. We’d multiply 45 by 2 to get the 90 then divide by 9 to isolate it

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3 years ago

Read 2 more answers

What is 3(x + 7) expanded

Alex

Distributive proeprty
a(b+c)=ab+ac
3(x+7)=3x+21

6 0

2 years ago