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Leya [2.2K]
2 years ago
12

I WILL GIVE BRAINLIEST TT

Mathematics
1 answer:
Vinil7 [7]2 years ago
3 0

Answer:

2

Step-by-step explanation:

Yeah its 2 I can explain but will be long but if want comment and I will post but most likely its 2

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A researcher wants to estimate the percentage of all adults that have used the Internet to seek pre-purchase information in the
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Answer:

The required sample size for the new study is 801.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

The margin of error is:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

25% of all adults had used the Internet for such a purpose

This means that \pi = 0.25

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

What is the required sample size for the new study?

This is n for which M = 0.03. So

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.03 = 1.96\sqrt{\frac{0.25*0.75)}{n}}

0.03\sqrt{n} = 1.96\sqrt{0.25*0.75}

\sqrt{n} = \frac{1.96\sqrt{0.25*0.75}}{0.03}

(\sqrt{n})^2 = (\frac{1.96\sqrt{0.25*0.75}}{0.03})^2

n = 800.3

Rounding up:

The required sample size for the new study is 801.

4 0
2 years ago
The probability of drawing an ace from a deck of cards is 1/13. If you drew one card at a time (and put the card back each time)
Digiron [165]

Answer:

31 aces

Step-by-step explanation:

To find the number of aces you would expect, take the probability of an aces and multiply by the number of tries

1/13 * 400 =30.76923

Rounding up, approximately 31 aces

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3 years ago
How many times can 4 go into 724
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Please help me, pic provide d
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Fill in the y of the bottom equation with the top equation because it gives you what Y equals. To answer the question it would be easier to input the top equation into the bottom one. Once you then get the equation for X go to the equation on top and fill in the x to get the y. X=0 Y=2
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