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otez555 [7]
3 years ago
11

The recomended daily allowance of niacin is 20 mg. One serving of certain breakfast cereal provides 15mg of niacin. What percent

of the recommended daily allowance of niacin does one serving of this cereal provide
Mathematics
2 answers:
PtichkaEL [24]3 years ago
6 0
75%. Just divide 15/20 and you will get .75 then multiply by 100 to turn the decimal to a percent

Sever21 [200]3 years ago
6 0
75% is the answer because its 3/4 of the recommended daily
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6.Suppose the Gallup Organization wants to estimate the population proportion of those who think there should be a law that woul
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A sample of n = (\frac{1.96\sqrt{0.3696*0.6304}}{E})^2 is needed, in which E is the desired margin of error, as a proportion. If we find a decimal value, we round up to the next whole number.

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.

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In a previous study of 1012 randomly chosen respondents, 374 said that there should be such a law.

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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.

How large a sample size is needed to be 95% confident with a margin of error of E?

A sample size of n is needed, and n is found when M = E.

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

E = 1.96\sqrt{\frac{0.3696*0.6304}{n}}

E\sqrt{n} = 1.96\sqrt{0.3696*0.6304}

\sqrt{n} = \frac{1.96\sqrt{0.3696*0.6304}}{E}

(\sqrt{n})^2 = (\frac{1.96\sqrt{0.3696*0.6304}}{E})^2

n = (\frac{1.96\sqrt{0.3696*0.6304}}{E})^2

A sample of n = (\frac{1.96\sqrt{0.3696*0.6304}}{E})^2 is needed, in which E is the desired margin of error, as a proportion. If we find a decimal value, we round up to the next whole number.

4 0
3 years ago
18 8/9 - 9 1/6 show your work
Genrish500 [490]

Answer:

8 17/18 or 8.945

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2/2x165/9-55/6x3/3

326/18-165/18

161/18= 8 17/18

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