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nika2105 [10]
3 years ago
14

Plz help.....................

Mathematics
1 answer:
sdas [7]3 years ago
8 0
Perfect square trinomial means that the constant term is a perfect square, and the middle term is the sum of it's factors.
Option 1: constant  64 sqrt = 8; 8 + 8 = 16 or -8 + -8 = -16 NO
Option 2: constant  36 sqrt = 6; 6 + 6 = 12 or -6 + -6 = -12 NO
Option 3: constant 64 sqrt = 8; 8 + 8 = 16 or -8 + -8 = -16 YES
(x - 8)^2 is the factored form
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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.

\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 of:

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

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

This means that n = 1012, \pi = \frac{374}{1012} = 0.3696

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.

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Answer:

what's the answer

ok

tell

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