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Ainat [17]
3 years ago
9

Verify the identity. Show your work. (1 + tan^2u)(1 - sin^2u) = 1

Mathematics
1 answer:
yuradex [85]3 years ago
7 0
\bf \textit{Pythagorean Identities}
\\\\
sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)
\\\\ 1+tan^2(\theta)=sec^2(\theta)\\\\
-------------------------------\\\\\
[1+tan^2(u)][1-sin^2(u)]=1
\\\\\\\
[sec^2(u)][cos^2(u)]\implies \cfrac{1}{cos^2(u)}\cdot cos^2(u)\implies \cfrac{cos^2(u)}{cos^2(u)}\implies 1
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What sample size is needed to give a margin of error within in estimating a population mean with 95% confidence, assuming a prev
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n = (\frac{1.96\sqrt{\pi(1-\pi)}}{M})^2

The sample size needed is n(if a decimal number, round up to the next integer), considering the estimate of the proportion \pi(if no previous estimate use 0.5) and M is the desired margin of error.

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}}

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Needed sample size:

The needed sample size is n. We have that:

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

M = 1.96\sqrt{\frac{\pi(1-\pi)}{n}}

\sqrt{n}M = 1.96\sqrt{\pi(1-\pi)}

\sqrt{n} = \frac{1.96\sqrt{\pi(1-\pi)}}{M}

(\sqrt{n})^2 = (\frac{1.96\sqrt{\pi(1-\pi)}}{M})^2

n = (\frac{1.96\sqrt{\pi(1-\pi)}}{M})^2

The sample size needed is n(if a decimal number, round up to the next integer), considering the estimate of the proportion \pi(if no previous estimate use 0.5) and M is the desired margin of error.

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