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UkoKoshka [18]
3 years ago
6

Express the confidence interval

Mathematics
1 answer:
marissa [1.9K]3 years ago
7 0

Answer:

\hat p = \frac{0.244+0.326}{2}=0.285

ME = \frac{0.326-0.244}{2}=0.041

0.285 \pm 0.041

Step-by-step explanation:

For this case we have a confidence interval given as a percent:

24.4\% \leq p \leq 32.6\%

If we express this in terms of fraction we have this:

0.244 \leq p \leq 0.326

We know that the confidence interval for the true proportion is given by:

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

And thats equivalent to:

\hat p \pm ME

We can estimate the estimated proportion like this:

\hat p = \frac{0.244+0.326}{2}=0.285

And the margin of error can be estimaed using the fact that the confidence interval is symmetrical

ME = \frac{0.326-0.244}{2}=0.041

And then the confidence interval in the form desired is:

0.285 \pm 0.041

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l^{2}=28^{2}+89^{2}-2 \cdot 28 \cdot 89 \cdot \cos 42^\circ

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77.  \cot^{6} x = \cot^{4} x \csc^{2}x - \cot^{4} xProved

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Step-by-step explanation:

77. Left hand side

= \cot^{6} x

= \cot^{4} x \times \cot^{2} x

= \cot^{4}x [\csc^{2}x - 1]  

{Since we know, \csc^{2} x - \cot^{2}x = 1}

= \cot^{4} x \csc^{2}x - \cot^{4} x  

= Right hand side (Proved)

78. Left hand side

= \sec^{4}x \tan^{2} x

= \sec^{2} x [1 + \tan^{2}x] \tan^{2} x  

{Since \sec^{2}x - \tan^{2}x = 1}

= \sec^{2}x [\tan^{2}x + \tan^{4}x ]

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= \cos^{3} x\sin^{2} x

= \cos x[1 - \sin^{2} x] \sin^{2} x

{Since \sin^{2}x + \cos^{2} x = 1}

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{Since \sin^{2}x + \cos^{2} x = 1}

= 1 - 2\cos^{2} x[1 - \cos^{2}x ]

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