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Daniel [21]
3 years ago
14

Options A. 1/49 B.1/14 C.49 D.14

Mathematics
1 answer:
grigory [225]3 years ago
5 0

\bf ~\hspace{7em}\textit{negative exponents} \\\\ a^{-n} \implies \cfrac{1}{a^n} ~\hspace{4.5em} a^n\implies \cfrac{1}{a^{-n}} ~\hspace{4.5em} \cfrac{a^n}{a^m}\implies a^na^{-m}\implies a^{n-m} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ 7^{-\frac{5}{6}}\cdot 7^{-\frac{7}{6}}\implies 7^{-\frac{5}{6}-\frac{7}{6}}\implies 7^{\frac{-5-7}{6}}\implies 7^{\frac{-12}{6}}\implies 7^{-2}\implies \cfrac{1}{7^2}\implies \cfrac{1}{49}

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Step-by-step explanation:

Let p be the population proportion of 8th graders are involved with some type of after school activity.

As per given , we have

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Significance level : \alpha= 1-0.98=0.02

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Then, the 98% confidence interval that estimates the proportion of them that are involved in an after school activity will be :-

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

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i.e. \approx0.84\pm 0.085

i.e. (0.84- 0.085,\ 0.84+ 0.085)=(0.755,\ 0.925)

Hence, the 98% confidence interval that estimates the proportion of them that are involved in an after school activity : a) (0.755, 0.925)

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3 years ago
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Some information is missing for #6.
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Data provided in the question:

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