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3 years ago

I need help with this problem

Mathematics

1 answer:

cupoosta [38]3 years ago

7 0

$\bf \begin{cases}
(n^4)^p=n^{12}\to &n^{4\cdot p}=n^{12}\to 4p=12
\\
&\qquad \uparrow \\
&\textit{same bases, thus}\\
&\qquad \downarrow 
\\
n^3\cdot n^q=n^6\to &n^{3+q}=n^6\to 3+q=6
\end{cases}
\\\\\\
thus\implies 
\begin{cases}
4p=12\implies p=\frac{12}{4}
\\\\
3+q=6\implies q=6-3
\end{cases}
\\\\

\\\\
then\implies p\cdot q=\boxed{?}$

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From the question we are told that

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The standard deviation is $\sigma = 0.85$

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$P(\= X \le 3.0 ) = P[\frac{\= X - \mu} {\frac{\sigma }{\sqrt{n} } } \le \frac{3.0 - \mu}{\frac{ \sigma}{\sqrt{n} } } ]$

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$P(\= X \le 3.0 ) = P[Z \le \frac{3.0 - 2.35}{\frac{ 0.85}{\sqrt{25} } } ]$

$P(\= X \le 3.0 ) = P[Z \le 2.06 ]$

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Thus

$P(\= X \le 3.0 ) =0.980$

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$P(2.65 \le \= X \le 3.00) = P [\frac{2.65 - \mu }{ \frac{\sigma }{\sqrt{n} } } < \frac{\= X - \mu }{ \frac{\sigma }{\sqrt{n} } } < \frac{3.0 - \mu }{ \frac{\sigma }{\sqrt{n} } }]$

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$P(2.65 \le \= X \le 3.00) = P[0 < Z< 2.06]$

$P(2.65 \le \= X \le 3.00) = P(Z < 2.06) - P(Z$

From the z-table

$P(2.65 \le \= X \le 3.00) = 0.980 - 0.50$

$P(2.65 \le \= X \le 3.00) = 0.480$

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Generally the critical value of z for a one tail test such as the one we are treating that is under the area 0.99 is $t_z = 2.33$ this is obtained from the critical value table

$t_z = \frac{3.0 - 2.35}{\frac{ 0.85}{\sqrt{n} } }$

$2.33 = \frac{3.0 - 2.35}{\frac{ 0.85}{\sqrt{n} } }$

=> $n = 32$

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