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yulyashka [42]
3 years ago
10

Simplify the expression ((81y^-8)/(16x^20))^-1/4

Mathematics
1 answer:
Darina [25.2K]3 years ago
6 0
\bf ~~~~~~~~~~~~\textit{negative exponents}
\\\\
a^{-n} \implies \cfrac{1}{a^n}
\qquad \qquad
\cfrac{1}{a^n}\implies a^{-n}
\qquad \qquad 
a^n\implies \cfrac{1}{a^{-n}}
\\\\
-------------------------------

\bf \left( \cfrac{81y^{-8}}{16x^{20}} \right)^{-\frac{1}{4}}\implies \left( \cfrac{16x^{20}}{81y^{-8}} \right)^{\frac{1}{4}}\implies \left( \cfrac{2^4x^{20}}{3^4y^{-8}} \right)^{\frac{1}{4}}
\\\\\\
\textit{and now we'll distribute the exponent}\qquad \left( \cfrac{2^{4\cdot \frac{1}{4}}x^{20\cdot \frac{1}{4}}}{3^{4\cdot \frac{1}{4}}y^{-8\cdot \frac{1}{4}}} \right)
\\\\\\
\cfrac{2^1x^5}{3^1y^{-2}}\implies \cfrac{2^1x^5y^2}{3^1}\implies \cfrac{2x^5y^2}{3}
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There were 800 math instructors at a mathematics convention. Forty instructors were randomly selected and given an IQ test. The
Leya [2.2K]

Answer:

95% confidence interval for the mean of the 800 instructors is [126.80 , 133.20].

Step-by-step explanation:

We are given that there were 800 math instructors at a mathematics convention.

Forty instructors were randomly selected and given an IQ test. The scores produced a mean of 130 with a standard deviation of 10.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

                           P.Q. = \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean score = 130

             s = sample standard deviation = 10

             n = sample of instructors = 40

             \mu = population mean of 800 instructors

<em>Here for constructing 95% confidence interval we have used One-sample t test statistics as we know don't about population standard deviation.</em>

So, 95% confidence interval for the population mean, \mu is ;

P(-2.0225 < t_3_9 < 2.0225) = 0.95  {As the critical value of t at 39 degree of

                                         freedom are -2.0225 & 2.0225 with P = 2.5%}  

P(-2.0225 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 2.0225) = 0.95

P( -2.0225 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 2.0225 \times {\frac{s}{\sqrt{n} } } ) = 0.95

P( \bar X-2.0225 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+2.0225 \times {\frac{s}{\sqrt{n} } } ) = 0.95

<u><em /></u>

<u><em>95% confidence interval for</em></u> \mu = [ \bar X-2.0225 \times {\frac{s}{\sqrt{n} } } , \bar X+2.0225 \times {\frac{s}{\sqrt{n} } } ]

                   = [ 130-2.0225 \times {\frac{10}{\sqrt{40} } } , 130+2.0225 \times {\frac{10}{\sqrt{40} } } ]

                   = [126.80 , 133.20]

Therefore, 95% confidence interval for the mean of the 800 instructors is [126.80 , 133.20].

5 0
3 years ago
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