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4 years ago

13

Product of (-2a^2+s)(5d^2-6s)

1 answer:

mr_godi [17]4 years ago

5 0

$(-2a^2+s)(5d^2-6s)=\\-10a^2d^2+12a^2s+5d^2s-6s^2$

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The Smith Family went out to to dinner.

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3 years ago

Given parallelogram ABCD, find x, y, length AB and Length BC.

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4 years ago

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In a sample of 60 electric motors, the average efficiency (in percent) was 85 and the standard deviation was 2. Section 05.01 Ex

jasenka [17]

Answer:

95% confidence interval for the mean efficiency is [84.483 , 85.517].

Step-by-step explanation:

We are given that in a sample of 60 electric motors, the average efficiency (in percent) was 85 and the standard deviation was 2.

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

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

where, $\mu$ = sample average efficiency = 85

$\sigma$ = sample standard deviation = 2

n = sample of motors = 60

$\mu$ = population mean efficiency

<em>So, 95% confidence interval for the mean efficiency, </em> $\mu$ <em> is ;</em>

P(-2.0009 < $t_5_9$ < 2.0009) = 0.95

P(-2.0009 < $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ < 2.0009 ) = 0.95

P( $-2.0009 \times {\frac{s}{\sqrt{n} }$ < ${\bar X - \mu}$ < $2.0009 \times {\frac{s}{\sqrt{n} }$ ) = 0.95

P( $\bar X -2.0009 \times {\frac{s}{\sqrt{n} }$ < $\mu$ < $\bar X +2.0009 \times {\frac{s}{\sqrt{n} }$ ) = 0.95

<u>95% confidence interval for</u> $\mu$ = [ $\bar X -2.0009 \times {\frac{s}{\sqrt{n} }$ , $\bar X +2.0009 \times {\frac{s}{\sqrt{n} }$ ]

= [ $85 -2.0009 \times {\frac{2}{\sqrt{60} }$ , $85 +2.0009 \times {\frac{2}{\sqrt{60} }$ ]

= [84.483 , 85.517]

Therefore, 95% confidence interval for the population mean efficiency is [84.483 , 85.517].

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3 years ago