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3 years ago

Number 10 only please and thank u

Mathematics

1 answer:

lapo4ka [179]3 years ago

4 0

$\bf \stackrel{\textit{small cabinets}}{11.5}\qquad \qquad \stackrel{\textit{large cabinets 2.3 times as long}}{2.3(11.5)\implies 26.45} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{board's length in feet}}{10\frac{1}{2}\implies 10.5}\implies \stackrel{\textit{length in inches}}{10.5(12)\implies 126} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{how many times does 26.45 go into 126?}}{126\div 26.45 ~~\approx ~~4.76}~\hfill \boxed{\stackrel{\textit{whole cabinets only}}{4}}$

bearing in mind that, a foot = 12 inches.

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Martha's mom is making party bags for all of her guests. She has 24 green gumballs and 33 blue gumballs. If she splits them even

Dmitry_Shevchenko [17]

Answer:

3 bags

Step-by-step explanation:

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3 years ago

The manufacturer of an electronic device claims that the probability of the device failing during the warranty period is 0.005.

Zielflug [23.3K]

Answer:

The value is $P(X \ge 15) = 0.5$

Step-by-step explanation:

From the question we are told that

The probability of the device failing during the warranty period is $p = 0.005$

The sample size is $n = 3000$

The random variable considered is x = 15

Generally this is distribution is binomial given the fact that there is only two out comes hence

X which is a variable representing a randomly selected selected electronic follows a binomial distribution i.e

$X \~ \ B(n , p)$

Now the mean is mathematically evaluated as

$\mu = n * p$

=> $\mu = 3000 * 0.005$

=> $\mu =15$

The standard deviation is mathematically represented as

$\sigma = \sqrt{np(1 -p )}$

=> $\sigma = \sqrt{3000 * 0.005 * (1 - 0.005 )}$

=> $\sigma = 3.86$

Now given that n is very large, then it mean that we can successfully apply normal approximation on this binomial distribution

$P(X \ge 15) = P( \frac{X - \mu}{\sigma } \ge \frac{x - \mu}{\sigma } )$

Now applying Continuity Correction we have

$P(X \ge (15-0.5)) = P( \frac{X - \mu}{\sigma } > \frac{(15 -0.5) - 15}{3.86 } )$

Generally $\frac{X - \mu}{\sigma } = Z (The \ standardized \ value \ of \ X)$

$P(X \ge (15-0.5)) = P(Z >-0.130 )$

From the z-table

$P(Z >-0.130 ) =0.5$

Thus

$P(X \ge 15) = 0.5$

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