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kondor19780726 [428]
3 years ago
9

A store had 4 boxes of video games. How many days in total would it take to sell all of the games if each day they sold 3/5 of a

box?
Mathematics
2 answers:
Klio2033 [76]3 years ago
7 0

Answer:

The answer would be 20. Since the store sells 1/5th of a box each day, it takes 5 days to sell one box. There are four boxes, so multiply 5 by 4 and you get 20.

Step-by-step explanation:

fomenos3 years ago
3 0
Ya mom is very fat ugly poop
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Step-by-step explanation:

Using the rules of exponents

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F(x)=4x+2 find f(8)
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Consider the probability that greater than 96 out of 153 DVDs will work correctly. Assume the probability that a given DVD will
ICE Princess25 [194]

Answer:

0.3594 = 35.94% probability that greater than 96 out of 153 DVDs will work correctly.

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

n = 153, p = 0.62

So

\mu = E(X) = np = 153*0.62 = 94.86

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{153*0.62*0.38} = 6

Probability that greater than 96 out of 153 DVDs will work correctly.

97 or more DVDs, which is 1 subtracted by the pvalue of Z when X = 97. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{97 - 94.86}{6}

Z = 0.36

Z = 0.36 has a pvalue of 0.6406

1 - 0.6406 = 0.3594

0.3594 = 35.94% probability that greater than 96 out of 153 DVDs will work correctly.

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