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3 years ago

What is the ratio between the LCM of 180 and 594 and the GCF of 180 and 594

1 answer:

5 0

$180:2=90\\90:2=45\\45:3=15\\15:3=5\\5:5=1\\\\180=\boxed{2}\cdot2\cdot\boxed{3}\cdot\boxed{3}\cdot5\\\\594:2=297\\297:3=99\\99:3=3\\33:3=11\\11:11=1\\\\594=\boxed{2}\cdot\boxed{3}\cdot\boxed{3}\cdot3\cdot11$

$LCM(180,\ 594)=\boxed{2}\cdot\boxed{3}\cdot\boxed{3}\cdot2\cdot3\cdot5\cdot11=5940\\\\GCF(180,\ 594)=\boxed{2}\cdot\boxed{3}\cdot\boxed{3}=18$

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podryga [215]

Answer: $6.24

Step-by-step explanation:

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2 years ago

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Consider the probability that greater than 94 out of 153 people will not get the flu this winter. Assume the probability that a

Hatshy [7]

Answer:

0.8212 = 82.12% probability that greater than 94 out of 153 people will not get the flu this winter.

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:

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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:

$p = 0.65, n = 153$ . So

$\mu = E(X) = 153*0.65 = 99.45$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{153*0.65*0.35} = 5.9$

Consider the probability that greater than 94 out of 153 people will not get the flu this winter

This probability is 1 subtracted by the pvalue of Z when X = 94. So

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

$Z = \frac{94 - 99.45}{5.9}$

$Z = -0.92$

$Z = -0.92$ has a pvalue of 0.1788

1 - 0.1788 = 0.8212

0.8212 = 82.12% probability that greater than 94 out of 153 people will not get the flu this winter.

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3 years ago

The manager of a warehouse would like to know how many errors are made when a product’s serial number is read by a bar-code read

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