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Inessa05 [86]
3 years ago
11

Marti brought 20 juice boxes to the party. The guests drank 6/8 of the juice boxes.

Mathematics
2 answers:
ololo11 [35]3 years ago
7 0
15. 

You simply multiply the ratio by the total number. 
emmainna [20.7K]3 years ago
4 0
The guest got 15 juice boxes a piece
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Assume that adults have IQ scores that are normally distributed with a mean of 96 and a standard deviation of 15.7. Find the pro
astraxan [27]

Answer:

4.05% probability that a randomly selected adult has an IQ greater than 123.4.

Step-by-step explanation:

Problems of normally distributed samples are 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.

In this question, we have that:

\mu = 96, \sigma = 15.7

Probability that a randomly selected adult has an IQ greater than 123.4.

This is 1 subtracted by the pvalue of Z when X = 123.4. So

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

Z = \frac{123.4 - 96}{15.7}

Z = 1.745

Z = 1.745 has a pvalue of 0.9595

1 - 0.9595 = 0.0405

4.05% probability that a randomly selected adult has an IQ greater than 123.4.

7 0
3 years ago
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