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Ira Lisetskai [31]
3 years ago
5

HELLLPPPPPPPP ASAP WILL GIVE BRAINLIESTTTTTTT ALONG WITH 98 POINTSSSS!!! Make a pie chart.

Mathematics
2 answers:
IgorLugansk [536]3 years ago
7 0
Substitute for x then you will get the answer
weeeeeb [17]3 years ago
6 0

this is real  easy to do just substitue for X you can do this man mark me as brainlest please!!!!!!!

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Compare the investment below to an investment of the same principal at the same rate compounded annually. ​
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How to change a precent to a decimals
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multiple choice in picture
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. If IQ scores are normally distributed with a mean of 100 and a standard deviation of 5, what is the probability that a person
bulgar [2K]

Answer:

2.28% probability that a person selected at random will have an IQ of 110 or greater

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 problem, we have that:

\mu = 100, \sigma = 5

What is the probability that a person selected at random will have an IQ of 110 or greater?

This is 1 subtracted by the pvalue of Z when X = 110. So

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

Z = \frac{110 - 100}{5}

Z = 2

Z = 2 has a pvalue of 0.9772

1 - 0.9772 = 0.0228

2.28% probability that a person selected at random will have an IQ of 110 or greater

5 0
3 years ago
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