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Art [367]
3 years ago
8

3(6k+5)+3(6k+4 The expression has to be equivalent.

Mathematics
2 answers:
DerKrebs [107]3 years ago
7 0
Hey
I’m not sure that’s correct, but here it is!

romanna [79]3 years ago
5 0

Step-by-step explanation:

3(6k+5)+ 3(6k+4)

open the bracket

18k+15+18k+12=0

collect like terms

18k+18k+15+12=0

36k + 27 =0

36k=-27

k=-27/-36

divide through by 3

k= -9/-12

still divide through by 3

k=-3/-4

minus cancel minus

k=3/4

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3 years ago
If IQ scores are normally distributed with a mean of 100 and a standard deviation of 5, what is the probability that a person se
notsponge [240]

Answer:

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

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 higher?

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

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

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Z = 2

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

5 0
3 years ago
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