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Dimas [21]
2 years ago
9

30×30+2=320×600÷345=answer please​

Mathematics
2 answers:
olasank [31]2 years ago
8 0

Answer:

30×30+2= 902

320×600÷345=556.52173913

Step-by-step explanation:

uysha [10]2 years ago
5 0

Answer:

  1. 902
  2. 556. 521739

hope it helps..

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I need help finding the location of x
KatRina [158]

Answer:

Position of point X on the number line must be X = -27

Step-by-step explanation:

Notice that there are 14 units between W and Y, since the absolute value of their difference is :

| -15-(-29) | = | -15 + 29 | = 14

Now let's write an equation that tells as what is the distance between W and X in terms of WY knowing that we want WX to be 1/7 of WY:

WX=\frac{1}{7} \,*\, 14\\WX= 2

This is telling us that the point X must be located two (2) units to the right of point W. That is W + 2 = -29 + 2 = -27

Position of point X on the number line must be X = -27

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What is 7/15 as a decimal
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Answer:

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Step-by-step explanation:

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Weights of American adults are normally distributed with a mean of 180 pounds and a standard deviation of 8 pounds. What is the
ahrayia [7]

Answer:

15.87% probability that a randomly selected individual will be between 185 and 190 pounds

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 = 180, \sigma = 8

What is the probability that a randomly selected individual will be between 185 and 190 pounds?

This probability is the pvalue of Z when X = 190 subtracted by the pvalue of Z when X = 185. So

X = 190

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

Z = \frac{190 - 180}{8}

Z = 1.25

Z = 1.25 has a pvalue of 0.8944

X = 185

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

Z = \frac{185 - 180}{8}

Z = 0.63

Z = 0.63 has a pvalue of 0.7357

0.8944 - 0.7357 = 0.1587

15.87% probability that a randomly selected individual will be between 185 and 190 pounds

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3 years ago
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