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Greeley [361]
3 years ago
13

Which expression shows the result of applying the distributive property to 14(−3n+32) ?

Mathematics
1 answer:
shusha [124]3 years ago
4 0

Answer:

-3/4n +3/8

Step-by-step explanation:

1/4(−3n+3/2)

Distribute the 1/4

1/4*-3n + 1/4*3/2

-3/4n +3/8

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Find the distance between the points (1, 3) and (-7, 2).<br> V65<br> V37<br> V17
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Answer:

\sqrt{65}

Step-by-step explanation:

distance between 2 points:

d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}\\ \\d=\sqrt{(-7-1)^{2}+(2-1)^{2}}\\\\d=\sqrt{(-8)^{2}+(1)^{2}}\\\\d=\sqrt{65}

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Square root of 80 minus the square root of 5
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3 years ago
The defect length of a corrosion defect in a pressurized steel pipe is normally distributed with mean value 28 mm and standard d
Misha Larkins [42]

Answer:

14.69% probability that defect length is at most 20 mm

Step-by-step explanation:

When the distribution is normal, we use 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 = 28, \sigma = 7.6

What is the probability that defect length is at most 20 mm

This is the pvalue of Z when X = 20. So

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

Z = \frac{20 - 28}{7.6}

Z = -1.05

Z = -1.05 has a pvalue of 0.1469

14.69% probability that defect length is at most 20 mm

5 0
4 years ago
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