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Marianna [84]
3 years ago
7

Simplify the following expression: 2x − 6y + 3x2 + 7y − 14x. 3x2 + 12x + y 3x2 − 12x − y 3x2 − 12x + y 3x2 − 12x − 13y

Mathematics
2 answers:
zheka24 [161]3 years ago
8 0
-6x-13y 
2x+(3x2)-14x+(3x2)+12x+(3x2)-12x+(3x2)-12x+(3x2)-12x
2x+6x-14x+6x+12x+6x-12x+6x-12x+6x-12x=-6
-6y+7y+1y-1y+1y-13y=-13y
if a letter is by itself then there is automatically a one in front of it
ra1l [238]3 years ago
8 0
<span>3x2 − 12x + y  would be your answer 

</span>
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Answer:

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The area of a right triangle is 270 square meters. The height of the right triangle is 15 meters. What is the length of the hypo
soldi70 [24.7K]

Answer:

\sqrt{549} ≈ 23.4

Step-by-step explanation:

A = bh

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Step-by-step explanation:

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A set of test scores is normally distributed with a mean of 130 and a standard deviation of 30 What score is necessary to reach
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A score of 150.25 is necessary to reach the 75th percentile.

Step-by-step explanation:

Problems of normal distributions can be 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.

A set of test scores is normally distributed with a mean of 130 and a standard deviation of 30.

This means that \mu = 130, \sigma = 30

What score is necessary to reach the 75th percentile?

This is X when Z has a pvalue of 0.75, so X when Z = 0.675.

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

0.675 = \frac{X - 130}{30}

X - 130 = 0.675*30

X = 150.25

A score of 150.25 is necessary to reach the 75th percentile.

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Answer:

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Step-by-step explanation:

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