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3 years ago

(2x-10)(x+3) Simplify each expression

2 answers:

7 0

Use distributive property:
$(a+b)(c+d)+ac+ad+bc+bd$

$(2x-10)(x+3)=2x\cdot x+2x\cdot3-10\cdot x-10\cdot3\\\\=2x^2+6x-10x-30=2x^2-4x-30$

irga5000 [103]3 years ago

5 0

2x^2-4x-30 would be your answer

(2x−10)(x+3)
=(2x+−10)(x+3)
=(2x)(x)+(2x)(3)+(−10)(x)+(−10)(3)
=2x2+6x−10x−30
=2x2−4x−30

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2 years ago

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3 years ago

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Using the normal distribution, it is found that there was a 0.9579 = 95.79% probability of a month having a PCE between $575 and $790.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

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

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The mean and the standard deviation are given, respectively, by:

$\mu = 700, \sigma = 50$ .

The probability of a month having a PCE between $575 and $790 is the p-value of Z when X = 790 subtracted by the p-value of Z when X = 575, hence:

X = 790:

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

$Z = \frac{790 - 700}{50}$

Z = 1.8

Z = 1.8 has a p-value of 0.9641.

X = 575:

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

$Z = \frac{575 - 700}{50}$

Z = -2.5

Z = -2.5 has a p-value of 0.0062.

0.9641 - 0.0062 = 0.9579.

0.9579 = 95.79% probability of a month having a PCE between $575 and $790.

More can be learned about the normal distribution at brainly.com/question/4079902

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