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valkas [14]
2 years ago
5

Which expressions are equivalent to 5 x minus 15? Select three options. 5 (x + 15) 5 (x minus 3) 4 x + 3 y minus 15 minus 3 y +

x Negative 7 y minus 6 x minus 8 y + x Negative 20 minus 3 x + 5 + 8 x
Mathematics
2 answers:
fgiga [73]2 years ago
8 0

Answer:

ion ima guess it is the second choice.

Step-by-step explanation:

Aleksandr [31]2 years ago
6 0

Answer:

B) 5 (x minus 3)

C) 4 x + 3 y minus 15 minus 3 y + x

E) Negative 20 minus 3 x + 5 + 8 x

tep-by-step explanation:

5x minus 15

5x - 15

5(x - 3)

4 x + 3 y minus 15 minus 3 y + x

4x + 3y - 15 - 3y + x

5x - 15

Negative 20 minus 3 x + 5 + 8 x

-20 - 3x + 5 + 8x

5x - 15

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Answer:

To use a Normal distribution to approximate the chance the sum total will be between 3000 and 4000 (inclusive), we use the area from a lower bound of 3000 to an upper bound of 4000 under a Normal curve with its center (average) at 3500 and a spread (standard deviation) of 160 . The estimated probability is 99.82%.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For sums, we have that the mean is \mu*n and the standard deviation is s = \sigma \sqrt{n}

In this problem, we have that:

\mu = 100*35 = 3500, \sigma = \sqrt{100}*16 = 160

This probability is the pvalue of Z when X = 4000 subtracted by the pvalue of Z when X = 3000.

X = 4000

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

By the Central Limit Theorem

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

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Z = \frac{3000 - 3500}{160}

Z = -3.13

Z = -3.13 has a pvalue of 0.0009

0.9991 - 0.0009 = 0.9982

So the correct answer is:

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