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adelina 88 [10]
2 years ago
6

Which expressions are equivalent to 5+(-3)(6x-5)5+(−3)(6x−5)5, plus, left parenthesis, minus, 3, right parenthesis, left parenth

esis, 6, x, minus, 5, right parenthesis ? Choose all answers that apply: Choose all answers that apply: (Choice A) A 18x-2018x−2018, x, minus, 20 (Choice B) B 3x-33x−33, x, minus, 3 (Choice C) C None of the above
Mathematics
1 answer:
almond37 [142]2 years ago
8 0

5+(-18x+15)5 + (-18x +15)5

5+ (-90x +75) + (-90x + 75)

-180x +155

let's see the alternatives:

a) -2000x - 2018 -> -200x - 201,8

b) -30x - 33 -> -180x -198

c) none of the above

so, c is the right one

<em>hope it helps :)</em>

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Answer:

0.8665 = 86.65% probability that the sample mean would be at least $39000

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 normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, 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.

Mean $39725 and standard deviation $7320.

This means that \mu = 39725, \sigma = 7320

Sample of 125:

This means that n = 125, s = \frac{7320}{\sqrt{125}} = 654.72

The probability that the sample mean would be at least $39000 is about?

This is 1 subtracted by the pvalue of Z when X = 39000. So

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

By the Central Limit Theorem

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

Z = \frac{39000 - 39725}{654.72}

Z = -1.11

Z = -1.11 has a pvalue of 0.1335

1 - 0.1335 = 0.8665

0.8665 = 86.65% probability that the sample mean would be at least $39000

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