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Montano1993 [528]
3 years ago
13

Use the distributive property to multiply -7 (2x - 4) I think it's -14x + 28

Mathematics
2 answers:
SashulF [63]3 years ago
5 0
-14x+28 is correct. Good job!
timofeeve [1]3 years ago
4 0
It is-14x + 28 way to go!
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A number is chosen at random from the set of two digits numbers from 10 to 99 inclusive. what is the probability the number cont
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Answer:

\frac{1}{5}

Step-by-step explanation:

There are 99 - 10 + 1 = 90 numbers between 10 and 99

There are 18 numbers that have at least one two digit.

\frac{18}{90}=\frac{1}{5}

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What is... 8/9 x 7/8
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0.7 repeating is the answer
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henry and 28 classmates go to the roller skating ring. each van can hold 11 students. if all of the vans are full except one, ho
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A tobacco company claims that the amount of nicotene in its cigarettes is a random variable with mean 2.2 and standard deviation
Aleksandr-060686 [28]

Answer:

0% probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true.

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 random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 2.2, \sigma = 0.3, n = 100, s = \frac{0.3}{\sqrt{100}} = 0.03

What is the approximate probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true?

This is 1 subtracted by the pvalue of Z when X = 3.1. So

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

By the Central Limit Theorem

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

Z = \frac{3.1 - 2.2}{0.03}

Z = 30

Z = 30 has a pvalue of 1.

1 - 1 = 0

0% probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true.

4 0
4 years ago
Evaluate (Two-fifths) cubed. Check all that apply.
Daniel [21]

Answer:

B.The expanded form is (Two-fifths) (two-fifths) (two-fifths).

E.Two-fifths)cubed = StartFraction 8 Over 125 EndFraction

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3 years ago
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