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-BARSIC- [3]
3 years ago
11

Which of the following number sentences is an example of the Distributive Property?

Mathematics
1 answer:
leonid [27]3 years ago
7 0

Answer:

d. 7 × (10 + 9) = (7 × 10) + (7 × 9)

Step-by-step explanation:

The format of writing numbers in distributive property is a(b+c) = (a×b)+(a×c)

We can see that in choice d. 7 × (10 + 9) = (7 × 10) + (7 × 9) it is in correct format of writing numbers in distributive property. Thus, choice d. is correct.

Verification:-

a(b+c) = (a×b)+(a×c)

7 × (10 + 9) = (7 × 10) + (7 × 9)

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Dominic puts his cars in 6 rows with 5 in each row. His sister changes them into 3 rows with the same number in each row. How ma
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3 years ago
Please solve the math question: |4x−4(x+1)|=4
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x has infinite solutions

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|4x−4(x+1)|=4

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|4x−4x+4|=4

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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
3 years ago
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