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Diano4ka-milaya [45]
4 years ago
5

Find the greatest common factor of 6a^2b^2c^2d and 2acd^2 A. acd B. 2acd C. 2acd^2 D. 2a^2b^2c^2d^2

Mathematics
2 answers:
babunello [35]4 years ago
8 0
The greatest common factor of 6a^2b^2c^2d and 2acd^2 is <span>2acd "B."</span>
STALIN [3.7K]4 years ago
5 0
Answer:

<span><span>2acd</span> </span>

Explanation:

Ironically it is the lowest index which is the Greatest common factor. (You cannot get more from a term than what is there.)

Make sure that a particular base is in both terms and use the lowest power for the GCF.

<span><span>2acd</span> </span>
There is no <span>b </span> factor in the second term, so it cannot be common.

The opposite happens when find the LOWEST common multiple . then it is the HIGHEST power which has to be used, because the LCM has to contain the whole of each term

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posledela

Using the <em>normal distribution and the central limit theorem</em>, it is found that the power of the test is of 0.9992 = 99.92%.

<h3>Normal Probability Distribution</h3>

In a normal distribution with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

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

  • It 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, which is the percentile of X.
  • By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation s = \frac{\sigma}{\sqrt{n}}.

In this problem:

  • The mean is \mu = 8.5.
  • The standard deviation is \sigma = 0.87.
  • A sample of 30 is taken, hence n = 30, s = \frac{0.87}{\sqrt{30}} = 0.1588.

The power of the test is given by the probability of a sample mean above 8, which is <u>1 subtracted by the p-value of Z when X = 8</u>, so:

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

By the Central Limit Theorem:

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

Z = \frac{8 - 8.5}{0.1588}

Z = -3.15

Z = -3.15 has a p-value of 0.0008.

1 - 0.0008 = 0.9992.

The power of the test is of 0.9992 = 99.92%.

To learn more about the <em>normal distribution and the central limit theorem</em>, you can check brainly.com/question/24663213

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