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nika2105 [10]
3 years ago
7

The fifth root of x to the fourth power times the fifth root of x to the fourth power

Mathematics
2 answers:
drek231 [11]3 years ago
8 0
Remember
(√a)(√b)=√(ab)
and
(x^m)(x^n)=x^{m+n}
so

( \sqrt[5]{x^4})( \sqrt[5]{x^4}  )=
\sqrt[5]{(x^4)(x^4)}
4+4=8
\sqrt[5]{x^8}=
( \sqrt[5]{x^5})( \sqrt[5]{x^3}  )=
x\sqrt[5]{x^3}

valentinak56 [21]3 years ago
8 0
5th root of x to the 4th times 5th root of x to the 4th = 5th root of x to the 8th
Reduces to x times 5th root of x to the 3rd
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Which is the equation of the line with slope 0 passing through the point (-3,-1)?
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Suppose a batch of metal shafts produced in a manufacturing company have a population standard deviation of 1.3 and a mean diame
lbvjy [14]

Answer:

54.86% probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.1 inches

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.

In this problem, we have that:

\mu = 208, \sigma = 1.3, n = 60, s = \frac{1.3}{\sqrt{60}} = 0.1678

What is the probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.1 inches

Lesser than 208 - 0.1 = 207.9 or greater than 208 + 0.1 = 208.1. Since the normal distribution is symmetric, these probabilities are equal, so we find one of them and multiply by 2.

Lesser than 207.9.

pvalue of Z when X = 207.9. So

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

By the Central Limit Theorem

Z = \frac{207.9 - 208}{0.1678}

Z = -0.6

Z = -0.6 has a pvalue of 0.2743

2*0.2743 = 0.5486

54.86% probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.1 inches

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3 years ago
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