How to solve 8 to the power of negative one third

Mathematics

1 answer:

wariber [46]3 years ago

8 0

$8^{-\frac{1}{3}}=\dfrac{1}{8^\frac{1}{3}}=\dfrac{1}{\sqrt[3]8}=\dfrac{1}{2}\\\\Used:\\\\a^{-n}=\dfrac{1}{a^n}\\\\a^\frac{1}{n}=\sqrt[n]{a}\\------------------------\\Other\ method\\\\8=2^3\\\\8^{-\frac{1}{3}}=(2^3)^{-\frac{1}{3}}=2^{3\cdot\left(-\frac{1}{3}\right)}=2^{-1}=\dfrac{1}{2}\\\\Used:\\\\(a^n)^m=a^{n\cdot m}\\\\a^{-n}=\dfrac{1}{a^n}$

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Dafna1 [17]

Answer:

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Step-by-step explanation:

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.

In this problem, we have that:

$\mu = 4.9, \sigma = 1.4$