All categories

3 years ago

(5/8)^4 Find the value of the power.

Mathematics

1 answer:

Nimfa-mama [501]3 years ago

8 0

Answer:

$625/4096$

Step-by-step explanation:

<u>Let's simplify the problem.</u>

=> $(5/8)^4$
=> $(5/8) (5/8)(5/8)(5/8)$
=> $(625/4096)$
=> $625/4096$

Hence, the solution to the problem is 625/4096.

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3 years ago

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Answer:

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Step-by-step explanation:

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ehidna [41]

Answer:

$P(X>1000)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X>1000)=P(\frac{X-\mu}{\sigma}>\frac{1000-\mu}{\sigma})=P(Z>\frac{1000-950}{50})=P(z>1)$

And we can find this probability using the complement rule and the normal standard table and we got:

$P(z>1)=1-P(z$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution problem

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

$X \sim N(950,50)$

Where $\mu=950$ and $\sigma=50$

We are interested on this probability

$P(X>1000)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X>1000)=P(\frac{X-\mu}{\sigma}>\frac{1000-\mu}{\sigma})=P(Z>\frac{1000-950}{50})=P(z>1)$

And we can find this probability using the complement rule and the normal standard table and we got:

$P(z>1)=1-P(z$

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3 years ago

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liubo4ka [24]

Answer:

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Step-by-step explanation:

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