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3 years ago

5

Evaluate this expression

lign="absmiddle" class="latex-formula">

2 answers:

shepuryov [24]3 years ago

6 0

Answer:

0.015625

Step-by-step explanation:

Rule: $a^{-n} = \dfrac{1}{a^n}$

$8^{-2} = \dfrac{1}{8^2} = \dfrac{1}{64} = 0.015625$

Gwar [14]3 years ago

6 0

0.015625 hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

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Expand and simplify (x - 2)(x - 1)

Whitepunk [10]

Answer:

x^2 -3x+2

Step-by-step explanation:

(x - 2)(x - 1)

FOIL

first: x*x = x^2

outer: -1*x = -x

inner: -2*x =-2x

Last: -2*-1 =2

Add together

x^2 -x-2x+2

Combine like terms

x^2 -3x+2

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3 years ago

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Step-by-step explanation:

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3 years ago

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Step-by-step explanation:

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3 years ago

Store A uses the newsvendor model to manage its inventory. Demand for its product is normally distributed with a mean of 500 and

ra1l [238]

Answer:

$P(X$

Step-by-step explanation:

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".

Let X the random variable that represent the Demand for its product on this case, and for this case we know the distribution for X is given by:

$X \sim N(\mu=500,\sigma=300)$

And let $\bar X$ represent the sample mean, the distribution for the sample mean is given by:

$\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})$

What is its in-stock probability if Store A’s order quantity is 800 units?

We are looking for this probability:

What is its in-stock probability if Store A’s order quantity is 800 units?

So we can find the following values:

$P(X>800)$ and $P(X$

Sor this problem we can use the z score formula given by:

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

If we find the z score for the value 800 we got:

$z=\frac{800-500}{300}=1$

And if we find:

$P(X$

And by the complement rule:

$P(X>800)=1-P(X$

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3 years ago