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Ilia_Sergeevich [38]
3 years ago
8

A manufacturer finds that the revenue generated by selling of cortan commodity is given by function R(x)=80x-0.2x^ 2 , where he

maximum reveremany should be manufactured to obtain this moximum units
Mathematics
1 answer:
Nina [5.8K]3 years ago
7 0

The maximum units is 200 and , Total revenue is $8,000

<u>Step-by-step explanation:</u>

Here we have , A manufacturer finds that the revenue generated by selling of cortan commodity is given by function R(x)=80x-0.2x^ 2 , where he maximum reveremany should be manufactured to obtain this maximum units .Let's find out:

We have following function as R(x)=80x-0.2x^ 2 . Let's differentiate this and equate it to zero to find value of x for which the function is maximum!

⇒ R(x)=80x-0.2x^ 2

⇒ \frac{d(R(x))}{dx}=\frac{d(80x-0.2x^ 2)}{dx}

⇒ \frac{d(R(x))}{dx}=\frac{d(80x)}{dx}-\frac{d(0.2x^ 2)}{dx}}

⇒ 0=80-2x(0.2)

⇒ \frac{80}{0.4}=x

⇒ x=200

Now , Value of function at x=200 is :

⇒ R(200)=80(200)-0.2(200)^ 2

⇒ R(200)=16000-8000

⇒ R(200)=8000

Therefore , The maximum units is 200 and , Total revenue is $8,000

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A distribution of values is normal with a mean of 60 and a standard deviation of 16. From this distribution, you are drawing sam
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Step-by-step explanation:

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Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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.

A distribution of values is normal with a mean of 60 and a standard deviation of 16.

This means that \mu = 60, \sigma = 16

Samples of size 25:

This means that n = 25, s = \frac{16}{\sqrt{25}} = 3.2

Find the interval containing the middle-most 76% of sample means.

Between the 50 - (76/2) = 12th percentile and the 50 + (76/2) = 88th percentile.

12th percentile:

X when Z has a p-value of 0.12, so X when Z = -1.175.

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

By the Central Limit Theorem

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

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X - 60 = -1.175*3.2

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88th percentile:

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

1.175 = \frac{X - 60}{3.2}

X - 60 = 1.175*3.2

X = 63.76

The interval containing the middle-most 76% of sample means is between 56.24 and 63.76.

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