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cupoosta [38]
3 years ago
6

Suppose you are starting your own company selling chocolate covered strawberries. You decide to sell the milk chocolate covered

strawberries for a profit of $2.25 $ 2.25 /box and the white chocolate covered strawberries at $2.50 $ 2.50 /box. Market tests and available resources, however, have given you the following constraints. The combined production level should not exceed 800 800 boxes per month. The demand for the white chocolate is no more than half the demand for milk chocolate strawberries. The production level for white chocolate should be less than or equal to 200 200 boxes.
Mathematics
1 answer:
Amanda [17]3 years ago
7 0

Answer:

$1850 per month

Step-by-step explanation

There are two types of chocolates that can be produced milk chocolate and strawberry covered chocolate. To find the profit we make following equation,

P = $2.25 SC + $2.50 WC

where SC is strawberry chocolate and WC is White milk chocolate.

The maximum production level can be 800 boxes per month and white chocolates can not exceed the 200 boxes per month so we assume making 600 boxes of Strawberry covered chocolates and 200 boxes of white chocolates.

Profit = 2.25 * 600 + 2.50 * 200

Profit = $1850

This is the maximum profit that can be earned after making combination of two types of chocolates.

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Paraphin [41]

Answer:

0.3557 = 35.57% probability that one selected subcomponent is longer than 118 cm.

Step-by-step explanation:

Normal Probability Distribution:

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

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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.

Normally distributed with a mean of 116 cm and a standard deviation of 5.4 cm.

This means that \mu = 116, \sigma = 5.4

Find the probability that one selected subcomponent is longer than 118 cm.

This is 1 subtracted by the pvalue of Z when X = 118. So

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