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irina [24]
3 years ago
8

A company sells boxes of duck calls (d) for $35 and boxes of turkey calls (t) for $45. they make batches of duck calls that fill

6 boxes and batches of turkey calls that fill 8 boxes. the company only has 42 boxes. which system of equations helps the company plan to make $300? 6d = 42 − 8t35d + 45t = 300 d = 42 − t6d + 8t = 42 d = 42 − t35d + 45t = 300 6d = 42 − 8t35d − 45t = 300
Mathematics
1 answer:
SIZIF [17.4K]3 years ago
8 0

Answer:

Option A is the correct choice.

Step-by-step explanation:

Let d be the number of boxes of duck calls and t be the number of boxes of turkey calls.

We have been given that a company sells boxes of duck calls for $35 and boxes of turkey calls (t) for $45, so the revenue earned from selling d boxes of duck and t boxes of turkey call will be 35d and 45t respectively.

Further, the company plan to make $300. We can represent this information as:

35d+45t=300...(1)

We are also told that they make batches of duck calls that fill 6 boxes and batches of turkey calls that fill 8 boxes. the company only has 42 boxes. We can represent this information as:

6d+8t=42...(2)

6d=42-8t...(2)

Therefore, our desired system of equation will be:

35d+45t=300...(1)

6d=42-8t...(2)  


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Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples can be 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:

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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.

Z-scores lower than -2 or higher than 2 are considered unusual.

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The Central Limit Theorem estabilishes that, for a random normally distributed variable X, with mean \mu and standard deviation \sigma, the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

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Which of the following mean costs would be considered unusual?

We have to find the z-score for each of them

A. $6350

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

By the Central Limit Theorem

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

Z = \frac{6350 - 5850}{251.56}

Z = 1.99

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

Z = \frac{6180 - 5850}{251.56}

Z = 1.31

Not unusual

C. $5180

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

Z = \frac{5180 - 5850}{251.56}

Z = -2.66

Unusual, and this is the answer.

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