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Ivenika [448]
3 years ago
5

A vendor prepares 100.00 hotdogs every day and sells at $20.00/piece. For each hot dog, he spends $12.00 in the raw material. Ad

ditionally he spends $1.00 for packing each hotdog and monthly $50.00, $20.00, $10.00 as food truck rent, electricity and other expenses respectively. Lost sale are taken as $1 per unhappy customer. Leftover hotdogs can be sold for $5.00/piece. On a particular day in June it rained heavily so the vendor was able to sell only 80.00 hot dogs. Determine the vendor’s profit for that day
Mathematics
1 answer:
astra-53 [7]3 years ago
6 0

Answer:

$397.34 (if he sold the 20 leftover hot dogs), $297.34 if he didn't.

Step-by-step explanation:

We are going to assume that a month has 30 days.

  • First, we are going to see how much money the vendor got from selling the 80 hot dogs. He sold 80 hot dogs at 20 dollars/piece = 1600 dollars.

  • We need to subtract the amount of money he spent in each hot dog (12 dollars in raw material plus one dollar for packing): 13 dollars x 100 hot dogs he prepared = 1300 dollars

  • He also spends a total of 80 dollars per month in truck rent, electricity and other expenses. If we divide this by the amount of days per month we have: 80/30 = 2.66

  • The problem doesn't tell us that there were unhappy customers that day so that amount is zero.

  • We are going to assume that the vendor sold the remaining 20 hot dogs at 5 dollars/piece. 20 x 5 = 100.

Thus, the profit for that day is:

1600 - 1300 - 2.66 + 100 = 397.34

<u>(</u><u>Note:</u><u> If the vendor did not sell the leftover hot dogs and he actually only sold 80 hot dogs, then the profit would be: 1600 - 1300 - 2.66 = 297.34)</u>

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antiseptic1488 [7]

Answer:

53.84% probability that for a group of 10 students, the mean number of times they go to the movies each year is between 14 and 18 times.

Step-by-step explanation:

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

Normal probability distribution

Problems of normally distributed samples are 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:

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 pvalue, we get the probability that the value of the measure is greater than X.

Central limit Theorem

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

In this problem, we have that:

\mu = 17, \sigma = 8, n = 10, s = \frac{8}{\sqrt{10}} = 2.53

What is the probability that for a group of 10 students, the mean number of times they go to the movies each year is between 14 and 18 times?

This probability is the pvalue of Z when X = 18 subtracted by the pvalue of Z when X = 14. So

X = 18

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

By the Central Limit Theorem

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

Z = \frac{18 - 17}{2.53}

Z = 0.4

Z = 0.4 has a pvalue of 0.6554

X = 14

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

Z = \frac{14 - 17}{2.53}

Z = -1.19

Z = -1.19 has a pvalue of 0.1170

0.6554 - 0.1170 = 0.5384

53.84% probability that for a group of 10 students, the mean number of times they go to the movies each year is between 14 and 18 times.

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Answer:

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Step-by-step explanation:

Given:

Lucy spends in a week babysitting for 4 hours.

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Now, to find whether Lily babysit for more than 4 hours or less than that.

Number of hours Lucy babysit = 4 hours.

So, to get the hours Lily babysit:

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<em>On simplifying we get:</em>

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Therefore, Lily babysit for less than 4 hours.

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