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Valentin [98]
3 years ago
5

A company is trying to determine how many Valentine’s Day cards to produce. The cost of printing x cards is $2 million +0.70x. F

or example, printing 1 million cards costs $2.7 million. Weekly demand for cards follows a normal random variable with a mean of 1.5 million and a standard deviation of 200,000 for a week before the day and a normal random variable with a mean of 0.5 million and a standard deviation of 300,000 for the week before that. The cards are sold for $4.00, and leftover cards have a value of $0.05. With the precision of 0.2 million, determine how many cards should the company produce to maximize its profit.
Mathematics
1 answer:
vladimir1956 [14]3 years ago
4 0
Jiajianka+b=cbunaianiajianja ok
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Anit [1.1K]

Answer:

Modification has not reduced the number of accidents.

Step-by-step explanation:

H_0: \mu\geq0\\H_a:\mu

                                 A  B  C   D   E   F  G   H

Before modification 5  7  6   4    8   9   8   10

After modification    3  7  7   0    4   6   8    2

Difference                -2 0  1   -4   -4   -3  0   -8

Mean of differences M_d = \frac{sum}{8}=\frac{2+0-1+4+4+3-0+8}{8}=2.5

Standard deviation of differences = \sqrt{\frac{\sum(x-\bar{x})^2}{n}}=2.9277

Formula of paired t test :

t=\frac{M_d}{\frac{s}{\sqrt{n}}}\\t=\frac{2.5}{\frac{2.9277}{\sqrt{8}}}\\t = 2.4152

Df = n-1 = 8-1 =7

t critical = t_{(df, \alpha)}=t_{7,0.01}=2.998

t critical> t calculated

So, We failed to reject null hypothesis

Hence modification has not reduced the number of accidents.

6 0
3 years ago
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8 0
4 years ago
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Akimi4 [234]

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Step-by-step explanation:

8 0
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