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mario62 [17]
2 years ago
7

Demand per hour for gasoline at a local station is normally distributed with a mean of 875 gallons and std deviation of 55 gallo

ns. What is the probability that demand is greater than 1800 gallons over a 2-hour period
Business
1 answer:
marusya05 [52]2 years ago
5 0

The probability that demand is greater than 1800 gallons over a 2 hour period is : 0.5

<u>Given data :</u>

Mean value of gasoline per hour = 875 gallons

Standard deviation = 55 gallons

<h3>Determine the probability of demand being greater than 1800 gallons over 2 hours </h3>

Demand for gas in 1 hour = X₁

Demand for gas in 2 hours = X₁ + X₂

Therefore ; ( X₁ + X₂) ~ N ( u₁+u₂, sd₁² + sd₂² )

In order to  calculate probabilities for normals apply the equation below

Z = ( X- u ) / sd

where : u = 1800, sd = √ ( 55² + 55² ) = 77.78

using the z-table

P( Y > 1800) = P( Z > ( 1800 - 1800 ) / 77.78)

                    = P( Z>0 ) = 0.5

Hence we can conclude that The probability that demand is greater than 1800 gallons over a 2 hour period is : 0.5.

Learn more about probability : brainly.com/question/24756209

#SPJ1

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The information is incomplete, but we can assume that the machine was sold at the fifth year for an X amount of money, so we should prepare the journal records. Since we are not given the sales amount, I will just use any number, like $50,000. You can adjust the calculation depending on the exact sales amount.

Explanation:

January 2, Year 1, purchase of machine:

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January 3, Year 1, additional expenses needed to put machine into service (electric wiring):

Dr Machinery 10,000

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Dr Machinery 2,000

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Accumulated depreciation during 5 years = $23,120 x 5 = $115,600, carrying value = $156,000 - $115,600 = $40,400

If the machine is sold at $50,000, the journal entries should be:

December 31, year 5, machine is sold:

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Answer:

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This can be seen in the example above.

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