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Marizza181 [45]
3 years ago
9

The daily demand for gasoline at a local gas station is normally distributed with a mean of 1200 gallons, and a standard deviati

on of 350 gallons.
If R is a random number between 0 and 1, then which of the following correctly models daily demand for gasoline?

a) 1200 + 350 R
b) 1200 + 350*NORMSDIST(R)
c) NORM.INV(R, 1200, 350)
d) Both b) and c) are correct.
Mathematics
1 answer:
seropon [69]3 years ago
5 0

Answer:

c) NORM.INV(R, 1200, 350)

Step-by-step explanation:

Given that the daily demand for gasoline at a local gas station is normally distributed with a mean of 1200 gallons, and a standard deviation of 350 gallons.

X = demand for gasolene at a local gas station is N(1200, 350)

R is any random number between 0 and 1.

Daily demand for gasolene would be

X = Mean + std deviation * z value, where Z = normal inverse of a value between 0 and 1.

The norm inv (R, 1200, 350) for R between 0 and 1 gives all the values of X

Hence correct choice would be

Option c) NORM.INV(R, 1200, 350)

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Answer:

Step-by-step explanation:

in order to solve this system of equation we would say that let

3x+2y=-8........................................................ equation 1

1 x - y = -1​ ........................................................ equation 2

from equation 2

1 x - y = -1​ ........................................................ equation 2

x = -1 + y....................................... equation 3

substitute for equation 3 in equation 1

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3( -1 + y) + 2y = 8

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collect the like terms

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divide both sides by the coefficient of y

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

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