The auto parts department of an automotive dealership sends out a mean of 6.3 special orders daily. What is the probability that, for any day, the number of special orders sent out will be exactly 3

0.0765 = 7.65% probability that, for any day, the number of special orders sent out will be exactly 3

Step-by-step explanation:

We have the mean, which means that the poisson distribution is used to solve this question.

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

In which

x is the number of sucesses

e = 2.71828 is the Euler number

is the mean in the given interval.

The auto parts department of an automotive dealership sends out a mean of 6.3 special orders daily.

This means that

What is the probability that, for any day, the number of special orders sent out will be exactly 3?

This is P(X = 3). So

0.0765 = 7.65% probability that, for any day, the number of special orders sent out will be exactly 3

$9.89

step-by-step explanation:

9.90 = p x .06 x .25 (3/12 changed to a decimal)

9.90 = p x 0.015

9.90 - 0.015 = p

9.89 = p

since no one answered yet i will give it a shot, not completely sure tho.

i think it should be y=cos(x+2pi), (third answer, answer c)

good luck.