We are interested in determining the probability that a retail store will meet its daily revenue goal of $100. Analysis of sales history indicates that daily demand, D is random and independent of the demand on other days. Assume D follows the distribution below P(D=d) =

0.3, d=0
0.3, d=1
0.2, d=2
0.1, d=3
0.1, d=4

Furthermore, due to a complicated discount structure, the shop has determined that their revenue per day can be modeled as R(s) = â100 cos(20s) + 100 where s is the number of jeans sold that day and R(s) is in dollars. Assume that this retail store will always have enough inventory to meet demand. What is the probability the retail store will generate a revenue of at least 100 dollars?

Assume D follows the distribution below

P(D=d) =

So the probability of getting revenue of 100 dollars is 0.

Step-by-step explanation:

Ans - The cumulative probability for demand is given below

       d01234

P(D=d)0.30.30.20.10.1

C(D=d)0.30.60.80.91.0

Now the Revenue function is

R(s-100cos(20s) 100

As per the question the retail store always has inventory to meet demand D

Now to get the revenue at least 100 dollars ---

1cos(20s) <0

As cos heta is a continuously decreasing function within 0 degree to 180 degree withcos(0)1 and cos 180)-

and cos (90)0.

Hence 20s90

s-4.5

As s is an integer s must be 5 to get revenue of at least 100 dollars.

Now the maximum demand for a particular day is 4.

So the probability of getting revenue of 100 dollars is 0.

the 1st digit is in the 100s n its 1

step-by-step explanation:

3,589÷18 = 199.389

so the first digit of the quotient is in the 100s n its 1