If a customer rolls the dice and rents a second movie every Thursday for 30 consecutive weeks, what is the approximate probability that the total amount paid for these second movies will exceed $15.00?

Complete Question:

Every Thursday, Matt and Dave's Video Venture has "roll-the-dice" day. A customer may choose to roll two fair dice and rent a second movie for an amount (in cents) equal to the numbers uppermost on the dice, with the larger number first. For example, if the customer rolls a two and a four, a second movie may be rented for $0.42. If a two and a two are rolled, a second movie may be rented for $0.22. Let X represent the amount paid for a second movie on roll-the-dice day. The expected value of X is $0.47 and the standard deviation of X is $0.15

0.14

Step-by-step explanation:

Given that :

Expected value = 0.47

Standard deviation = 0.15

Sample size, n = 30

The mean = expected value * sample size

Mean = 0.47 * 30 = 14.1

Standard deviation of sample = σ / √n

0.15 = σ/√30

σ = 0.8216

P(x > 15) :

Z = (x - mean) /standard deviation

Z = (15 - 14.1) / 0.8216 = 1.095

P(Z > 1.095) = 0.1367

P(Z > 1.095) = 0.14

6 on both sides and twelve on the top and the bottom

step-by-step explanation:

32/80  

91/140

the last one 54 doesn't go into 45. 54/100 ?

step-by-step explanation: