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 se
cond 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 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.
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?
I saw other questions for this but why is the standard deviation 0.15*sqrt of 30 and not 0.15*30 because I thought this is a linear transformation
1 answer:
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Answer:
It is about 537.2
Step-by-step explanation:
V= π · r^2 · h
V= π · 3^2 · 19
V≈ 537.2
<span>[2×(10+5)]-5
[2*15]-5
30-5
25</span>
Option A will be the correct answer
<h3>
Answer: 40</h3>
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Explanation:
Assuming the function is f(x) = 5(2)^x, then we replace every x with 3. Then we use the order of operations PEMDAS to simplify. Or we can use a calculator to simplify in one step.
f(x) = 5(2)^x
f(3) = 5(2)^3
f(3) = 5(8)
f(3) = 40
Answer:
z is a variable so it can be any number.
z=-3
if you do -8 times -3 that equals 24.
the answer is -3
Step-by-step explanation: