It already is in simplest form because the numerator and denonimator don't have a common factor to then divide them by.
\left(\mathrm{Decimal:\quad }x=-0.75\right)
hope it helps :P
Answer:
P(z>1.3) = 0.9032
Step-by-step explanation:
We are given:
Mean = 5000
Standard deviation = 1000
x = 6300
P(x>6300)=?
z-score =?
z-score = x- mean/standard deviation
z-score = 6300 - 5000/1000
z- score = 1300/1000
z-score = 1.3
So, P(x>6300) = P(z>1.3)
Looking at the z-probability distribution table and finding value:
P(z>1.3) = 0.9032
So, P(z>1.3) = 0.9032
We have the following given
p1 - probability for outcome 1
p2 - probability for outcome 2
p3 - probability for outcome 3
v1 - amount of money that you will win or lose for outcome 1
v2 - amount of money that you will win or lose for outcome 2
v3 - amount of money that you will win or lose for outcome 3
Therefore,
p1v1 + p2v2 + p3v3 is the average money you win or lose in playing the game.
