Answer:
4 - 52 i
Step-by-step explanation:
Answer:
77%
Step-by-step explanation:
Given the following :
Probability of winning both games = 50%
Probability of winning just the first game = 65%
Let the probability of winning the ;
First game = p(A) = 65%
Second game = p(B)
Both games = p(A and B) = 50%
What is the probability that the team will win the 2nd game given that they have already won the first game
The above question is a conditional probability question :
Probability of winning the second Given that they've already won the first = p(B | A)
p(B | A) = (A and B) / p(A)
p(B | A) = 50% / 65%
p(B | A) = 0.5 / 0.65
p(B | A) = 0.7692307
= 76.9% = 77%
Answer:
approximately 9 weeks
Step-by-step explanation:
I can't really show the work on here, 1,000,000÷15,000= 66. (a whole lot of numbers, but only 66 is important). There is 7 days in a week, so 66 days ÷7 equals 9.(again a whole lot of useless numbers).
It’s a good game but it is not worth it for me to get it out but it still crashes after the game is so good
Answer:
Step-by-step explanation:
