In a chess club the probability that Shaun will beat Mike is 3/8 . The probability that Shaun will beat Tim is 5/7 .
(Assume all the games are independent of one another)
If Shaun plays 1 game with Mike and then 1 game with Tim, what is the probability that Shaun loses both games?

11%

Step-by-step explanation:

Shaun's percentage of winning against Mike is 8/11 or 73%

Shaun's percentage of winning against Tim is 7/12 or 58%

Shaun's percentage of losing against Mike is 100-73 = 27%

Shaun's percentage of losing against Tim is 100-58 = 42%

Shaun's percentage of losing against Mike and Tim is .27 x .42 which is approximately 11%

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

answer:

f(5/2) = 10/3

i hope this you!