A committee of four is chosen at random from a group of 66 women and 55 men. Find the probability that the committee contains at least one man.

 There are 13 C 4 = 715 ways to choose a group of men and women.

So there are 715 - 70 = 645 ways to choose a group with at least one man.

This means the probability is 645/715 = 129/143 = 0.9020979 (so roughly a 90.20979% chance)

Step-by-step explanation:  There are 13 C 4 = 715 ways to choose a group of men and women.

So there are 715 - 70 = 645 ways to choose a group with at least one man.

This means the probability is 645/715 = 129/143 = 0.9020979 (so roughly a 90.20979% chance)

i'm assuming it's looking for the missing length, which should be 4cm

hope this !

r(0,90)(x,y) . t(-5,0)

last option

step-by-step explanation:

∆def to ∆d"e"f"

- rotate 90 counterclockwise degrees

- then translated 5 to the left (x --> x - 5)

answer is the last option