Any of the 11 man can be chosen, and combined with any of the 8 women.
Assume we select man1. The selected committee can be:
(m1,w1), (m1,w2), (m1,w3), (m1,w4), (m1,w5), (m1,w6), (m1,w7), (m1,w8),
so there are 8 committees selections with man1 in them.
we could repeat the same procedure for the remaining 10 men, and get 8 committees where each of them is a member.
so there are 11*8=88 ways of choosing 1 man and 1 woman.
Answer: 88
Answer:
1. 2508.96
2. 40123.050624
Step-by-step explanation:
28.26 x 28.26 x 3.14 = 2508.96
113.04 x 113.04 x 3.14 = 40123.050624
Answer:
-2
Step-by-step explanation:
<span>5d^2 + 4d - 3 - (3d^2 - 2d + 4)
=</span>5d^2 + 4d - 3 - 3d^2 + 2d - 4
= 2d^2 + 6d - 7
