The answer is 30
7(8)+4(2)-2(17)=30
Total number of cards = 52
Number of odd cards in one set = 4
Number of odd cards in total 4 sets= 4*4=16
52-16 = 36 = number of not odd cards
36/52 = 9/13
<span>If there has to be 2 men and 2 women, we know
that we must take a group of 2 men out of the group of 15 men and a group of 2
women out of the group of 20 women. Therefore, we have:
(15 choose 2) x (20 choose 2)
(15 choose 2) = 105
(20 choose 2) = 190
190*105 = 19950
Therefore, there are 19950 ways to have a group of 4 with 2 men and 2women.</span>
<span>If there has to be 1 man and 3 women, we know
that we must take a group of 1 man out of the group of 15 men and a group of 3
women out of the group of 20 women. Therefore, we have:
(15 choose 1) x (20 choose 3)
(15 choose 1) = 15
(20 choose 3) = 1140
15*1140 = 17100
Therefore, there are 17100 ways to have a group of 4 with 3 women and 1 man.</span>
<span>We now find the total outcomes of having a group
with 4 women.
We know this is the same as saying (20 choose 4) = 4845</span>
Therefore, there are 4845 ways to have a group of
4 with 4 women.
We now add the outcomes of 2 women, 3 women, and
4 women and get the total ways that a committee can have at least 2 women.
19950 + 17100 + 4845 = 41895 ways that there will
be at least 2 women in the committee
Answer:
-17
Step-by-step explanation:
15=-3(h+12)
1. Do distributive property
15=-3h-36
-3h=51
H=-17
