Answer:
0.234
Step-by-step explanation:
Answer:
I don`t see the attachment
Step-by-step explanation:
<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: you sold 10 adult tickets and 11 children tickets.
Step-by-step explanation:
Let x represent the number of adult tickets that you sold.
Let y represent the number of children tickets that you sold.
You sold a total of 21 tickets. This means that
x + y = 21
Admission prices are six dollars for adults and four dollars for children. You sell tickets for admission to your school play and collect a total of $104. This means that
6x + 4y = 104 - - - - - - - - - -1
Substituting x = 21 - y into equation 1, it becomes
6(21 - y) + 4y = 104
126 - 6y + 4y = 104
- 6y + 4y = 104 - 126
- 2y = - 22
y = - 22/ -2 = 11
Substituting y = 11 into x = 21 - y, it becomes
x = 21 - 11 = 10
