3:4 = G:B
1:5 = Mr. Smith's class: 7th Grade
2:7= 7th: Middle school
12 girls = 3 units
1 unit = 12/3= 4
Boys = 4x4= 16
<em>Whole class = 28 students</em>
Class : Grade = 1:5 <em> 7 = number of units in Mr. Smith's class</em>
28 = 1 unit
5 units= 28x5= 140 <em>There are 140 kids in the grade</em>
140 = 2 units
1 unit = 140/2= 70
70x7=490
<u><em>There are 490 students in the whole grade</em></u>
We can check to see if (x - 2)(x - 9)(x - 1) is the factored form of x^3 + 8x^2 - 11x - 18 by using foil.
(x - 2)(x - 9)(x - 1)
(x - 2)x^2 - x - 9x + 9
Combine like terms.
(x - 2)x^2 - 10x + 9
FOIL again.
x^3 - 10x^2 + 9x - 2x^2 + 20x - 18
Combine like terms.
x^3 - 12x^2 + 29x - 18
<h3><u>The listed factors are not true factors of the given polynomial.</u></h3>
<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:
The correct answer is "0.9332".
Step-by-step explanation:
Given that:
So,
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I think that the answer will be 63
