you have the hypotenuse and the opposite side so you will use sin
sin58= (30/q)
multiply both sides by q
q(sin58)=30
divide by sin58 in your calculatior
q=(30/sin58)
^ I don't have a calculator with me so you will need to plug that in
Answer:
-144m^11n^3
Step-by-step explanation:
-(6m^3)^2 x 4m^5n
-36m^6n^2 x 4m^5n
Answer:
-144m^11n^3
Answer:
First then second answer below
Step-by-step explanation:
-x+y 4
-1/-1=1 x=1 y=1
y/1=1/1+4/1
y=1x+4
-x+7y 21
-1/-1=1
7y/7=1/7+21/7
y=1/7x+3
Answer:
-6/17
Step-by-step explanation:
-60/170 can be simplified to -6/17 when dividing the numerator and denominator by 10. If you want a decimal, it is -0.352941164705882 with a bar over all the decimal.
<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
