Answer:
C.No, because each lime will cost a bit more than 30¢, so 4 limes will cost a bit more than $1.20.Step-by-step explanation:
The perimeter is 35. If we were to change the width, which is one of the dimensions of the flower bed, The perimeter will change. This means that perimeter will no longer be 35. So in order to keep the perimeter as it is, if we change one dimension, we must also change the other. Let's solve for the length, using the formula to see how much the length changes from. p = 2l + 2w 35 = 2l + 2(15) 35 = 2l + 30 5 = 2l 2.5 = l We must increase the length from 2.5 feet. This is because decreasing one dimension will decrease the perimeter. But if we increase the other dimension as well, it will restore the perimeter to where is was initially.
Answer:
y=2.9
Step-by-step explanation:
3(2y - 0.3) = 19.4 - y
6y-0.9=19.4
6y+y-0.9=19.4
6y+y=19.4+0.9
7y=19.4+0.9
7y=20.3
7y/7=20.3/7 (divid each one by seven. that's what /7 means)
y=2.9
35. To find the solution to this problem, we need to add up all of the juice.
5 3/4 + 4 1/2 + 2 3/4 = 13
The answer for this question would be C.
36. Since each square tile is 1 cm by 1 cm:
7 * 12 = 84
84/1 = 84
The answer is 84.
Hope this helps!
<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
