Given:
Quadrilateral QRST has diagonals that do NOT bisect each other, are congruent and are not perpendicular.
By checking the options:
Rhombus, rectangle and the square each of them the diagonals bisect each other
So, the answer will be option 4) Isosceles Tra
The answer is £287.98 is the correct answer
Answer:
The number of times Ellis get to bat is 558.
Step-by-step explanation:
Here, let us assume the number of times Ellis bat = m
So, the number of times Dwight bat = 17 times fewer than Ellis
= m - 17
Also, number of times Wade got to bat = 10 more times than Dwight
= ( m- 17 )+ 10
Total number of bat times = 1650
So, the number of times ( Wade + Dwight + Ellis) bat together = 1650
⇒ ( m- 17 )+ 10 + (m - 17) + m = 1650
or, 3 m - 34 + 10 = 1650
or, 3 m = 1674
⇒ m = 1674/3 = 558 , or m = 558
Hence, the number of times Ellis get to bat = m = 558.
X^2+4x+4+y^2-10y+25=-13+25+4
(x+2)^2 + (y-5)^2= 16
Third option
<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