First we will find the 11th term
an = a1 + (n-1) * d
a11 = 12 + (11 - 1) * 5
a11 = 12 + 10 * 5
a11 = 12 + 50
a11 = 62
now we use the sum formula...
Sn = (n (a1 + an)) / 2
S11 = (11 (12 + 62)) / 2
S11 = (11 (74)) / 2
S11 = 814/2
S11 = 407
Use the hypergeometric distribution.
M=number of Men=5
F=number of women=4
m=number of men elected=2
f=number ow women elected=2.
Assuming equal chance to get elected, then
P(2M,2F)=C(M,m)*C(F,f)/C(M+F,m+f)
=C(5,2)*C(4,2)/C(9,4)
=10*6/126
=10/21
Reference: Hypergeometric distribution.
I used Desmos to graph it but here are the coordinates used to graph it in the pictures below.
Answer: I'm guessing it would be a parabola, with the line going through -6 on the y-axis and passing through 2 and 6 on the x-axis, but we cannot see any answers, so therefore we can't answer it accurately.
Step-by-step explanation: