An alternate method would be to take out 3 first
(since it is common to each term)
3(2x + x - 21)
now break down the middle terms
3(2x - 6x + 7x - 21)
now you can proceed to factoring your equation.
If you would like to simplify the product using the distributive property, you can do this using the following steps:
<span>(5h – 5) * (5h – 6) = 5h * 5h - 5h * 6 - 5 * 5h + 5 * 6 = 25h^2 - 30h - 25h + 30 = 25h^2 - 55h + 30
</span>
The correct result would be <span>a. 25h^2 – 55h + 30</span>.
Solution :
Given :
P (never married) = P = 0.44
Sample size, n = 15
Let
(n = 15, P = 0.44)
P(X = x) = ![$^nC_xP^x(1-P)^{n-x}; x=0, 1,....n$](https://tex.z-dn.net/?f=%24%5EnC_xP%5Ex%281-P%29%5E%7Bn-x%7D%3B%20x%3D0%2C%201%2C....n%24)
a). The probability that exactly two of them have been married is
P(X=2) = ![$^{15}C_2(0.44)^2(1-0.44)^{15-2}$](https://tex.z-dn.net/?f=%24%5E%7B15%7DC_2%280.44%29%5E2%281-0.44%29%5E%7B15-2%7D%24)
![$=\frac{15!}{2!(15-2)!}(0.44)^2(0.56)^{13}$](https://tex.z-dn.net/?f=%24%3D%5Cfrac%7B15%21%7D%7B2%21%2815-2%29%21%7D%280.44%29%5E2%280.56%29%5E%7B13%7D%24)
![$=\frac{15\times14}{2}\times (0.44)^2(0.56)^{13}$](https://tex.z-dn.net/?f=%24%3D%5Cfrac%7B15%5Ctimes14%7D%7B2%7D%5Ctimes%20%280.44%29%5E2%280.56%29%5E%7B13%7D%24)
= 0.0108
b). That at most two of them have never been married.
![$P(X \leq 2)= P(X=0)+P(X+1)+P(X=2)$](https://tex.z-dn.net/?f=%24P%28X%20%5Cleq%202%29%3D%20P%28X%3D0%29%2BP%28X%2B1%29%2BP%28X%3D2%29%24)
![$=^{15}C_0(0.44)^0(1-044)^{15-0}+^{15}C_1(0.44)^1(1-0.44)^{15-1}+^{15}C_2(0.44)^2(1-0.44)^{15-2}$](https://tex.z-dn.net/?f=%24%3D%5E%7B15%7DC_0%280.44%29%5E0%281-044%29%5E%7B15-0%7D%2B%5E%7B15%7DC_1%280.44%29%5E1%281-0.44%29%5E%7B15-1%7D%2B%5E%7B15%7DC_2%280.44%29%5E2%281-0.44%29%5E%7B15-2%7D%24)
![$=(1)(1)(0.56)^{15}+(15)(0.44)(0.56)^{14}+(105)(0.44)^2(0.56)^{13}$](https://tex.z-dn.net/?f=%24%3D%281%29%281%29%280.56%29%5E%7B15%7D%2B%2815%29%280.44%29%280.56%29%5E%7B14%7D%2B%28105%29%280.44%29%5E2%280.56%29%5E%7B13%7D%24)
= 0.000167+0.001969+0.010828
= 0.012964
= 0.0130
c). That at least 13 of them have been married.
P(married) = 1 - P(never married)
= 1 - 0.44
= 0.56
![$P(X \geq 13)= P(X=13)+P(X=14)+P(X+15)$](https://tex.z-dn.net/?f=%24P%28X%20%5Cgeq%2013%29%3D%20P%28X%3D13%29%2BP%28X%3D14%29%2BP%28X%2B15%29%24)
![$=^{15}C_{13}(0.56)^{13}(1-0.56)^{15-13}+^{15}C_{14}(0.56)^{14}(1-0.56)^{15-14}+^{15}C_{15}(0.56)^{15}(1-0.56)^{15-15}$](https://tex.z-dn.net/?f=%24%3D%5E%7B15%7DC_%7B13%7D%280.56%29%5E%7B13%7D%281-0.56%29%5E%7B15-13%7D%2B%5E%7B15%7DC_%7B14%7D%280.56%29%5E%7B14%7D%281-0.56%29%5E%7B15-14%7D%2B%5E%7B15%7DC_%7B15%7D%280.56%29%5E%7B15%7D%281-0.56%29%5E%7B15-15%7D%24)
![$=(105)(0.56)^{13}(0.44)^{2}+(15)(0.56)^{14}(0.44)+(1)(0.56)^{15}(1)$](https://tex.z-dn.net/?f=%24%3D%28105%29%280.56%29%5E%7B13%7D%280.44%29%5E%7B2%7D%2B%2815%29%280.56%29%5E%7B14%7D%280.44%29%2B%281%29%280.56%29%5E%7B15%7D%281%29%24)
![$=0.010828+0.001969+0.000167$](https://tex.z-dn.net/?f=%24%3D0.010828%2B0.001969%2B0.000167%24)
= 0.012964
=0.0130