<span>The solution:
= 40, p = q = 0.5
P[x] = nCx *p^x *q^(n-x)
when p = q = 0.5, the formula simplifies to
P[x] = nCx/2^n = 40Cx/2^40
at least 18 of each type means 18 to 22 of (say) type I
P(18 <= X <= 22) = 0.5704095 <-------
qb
mean = 40*0.5 = 20
SD = sqrt(npq) = sqrt(40*0.5*0.5) = 3.1623
z1= (18-20)/3.1623 = -0.63 , z2 = (22-20)/3.1623 = 0.63
P(-0.63 < z < 0.63) = 0.4713 <-------</span>
The answer for number 7 is x=3
Since x=-10, plug that into the equation.
f(x) = 2(-10) + 11
= -20 + 11
y = -9
So the ordered pair in (x,y) terms is (-10,-9).
Answer:
Step-by-step explanation:
The horizontal distance from points (5,-18) and (8,-17) is 3 because it is 3 units from 5 to 8. The vertical distance is 1 since it is one unit from -18 to -17. Now we can use the equation 
where a=3 and b=1 and c is the distance that you are looking for:
270 = 27 * 10
= 3*9 * 5*2
= 3 * 3*3 * 5*2
in order from smallest to largest
2*3*3*3 *5
