Add 2 to both sides.
d=16
<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>
This is a basic addition/subtraction question. It's saying that they started at the 50 yard line. On the play, they gained 7 yards. So, we'll add 50+7 = 57.
They're at the 57 yard line at this point. On the second play they lost 10 yards (how unfortunate). Thereby, we will subtract 57-10=47.
50 yards +7 yards. -10 yards.
The answer will be that they'll be on the 47 yard line on the next play.
When you raise a number to a negative exponent, change the exponent to a positive and make the complete number a fraction with 1 as the numerator.
5 ^-2 = 1/5^2
The answer is C.
Answer:
(6 , -1)
Step-by-step explanation:
2x+ y = 11 ----------------(I)
y = 11 - 2x --------------------(II)

Multiply the whole equation by 2

x - 10y = 16 --------------------(III)
Substitute y = 11- 2x in equation (III)
x - 10(11 - 2x) = 16
x - 110 + 20x = 16
21x - 110 = 16
21x = 16 +110
21x = 126
x = 126/21
x = 6
Plugin x = 6 in equation (II)
y = 11 - 2*6
y = 11 - 12
y = -1