Answer:
Probability of missing two passes in a row is 0.08.
Step-by-step explanation:
Event E = A football player misses twice in a row.
P(E) = ?
Event X = Football player misses the first pass
P(X) = 0.4
Event Y = Football player misses just after he first miss
P(Y) = 0.2
Both the events are exclusive so the probability of occuring of these two events can be calculated by the formula:
P(E) = P(X).P(Y)
P(E) = 0.4*0.2
P(E) = 0.08
-4x +12 -12 = -4 -12
-4x = -16
x = 4

You pass the free terms wich don't contain 'x' in the othe part of the equal with changed signe ans the same for the terms wich contain x.
Answer:
Step-by-step explanation:
2x² -x - 6 = 2x² - 4x + 3x - 3*2
= 2x*(x - 2) + 3*(x - 2)
= (x-2)(2x + 3)
A = 2x² -x - 6
A = (x - 2)(2x + 3)
length*width = (x - 2)(2x + 3)
length * (2x +3) = (x - 2)(2x + 3)

length = x - 2
Answer:
D) -31
Step-by-step explanation:
use PEMDAS
3^2-(5 x 2^3)
3^2-(5 x 8)
3^2-40
9-40
-31