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
The answer is B) it shifts up and to the right
3f + 2g = 30 . . .(1)
f + 2g = 26 . . . .(2)
(1) - (2) => 2f = 4 => f = 4/2 = 2
From (2), 2 + 2g = 26 => 2g = 26 - 2 = 24 => g = 24/2 = 12
Therefore, f = 2 and g = 12
