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
Answer:
you only have the answer choices not the question itself
could you please attach the question so i can help?
Answer:
17
Step-by-step explanation:
ifhrFj,iyjgztizzkyraykeLyrgodyyixynj
You could convert them into improper fractions first. 7 7/8 should become 63/8 and 3 1/4 should become 13/4. Then, change both fractions so they share a common denominator. I'll use 8.
63/8-26/8 = 37/8, which is then converted into 4 5/8.
Hope that helped you.
