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 percentage method aids in calculating FIT.
-5 ≤ 3m + 1 < 4
- 1 - 1 - 1
-6 ≤ 3m < 3
3 3 3
-2 ≤ m < 1
Solution Set: {m|-2 ≤ m < 1}, {m|m ≥ -2 and m < 1}, [-2, 1)
The answer is A.
Answer:
dy/dx = 
Step-by-step explanation:
-y + 2x²- x = -5y³
2x² - x = y - 5y³
4x - 1 = dy/dx - 15y².dy/dx
dy/dx(1 - 15y²) = 4x - 1
dy/dx =
Answer:
They're all like terms (x)
