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:
a = 4 + -1.333333333b
Step-by-step explanation:
Simplifying
3a + 4b = 12
Solving
3a + 4b = 12
Solving for variable 'a'.
Move all terms containing a to the left, all other terms to the right.
Add '-4b' to each side of the equation.
3a + 4b + -4b = 12 + -4b
Combine like terms: 4b + -4b = 0
3a + 0 = 12 + -4b
3a = 12 + -4b
Divide each side by '3'.
a = 4 + -1.333333333b
Simplifying
a = 4 + -1.333333333b
If I'm correct the answer would be C.) hope this help
We can solve this problem by using PEMDAS
PEMDAS stands for
Parenthesis
Exponents
Multiply
Add
Subtract
So start with (P+3) () is a parenthesis
