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:
0.675
Step-by-step explanation:
27/40
- Divide each number by 10
2.7/4
- Multiply each number by 25
67.5/100
Convert that into a decimal. :)
Of means multiply in math (most of the time).
So multiply.
2/5 x 7/10 = 14/50 divided by 2/2 = 7/25
3/5 of 1/6 is written as
3/5 x 1/6
Multiply both the numerator and denominators together:
3x1 / 5*6 = 3/30, reduces to 1/10.
The answer is 1/10.
Area will change by a factor of 3.5² = 12.25