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 factored form for this question is (5x-1)(x-5)
Answer:
4 seconds.
Step-by-step explanation:
The function f(x)=-10(x)(x-4) ........ (1), represents the approximate height of a projectile launch on the ground into the air as a function of time in seconds x.
Now, we are asked that for how long from the launch does the projectile stays in the air.
Therefore, we have to solve the equation (1) making f(x) as zero.
Hence, 10x(x - 4) = 0
⇒ x = 0 or x = 4
(As x can not be zero since at x = 0 sec, the projectile was at the ground.}
Hence, x = 4 seconds.
Therefore, the projectile was in the air for 4 seconds. (Answer)
Answer:
125
Step-by-step explanation:
-125 x -1
Let the number of hamburgers sold be "h" and the number of cheeseburgers be "2h" as it was two time as the humbugers .form equation frome it '2h+h=387".comp the like terms which gives you 3h=387 divide each side by 3 which gives you h=129 which is equal to the number hambyrgers .multiply 129 by 2 to get the number of cheeseburgers which is 258
