Answer:
a) 33.33%)
b) 135 minutes
c) 8.66 min
d) 50%
Step-by-step explanation:
a) the probability for a uniform distribution is
P(b<X<a) = (a-b)/(c-d) , where c and d are the maximum and minimum values
therefore the probability that the flight is more than 140 minutes ( and less than 150 since it is the maximum value)
P(140<X<150) = (a-b)/(c-d) = (150-140)/(150-120) = 10/30 = 1/3 (33.33%)
b) the mean (expected value) for a uniform probability distribution is
E(X) = (c+d)/2 = (120+150)/2 = 135 minutes
c) the standard deviation for a uniform probability distribution is
σ²(X)= (c-d)²/12 = (150-120)²/12 = 75 min²
σ = √75 min² = 8.66 min
b) following the same procedure as in a)
P(120<X<135) = (a-b)/(c-d) = (135-120)/(150-120) = 15/30 = 1/2 (50%)
Answer:
option (D) is the right answer
Answer:
A
Step-by-step explanation:
Just add both matrices
60, I think this is the right answer !
33 child tickets were sold that day.
adult tickets = a
child tickets = c
9.2a + 5.4c = 785.4
a = 2c
Substitute a into the first equation
9.2(2c) + 5.4c = 785.4
Simplify
18.4c + 5.4c = 785.4
Combine like terms by adding
23.8c = 785.4
Divide both sides by 23.8 to isolate the variable c
c = 33
