Answer:
0.0143
Step-by-step explanation:
In this question, we are asked to use the binomial distribution to calculate the probability that 10 or fewer passengers from a sample of MIT data project sample were on American airline flights.
We proceed as follows;
The probability that a passenger was an American flight is 15.5%= 15.55/100 = 0.155
Let’s call this probability p
The probability that he/she isn’t on the flight, let’s call this q
q =1 - p= 0.845
Sample size, n = 155
P(X < A) = P(Z < (A - mean)/standard deviation)
Mean = np
= 125 x 0.155
= 19.375
Standard deviation = √npq
= √ (125 x 0.155x 0.845)
= 4.0462
P(10 or fewer passengers were on American Airline flights) = P(X \leq 10)
= P(Z < (10.5 - 19.375)/4.0462)
= P(Z < -2.19)
= 0.0143
Answer:
8 cm
Step-by-step explanation:
... BO/OD = AO/OC = 3/1
Written another way, this is ...
... OD : BO = 1 : 3
Now, BD = OD + BO, so we have
... BD : BO = (OD +BO) : BO = (1 +3) : 3 = 4 : 3
That is, BD = 4/3 × BO
... BD = 4/3 × 6 cm = 8 cm
Number of combinations = (6x5) ÷ 2 = 15
Answer:There are 15 combinations
Answer:
1cy
Step-by-step explanation:
c(8)+4y=12
8c+4y=12
12cy /12cy=12 /12cy
1cy
