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
The diameter would be twice 5, or 10
The length of CD is 7-3=4.
Therefore, if we dilate by a scale factor of 3, the length is 4(3) = 12
You have to start by using the distributive property on the left side.
3(2x-7)=5x+4
6x-21=5x+4 Then remove and combine alike terms.
+21 +21
6x=5x+25
-5x -5x
x=25
