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 answer on APEX would be $2430
The graph shows solutions to be ...
(x, f(x)) = (x, g(x)) = (-2, 5), (2, -3)
_____
Analytically, it works well to find
g(x) -f(x) = 0
(x^2 -2x -3) -(-2x +1) = 0
x^2 -4 = 0
x^2 = 4
x = ±√4 = ±2
Then
f(-2) = -2(-2) +1 = 5
f(2) = -2(2) +1 = -3
Answer:
He needs to take 5.5 grams of medications.
Step-by-step explanation:
I added 2.1 to 3.4 and got 5.5. *booyah*