Answer:
The probability that all the five flights are delayed is 0.2073.
Step-by-step explanation:
Let <em>X</em> = number of domestic flights delayed at JFK airport.
The probability of a domestic flight being delayed at the JFK airport is, P (X) = <em>p</em> = 0.27.
A random sample of <em>n</em> = 5 flights are selected at JFK airport.
The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.
The probability mass function of <em>X</em> is:
![P(X=x)={5\choose x}0.27^{x}(1-0.27)^{5-x};\ x=0,1,2...](https://tex.z-dn.net/?f=P%28X%3Dx%29%3D%7B5%5Cchoose%20x%7D0.27%5E%7Bx%7D%281-0.27%29%5E%7B5-x%7D%3B%5C%20x%3D0%2C1%2C2...)
Compute the probability that all the five flights are delayed as follows:
![P(X=5)={5\choose 5}0.27^{5}(1-0.27)^{5-5}=1\times 1\times 0.207307=0.2073](https://tex.z-dn.net/?f=P%28X%3D5%29%3D%7B5%5Cchoose%205%7D0.27%5E%7B5%7D%281-0.27%29%5E%7B5-5%7D%3D1%5Ctimes%201%5Ctimes%200.207307%3D0.2073)
Thus, the probability that all the five flights are delayed is 0.2073.
Answer:
imagine
Step-by-step explanation:
Answer:
24 mm squared
Step-by-step explanation:
3 x 4 x 1/2 = area of the triangle
4 x 3 = area of the rectangular top left
1 x 6 = area of the bottom rectangle
3 x 4 = 12 x 1/2 = 6
4 x 3 = 12
6 x 1 = 6
6 + 12 + 6 =24
Answer:
m=8
Step-by-step explanation:
To find the GCF of the two terms, continuous division must be done.
What can be used to divide both terms such that there is not a remainder?
Start small, let's take 2. It could be a GCF.
Move up higher, say 3. Yes, it can be a GCF.
To see if there might be a greater common factor, divide the constants by 3.
48/3 = 16
81/3 = 27
Upon inspection and contemplation, there is no more common factor between 16 and 27. So, 3 is the GCF.
Moving on, when it comes to variables. The variable with the least exponents is easily the GCF. For the variable m, the GCF is m2 and for n, the GCF is n.
Combining the three, we have the overall GCF = 3m2n