Answer:
i think its not a polynomial
Step-by-step explanation:
Answer:
The probability is 
Step-by-step explanation:
From the question we are told that
The capacity of an Airliner is k = 300 passengers
The sample size n = 320 passengers
The probability the a randomly selected passenger shows up on to the airport

Generally the mean is mathematically represented as
=>
=>
Generally the standard deviation is

=> 
=> 
Applying Normal approximation of binomial distribution
Generally the probability that there will not be enough seats to accommodate all passengers is mathematically represented as

Here 
=>
Now applying continuity correction we have
=> ![P(X >300 ) = P(Z > \frac{[300.5] - 307.2}{3.50} )](https://tex.z-dn.net/?f=P%28X%20%20%3E300%20%29%20%3D%20%20P%28Z%20%3E%20%20%5Cfrac%7B%5B300.5%5D%20-%20307.2%7D%7B3.50%7D%20%29)
=> 
From the z-table

So

Answer:
You can choose which equation below suits your taste but I recommend the 2n + 1 = -33.
Step-by-step explanation:

Answer: 0.79
Step-by-step explanation:
I will suppose that this is not a continuos probability, as the individual probabilites add up to 100%.
If you want to obtain the probability that x ≤ 0, then you need to add the probability for the cases x= 0, x = -1, x = -2 .... etc
This is:
x = 0, p = .16
x = -2, p = .33
x = -3, p = .13
x = -5, p = .17
Then, the probability where x takes a negative value or zero {-5, -3, -2, 0} is:
P = 0.16 + 0.33 + 0.13 + 0.17 = 0.79