Answer:
0.2081 = 20.81% probability that at least one particle arrives in a particular one second period.
Step-by-step explanation:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
In which
x is the number of sucesses
e = 2.71828 is the Euler number
is the mean in the given interval.
Over a long period of time, an average of 14 particles per minute occurs. Assume the arrival of particles at the counter follows a Poisson distribution. Find the probability that at least one particle arrives in a particular one second period.
Each minute has 60 seconds, so 
Either no particle arrives, or at least one does. The sum of the probabilities of these events is decimal 1. So
We want 
. So
In which
0.2081 = 20.81% probability that at least one particle arrives in a particular one second period.
Answer:
b
Step-by-step explanation:
D) because only 4 is the common factor in the equation.
Answer:
f(n)=f(n-1)+f(n-2)
f(1)=1x
f(2)=1x
Step-by-step explanation:
This is the fibonacci sequence with each term times x.
Notice, you are adding the previous two terms to get the third term per consecutive triples of the sequence.
That is:
1x+1x=2x
1x+2x=3x
2x+3x=5x
3x+5x=8x
So since we need the two terms before the third per each consecutive triple in the sequence, our recursive definition must include two terms of the sequence. People normally go with the first two.
f(1)=1x since first term of f is 1x
f(2)=1x since second term of f is 1x
Yes, I'm naming the sequence f.
So I said a third term in a consecutive triple of the sequence is equal to the sum of it's two prior terms. Example, f(3)=f(2)+f(1) and f(4)=f(3)+f(2) and so on...
Note, the term before the nth term is the (n-1)th term and the term before the (n-1)th term is the (n-2)th term. Just like before the 15th term you have the (15-1)th term and before that one you have the (15-2)th term. That example simplified means before the 15th term you have the 14th and then the 13th.
So in general f(n)=f(n-1)+f(n-2).
So the full recursive definition is:
f(n)=f(n-1)+f(n-2)
f(1)=1x
f(2)=1x
