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:
20
√
5
Step-by-step explanation:
1. You'll need to simplify each term. You'll get 32
√
5-12
√
5.
2. After you subtract 12
√
5 from 32
√
5, you'll get 20
√
5.
When n = 1, 2(1) + 1 = 3
When n = 2, 2(2) + 1 = 5
When n = 3, 2(3) + 1 = 7
When n = 4, 2(4) + 1 = 9
When n = 5, 2(5) + 1 = 11
Sum = 3 + 5 + 7 + 9 + 11 = 35
----------------------------------------------------
Answer: 35
----------------------------------------------------
Answer:
what does d equal?
Step-by-step explanation:
You would multiply by -4/3. That's cuz to get to 1, you have to cancel out the negative, with another negative. You then have to multiply by the reciprocal.
Hope this helps!