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:
3ab^2
Step-by-step explanation:
Factor 6a^2b^3
= 2 * 3 * a * a * b * b * b
Factor 15ab^5
= 3 * 5 * a * b * b * b * b * b
Factor 9a^3b^2
3 * 3 * a * a * a * b * b
Common Factor
= 3 * a * b * b
Simplify
= 3ab^2
.1.Scale Factor of Triangle STU to Triangle PQR= 1.5
2.Scale Factor of Trapezoid EFGH to Trapezoid JKLM =2
3. about 90 armadillo
1.Side ST: 15 ÷ Side PQ: 10 = 1.5
2.Side JM: 14 ÷Side EH: 7 = 2
3.843/7=120.428571429
843/4=210.75
210.75-120.428571429=90.321428571
This rounds to 90 armadillo
Hope this helps :3 ❤