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:
1/2
Step-by-step explanation:
Given:
(2 cubed) (2 superscript negative 4)
= (2³)(2^-4)
= (2³) (1 / 2⁴)
= (2³ * 1) / 2⁴
= 2³ / 2⁴
Both numerator and denominator has the same base. Thus, pick one of the bases
Also, in indices, division sign can be translated to subtraction
Therefore,
2³ / 2⁴
= 2^3-4
= 2^-1
= 1/2¹
= 1/2
(2³)(2^-4) = 1/2
Another one where there's a danger of overthinking it !
The sum of ANY two integers is odd only if one is even and one is odd.
If they're both odd OR both even, then their sum is even.
So if their sum is odd, then one is odd and the other is even.
The probability is ' 1 ' (100%) .
Answer:
r = 18
Step-by-step explanation:
r -8 = 10
r -8 + 8 = 10 + 8
r = 18
Answer:
about .24 mile per hour
1 mile is 5280 foot
Step-by-step explanation: