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:
3/20 is the answer.......
Answer:
Step-by-step explanation:
After removing one black marble:
Consider the digit expansion of one of the numbers, say,
676₉ = 600₉ + 70₉ + 6₉
then distribute 874₉ over this sum.
874₉ • 6₉ = (8•6)(7•6)(4•6)₉ = (48)(42)(24)₉
• 48 = 45 + 3 = 5•9¹ + 3•9⁰ = 53₉
• 42 = 36 + 6 = 4•9¹ + 6•9⁰ = 46₉
• 24 = 18 + 6 = 2•9¹ + 6•9⁰ = 26₉
874₉ • 6₉ = 5(3 + 4)(6 + 2)6₉ = 5786₉
874₉ • 70₉ = (8•7)(7•7)(4•7)0₉ = (56)(49)(28)0₉
• 56 = 54 + 2 = 6•9¹ + 2•9⁰ = 62₉
• 49 = 45 + 4 = 5•9¹ + 4•9⁰ = 54₉
• 28 = 27 + 1 = 3•9¹ + 1•9⁰ = 31₉
874₉ • 70₉ = 6(2 + 5)(4 + 3)10₉ = 67710₉
874₉ • 600₉ = (874•6)00₉ = 578600₉
Then
874₉ • 676₉ = 578600₉ + 67710₉ + 5786₉
= 5(7 + 6)(8 + 7 + 5)(6 + 7 + 7)(0 + 1 + 8)(0 + 0 + 6)₉
= 5(13)(20)(20)(1•9)6₉
= 5(13)(20)(20 + 1)06₉
= 5(13)(20)(2•9 + 3)06₉
= 5(13)(20 + 2)306₉
= 5(13)(2•9 + 4)306₉
= 5(13 + 2)4306₉
= 5(1•9 + 6)4306₉
= (5 + 1)64306₉
= 664306₉
Answer:
#1 is 9 times 10^15
#2 is 3 times 10^8
Step-by-step explanation:
