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OverLord2011 [107]
2 years ago
14

HEEEEEEEELLLLLLPPPPPPP PLLLLLZZZZ T^T

Mathematics
2 answers:
kiruha [24]2 years ago
7 0
The first one is (5,-6)
The second one is (-5,6)
The 3rd one is the answer with all negatives and 4 coordinates
the 4th one is the answer that has 4 coordinates and some positive and some negative
Marina CMI [18]2 years ago
4 0

Answer:

i dont know sorry

Step-by-step explanation:

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Answer:

umm 11

Step-by-step explanation:

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A Geiger counter counts the number of alpha particles from radioactive material. Over a long period of time, an average of 14 pa
UkoKoshka [18]

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:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu 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 \mu = \frac{14}{60} = 0.2333

Either no particle arrives, or at least one does. The sum of the probabilities of these events is decimal 1. So

P(X = 0) + P(X \geq 1) = 1

We want P(X \geq 1). So

P(X \geq 1) = 1 - P(X = 0)

In which

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-0.2333}*(0.2333)^{0}}{(0)!} = 0.7919

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.7919 = 0.2081

0.2081 = 20.81% probability that at least one particle arrives in a particular one second period.

8 0
3 years ago
This question has two parts.
LUCKY_DIMON [66]

Answer:

Part A: $6.76

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Step-by-step explanation:

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Lesechka [4]
The answer is (2,2)
5 0
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