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Margarita [4]
3 years ago
7

Emmie rolls a dice three times. What is the probability that she will only roll numbers greater than two on all of her rolls?

Mathematics
1 answer:
Likurg_2 [28]3 years ago
6 0
12/18 or 2/3 chance simplified
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Additon and subtraction and the whole no​
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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
4 years ago
Solve the system of linear equations by substitution 2x-y=6 x=y-1
Oksana_A [137]

substitute x for y-1
2(y-1)-y=6 distribute
2y-2-y=6 subtract y in 2y
y-2=6 add 2 to get y by itself
y=8

Put y back into equation x=y-1
x=8-1
X=7
6 0
3 years ago
What is 4.059 divided by 1,000
aev [14]

Answer:

0.004059

Step-by-step explanation:

You can use a calculator and if it comes out in fraction form, press the fraction to decimal button.

4 0
3 years ago
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