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cluponka [151]
3 years ago
14

HELP PLSS

Mathematics
1 answer:
masya89 [10]3 years ago
3 0

Answer:

the bacteria population increases at a constant rate

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True or False? If a radius of a circle intersects a chord, it is perpendicular to that chord.
kifflom [539]
If a radius of a circle intersects a chord, it would be unlikely for it to be perpendicular to that chord, so your answer choice for this question should be false.
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Find the missing number according to the associative property <br><br> (20×5)×11= 20×(11×______)
Ede4ka [16]

<u><em>5 is the missing number according to the associative property is the final answer.  </em></u>

(20*5)=100

100*11=1100

(20*5)*11=20(11*5)

Hope this helps! And thank you for posting your question at here on brainly, and have a great day. -Charlie

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The seismic activity density of a region is the ratio of the number of earthquakes during a given time span to the land area aff
vlada-n [284]
It is given that seismic activity d<span>ensity of a region is the ratio of the number of earthquakes during a given time span to the land area affected. so if the there 424 earthquakes and the land area is 71,300 sq miles.


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3 years ago
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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
What’s the correct answer
Lisa [10]
A would be your answer I believe
6 0
3 years ago
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