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amm1812
2 years ago
13

The ruby- throated hummingbird has a wing beat of about 200 beats per second. About how many wings beats would a huming bird hav

e in 3 minutes
Mathematics
2 answers:
Ede4ka [16]2 years ago
7 0
There are 60 seconds in one minute
200 beats per second
200 x 60 = 12000 beats per minute

Then multiply 12000 by 3

36000 wing beats in 3 minutes

hope this helps
marishachu [46]2 years ago
5 0
Well 200 beats per second is 12,000 beats a minute times 3 for 3 minutes is 36,000 beats every three minutes

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What is the least common denominator (LCD) of 1/3 and 3/4
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The graphs of the function f (given in blue) and g (given in red) are plotted above. Suppose that u(x)=f(x)g(x) and v(x)=f(x)/g(
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Which graph represents the solution set for the quadratic inequality x2 + 2x + 1 > 0?
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A bank gets an average of 12 customers per hour. Assume the variable follows a Poisson distribution. Find the probability that t
Alina [70]

Answer:

75.76% probability that there will be 10 or more customers at this bank in one hour.

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.

A bank gets an average of 12 customers per hour.

This means that \mu = 12

Find the probability that there will be 10 or more customers at this bank in one hour.

Either there are less than 10 customers, or there are 10 or more. The sum of the probabilities of these events is 1. Then

P(X < 10) + P(X \geq 10) = 1

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P(X < 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9)

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

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P(X = 2) = \frac{e^{-12}*12^{2}}{(2)!} = 0.0004

P(X = 3) = \frac{e^{-12}*12^{3}}{(3)!} = 0.0018

P(X = 4) = \frac{e^{-12}*12^{4}}{(4)!} = 0.0053

P(X = 5) = \frac{e^{-12}*12^{5}}{(5)!} = 0.0127

P(X = 6) = \frac{e^{-12}*12^{6}}{(6)!} = 0.0255

P(X = 7) = \frac{e^{-12}*12^{7}}{(7)!} = 0.0437

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P(X = 9) = \frac{e^{-12}*12^{9}}{(9)!} = 0.0874

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Then

P(X \geq 10) = 1 - P(X < 10) = 1 - 0.2424 = 0.7576

75.76% probability that there will be 10 or more customers at this bank in one hour.

3 0
2 years ago
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