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HACTEHA [7]
3 years ago
7

Starting with the number 3, build a sequence of 5 numbers.

Mathematics
1 answer:
just olya [345]3 years ago
8 0

Answer:

3, 6, 9, 12, 15

adding 3 each time

or

3, 9, 27, 81, 243

multiplying by 3 each time

Step-by-step explanation:

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The probability that a call received by a certain switchboard will be a wrong number is 0.02. Use the Poisson distribution to ap
MAXImum [283]

Answer:

0.2008 = 20.08% probability that among 150 calls received by the switchboard, there are at least two wrong numbers.

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.

The probability that a call received by a certain switchboard will be a wrong number is 0.02.

150 calls. So:

\mu = 150*0.02 = 3

Use the Poisson distribution to approximate the probability that among 150 calls received by the switchboard, there are at least two wrong numbers.

Either there are less than two calls from wrong numbers, or there are at least two calls from wrong numbers. The sum of the probabilities of these events is 1. So

P(X < 2) + P(X \geq 2) = 1

We want to find P(X \geq 2). So

P(X \geq 2) = 1 - P(X < 2)

In which

P(X < 2) = P(X = 0) + P(X = 1)

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

P(X = 0) = \frac{e^{-3}*3^{0}}{(0)!} = 0.0498

P(X = 1) = \frac{e^{-3}*3^{1}}{(1)!} = 0.1494

P(X < 2) = P(X = 0) + P(X = 1) = 0.0498 + 0.1494 = 0.1992

Then

P(X \geq 2) = 1 - P(X < 2) = 1 - 0.1992 = 0.2008

0.2008 = 20.08% probability that among 150 calls received by the switchboard, there are at least two wrong numbers.

6 0
3 years ago
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