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shtirl [24]
3 years ago
15

Graph the following and state the domain, range, and end behavior y=2x-4

Mathematics
2 answers:
jasenka [17]3 years ago
8 0

Answer:

3

Step-by-step explanation:

Kamila [148]3 years ago
7 0
The answer to your question is three
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64 boards:)

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Evaluate 4 + (-2) - (-3) – 6.
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The number of typing errors made by a typist has a Poisson distribution with an average of three errors per page. If more than t
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Answer:

0.6472 = 64.72% probability that a randomly selected page does not need to be retyped.

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.

Poisson distribution with an average of three errors per page

This means that \mu = 3

What is the probability that a randomly selected page does not need to be retyped?

Probability of at most 3 errors, so:

P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

In which

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) = \frac{e^{-3}*3^{2}}{(2)!} = 0.2240

P(X = 3) = \frac{e^{-3}*3^{3}}{(3)!} = 0.2240

Then

P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0498 + 0.1494 + 0.2240 + 0.2240 = 0.6472

0.6472 = 64.72% probability that a randomly selected page does not need to be retyped.

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