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Readme [11.4K]
3 years ago
8

A random experiment consists of tossing a fair coin five times and recording the number of tails. What is the sample space?

Mathematics
1 answer:
Lena [83]3 years ago
8 0

Answer:

a=54iiiipoiujhghnmn n

Step-by-step explanation:

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To show that TRK= TUK by the SAS postulate, what additional information is necessary?
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Answer:

RTK = UTK

Step-by-step explanation:

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A regular paper clip is 1 1/4 inches long, and a jumbo paper clip is 1 7/8 inches long. How many times longer is the jumbo paper
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It is 6.4 times longer
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Thu wants to play "guess my number".she states,"when I triple my number and add five,I get twenty-six. What's my number?" what i
Mama L [17]
Let's break this up:
let x = the number
"when I triple my number" is the same as
3x
"and add five" is the same as
+5
We combine it to get
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"I get twenty-six"
=26
combine to get
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Now we just solve:
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3 years ago
Ok last one lets get it!
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No yes no yes no

Step-by-step explanation:

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3 years ago
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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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