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kvasek [131]
3 years ago
10

Please help, thanks! Ok so, you pick a card at random, put it back, and then pick another card at random. There are FIVE cards a

nd they are numbered like this: 1, 2, 3, 4, 5. What is the probability of picking a number greater than 1 and then picking a number less than 2? Write the answer as a fraction or whole number. Thank you!
Mathematics
1 answer:
tatyana61 [14]3 years ago
6 0

Answer:

P(x > 1\ and\ x < 2) = \frac{4}{25}

Step-by-step explanation:

Given

S = \{1,2,3,4,5\}

n(S) = 5

Required

P(x > 1\ and\ x < 2)

P(x > 1\ and\ x < 2) is calculated as:

P(x > 1\ and\ x < 2) = P(x > 1) * P(x < 2)

Since it is a probability with replacement, we have:

P(x > 1\ and\ x < 2) = \frac{n(x > 1)}{n(S)} * \frac{n(x < 2)}{n(S)}

For x > 1, we have:

x > 1 = \{2,3,4,5\}\\

n(x > 1) = 4

For x < 2, we have:

x < 2 = \{1\}

n(x < 2) = 1

P(x > 1\ and\ x < 2) = \frac{n(x > 1)}{n(S)} * \frac{n(x < 2)}{n(S)}

becomes

P(x > 1\ and\ x < 2) = \frac{4}{5} * \frac{1}{5}

P(x > 1\ and\ x < 2) = \frac{4}{25}

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The number of errors in a textbook follow a Poisson distribution with a mean of 0.03 errors per page. What is the probability th
maksim [4K]

Answer:

The probability that there are 3 or less errors in 100 pages is 0.648.        

Step-by-step explanation:

In the information supplied in the question it is mentioned that the errors in a textbook follow a Poisson distribution.

For the given Poisson distribution the mean is p = 0.03 errors per page.

We have to find the probability that there are three or less errors in n = 100 pages.

Let us denote the number of errors in the book by the variable x.

Since there are on an average 0.03 errors per page we can say that

the expected value is, \lambda = E(x)

                                       = n × p

                                       = 100 × 0.03

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Therefore the we find the probability that there are 3 or less errors on the page as

     P( X ≤ 3) = P(X = 0) + P(X = 1) + P(X=2) + P(X=3)

                 

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Therefore P( X ≤ 3) = \frac{e^{-3} 3^0}{0!} + \frac{e^{-3} 3^1}{1!} + \frac{e^{-3} 3^2}{2!} + \frac{e^{-3} 3^3}{3!}

                                 = 0.05 + 0.15 + 0.224 + 0.224

                                 = 0.648

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7 0
3 years ago
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hram777 [196]

Answer:

<h2>Function is y = x.</h2><h2>Domain: -3 \leq x \leq 2.</h2><h2>Range: -5 \leq y \leq 4.</h2>

Step-by-step explanation:

In the given image, the line passes through (-1, -1) and (1, 1).

Let the equation of the line is y = mx + c, where m is the tangent of the line and c is a constant.

Putting the co-ordinates of the points in the equation, we get -1 = -m + c and 1 = m + c

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Hence, the function is y = x.

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5 0
3 years ago
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