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Harman [31]
3 years ago
5

Here is a table showing all 52 cards in a standard deck.

Mathematics
1 answer:
stepan [7]3 years ago
8 0

We want to find the probability of drawing 3 red cards from a deck of 52 cards.

We will find that the probability is P = 0.1176

Let's see how to get that probability.

We know that a standard deck of 52 cards has:

  • 26 red cards
  • 26 black cards

And we assume that all the cards have the same probability of being drawn. Then the probability of drawing a red card in the first draw is just the quotient between the number of red cards and the total number of cards, we get:

p₁ = 26/52

For the second draw, we compute the probability in the same way, but now there are 25 red cards in the deck and 51 cards in total (because one is already drawn). Then the probability in this case is:

p₂ = 25/51

Finally, for the third card, we have 24 red cards and 50 total cards, then the probability will be:

p₃ = 24/50

The joint probability (the probability of drawing the 3 red cards in the same event) is the product of the individual probabilities.

P = p₁*p₂*p₃ = (26/52)*(25/51)*(24/50) = 0.1176

If you want to learn more, you can read:

brainly.com/question/10224828

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Answer:

53/5

Step-by-step explanation:

since 10 is the denominator of the denominator we can move it to the numerator, giving us 10n/5-3n = 1/5

multiply all sides by 5-3n, giving us 10n = (5-3n)/5

now multiply by 5 to get rid of the final denomintor, 50n = 5-3n

move 3n to the other side 53n = 5, n = 53/5

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What is 1.1865 rounded to the nearest cent.​
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Write 0.27 as a fraction
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572 cars were parked in a parking garage. The same number of cars was parked on each floor. If there were 4 floors, how many car
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If f(x) = |(x2 − 9)(x2 + 1)|, how many numbers in the interval [−1, 1] satisfy the conclusion of the mean value theorem?
ivann1987 [24]

Answer:

only one number c=0 in the interval [-1,1]

Step-by-step explanation:

Given : Function f(x) = |(x^2-9)(x^2 + 1)|   in the interval [-1,1]

To find : How many numbers in the interval [−1, 1] satisfy the conclusion of the mean value theorem.

Mean value theorem : If f is a continuous function on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists a point 'c' in (a,b) such that f'(c)=\frac{f(b)-f(a)}{b-a}

Solution : f(x) is a function that satisfies all of the following :

1) f(x)  is continuous on the closed interval [-1,1]  

\lim_{x\to a} f(x)=f(a)

2) f(x) is differentiable on the open interval  (-1,1)

Then there is a number  c such that  f'(c)=\frac{f(b)-f(a)}{b-a}

f(a)=f(-1) = |(-1^2-9)(-1^2 + 1)|=|(-8)(2)|=16

f(b)=f(1) = |(1^2-9)(1^2 + 1)|=|(8)(2)|=16

f'(c)=\frac{f(b)-f(a)}{b-a}=\frac{16-16}{2}=0

f'(c)=0 ........[1]

Now, we find f'(x)

f(x) = |(x^2-9)(x^2 + 1)|

f(x) =x^4-8x^2-9

Differentiating w.r.t  x

f'(x) =4x^3-16x

In place of x we put x=c

f'(c) =4c^3-16c

f'(c) =4c^3-16c=0  (by [1], f'(c)=0)

4c(c^2-4)=0

4c=0,c^2-4=0

either c=0 or  c^2-4=0\rightarrow c=\pm2

we cannot take c=\pm2  because they don't lie in the interval [-1,1]

Therefore, there is only one number c=0 which lie in interval [-1,1] and satisfying the conclusion of the mean value theorem.



3 0
3 years ago
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