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Anastaziya [24]
3 years ago
12

You draw 4 cards from a deck of 52 cards with replacement. What are the probabilities of drawing a black card on each of your fo

ur trials?
Mathematics
2 answers:
lbvjy [14]3 years ago
7 0
1/2 ( or 2/4) because half of the cards are black
lawyer [7]3 years ago
4 0

Solution:

As, in a pack of cards, there are 26 black cards and 26 red cards.

Probability of an event = \frac{\text{total favorable outcome}}{\text{total possible outcome}}

Probability of drawing a black card from a pack of 52 cards = \frac{26}{52}=\frac{1}{2}

As each card drawn is replaced,Each black card draw is an independent with another black card draw.

Probabilities of drawing a black card on each of  four trial=

  \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}=[\frac{1}{2}]^4

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Given g(x)=3x-3, solve for x when g(x)=6.
lana [24]

Answer:

3

Step-by-step explanation:

g ( x ) = 6

g ( x ) = 3x - 3

3x - 3 = 6

3x = 6 + 3

3x = 9

x = 9 / 3

x = 3

7 0
2 years ago
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Step-by-step explanation:

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3 years ago
2+15+12+85+666+123*322158%5556
bonufazy [111]
The answer for this question is 822
5 0
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A farmer produces both beans and corn on her farm. If she must give up 16 bushels of corn to be able to get 6 bushels of beans,
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3 years ago
Solve this identifying holes, vertical asymptotes, and horizontal asymptotes for
USPshnik [31]

x^2+7x+12=x^2+4x+3x+12=x(x+4)+3(x+4)=(x+4)(x+3)\\\\-x^2-3x+4=-(x^2+3x-4)=-(x^2+4x-x-4)\\\\=-[x(x+4)-1(x+4)]=-(x+4)(x-1)\\\\f(x)=\dfrac{x^2+7x+12}{-x^2-3x+4}=\dfrac{(x+4)(x+3)}{-(x+4)(x-1)}\\\\\text{Vertical asymptotes:}\\\\(x+4)(x-1)=0\iff x+4=0\ \vee\ x-1=0\\\\\boxed{x=-4\ and\ x=1}\\\\\text{Horizontal asymptotes:}

\lim\limits_{x\to\pm\infty}\dfrac{x^2+7x+12}{-x^2-3x+4}=\lim\limits_{x\to\pm\infty}\dfrac{x^2\left(1+\dfrac{7}{x}+\dfrac{12}{x^2}\right)}{x^2\left(-1-\dfrac{3}{x}+\dfrac{4}{x^2}\right)}=\dfrac{1}{-1}=-1\\\\\boxed{y=-1}

6 0
4 years ago
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