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Free_Kalibri [48]
3 years ago
15

You have a set of 10 cards—five red and five blue. Each group of five colored cards is numbered one through five.What is the pro

bability of drawing a red four and then a blue four, while replacing the card between the drawings?
Mathematics
2 answers:
MatroZZZ [7]3 years ago
7 0
The probability of drawing the Red 4 is 1 in 10. So, the chances of getting it would be 1/10.The probability is also the same for drawing the Blue 4, so it would be 1/10 also.

The probability assignment for this problem is the classical method which means that experimental outcomes are equally likely. So, just multiply the probability of the first event to the second event.
= 1/10 * 1/10 = 1/100
The probability of getting the blue and red four is 1/100 or 0.01.
ivann1987 [24]3 years ago
4 0
The answer is 1/100.
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Answer:

\displaystyle x_1=-1+\sqrt{3}i

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Step-by-step explanation:

<u>Second-Degree Equation</u>

The second-degree equation or quadratic equation has the general form

ax^2+bx+c=0

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There are many methods to solve the equation, one of the most-used is by using the solver formula:

\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

The equation of the question has the values: a=1, b=2, c=4, thus the values of x are

\displaystyle x=\frac{-2\pm \sqrt{2^2-4\cdot 1\cdot 4}}{2\cdot 1}

\displaystyle x=\frac{-2\pm \sqrt{-12}}{2}

Since the square root has a negative argument, both solutions for x are imaginary or complex. Simplifying the radical

\displaystyle x=\frac{-2\pm 2\sqrt{-3}}{2}=-1\pm\sqrt{3}i

The solutions are

\displaystyle x_1=-1+\sqrt{3}i

\displaystyle x_2=-1-\sqrt{3}i

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