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3 years ago

2x+3y+2z=13 2y+z=1 z=3

Mathematics

1 answer:

Igoryamba3 years ago

6 0

<h3>Given</h3>

z = 3

2y +z = 1

2x +3y +2z = 13

x, y, z

<h3>Solution</h3>

... z = 3 is given

Substitute that into the second equation.

... 2y + 3 = 1

... 2y = -2 . . . . . subtract 3

... y = -1 . . . . . . . divide by the coefficient of y

Substitute these two solutions into the third equation.

... 2x +3(-1) +2(3) = 13

... 2x = 10 . . . . . . . . . . . . collect terms, subtract 3

... x = 5 . . . . . . . . . . . . . . divide by the coefficient of x

The variable values are: x = 5, y = -1, z = 3

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3 years ago

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3 years ago

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A deck of 52 cards contains 12 picture cards. If the 52 cards are distributed in a random manner among four players in such a wa

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The probability that each player will receive three picture cards = 0.0324

Step-by-step explanation:

As given,

A deck of 52 cards contains 12 picture cards

Remaining card = 52 - 12 = 40

So,

Total number of ways in which 12 picture card is distributed = $\frac{12!}{3! 3! 3! 3!}$

Now,

The Total number of ways in which Remaining cards are distributed = $\frac{40!}{10! 10! 10! 10!}$

So,

Total number of ways of getting 3 picture card and remaining card = $\frac{12!}{3! 3! 3! 3!}$ × $\frac{40!}{10! 10! 10! 10!}$

= $\frac{12! 40!}{(3!)^{4} (10!)^{4} }$

Now,

Total number of ways to distribute 52 cards so that each people get 13 card = $\frac{52!}{13! 13! 13! 13!} = \frac{52!}{ (13!)^{4} }$

∴ The probability = $\frac{\frac{12! 40!}{(3!)^{4} (10!)^{4} }}{\frac{52!}{(13!)^{4} }}$

= $\frac{12! 40!}{(3!)^{4} (10!)^{4} }$ × $\frac{(13!)^{4} }{ 52! }$

= $\frac{12! 40!}{(3!)^{4} (10!)^{4} }$ × $\frac{(13.12.11.10!)^{4} }{ 52.51.50.49.48.47.46.45.44.43.42.41.40! }$

= $\frac{12!}{(3!)^{4} }$ × $\frac{(13.12.11)^{4} }{ 52.51.50.49.48.47.46.45.44.43.42.41}$

= $\frac{479,001,600}{(6)^{4} }$ × $\frac{(1716)^{4} }{ 52.51.50.49.48.47.46.45.44.43.42.41}$

= 0.0324

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