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3 years ago

5

14 3/5 - 3 ÷ 5 11/12

1 answer:

dalvyx [7]3 years ago

5 0

Answer:

33/355

not sure though

Step-by-step explanation:

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For the function y=8−5x, what is the ordered pair when x=7?

zzz [600]

Answer:

Ordered pairs are (0,7) , (1,8) , (-1,6) and (-7,0)

Step-by-step explanation:

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3 years ago

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ohaa [14]

Answer:

$(2,2)~ \text{and}~ (-2,-2)$

Step by step explanation:

$x^2 +y^2 =8~~~~~~...(i)\\\\x-y=0~~~~~~~~~...(ii)\\\\\text{From (ii):}\\\\x-y = 0 \implies x = y\\\\\text{Substitute x = y in eq (i):}\\\\~~~~~~y^2 +y^2 =8\\\\\implies 2y^2 = 8\\ \\\implies y^2 = \dfrac 82\\\\\implies y^2 = 4\\\\\implies y = \pm\sqrt 4\\\\\implies y = \pm 2\\\\\text{Substitute y = x in equation (i):}\\ \\~~~~~~x^2 +x^2 = 8\\ \\\implies 2x^2 = 8\\\\\implies x^2 = \dfrac 82\\\\\implies x^2 = 4\\\\\implies x = \pm\sqrt 4\\\\\implies x = \pm2\\\\$

$\text{Hence,}~ (x,y) = (\pm2, \pm 2)$

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2 years ago

What's total area of cube and cuboid?

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2 years ago

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Your social media ad costs $2.50 per thousand impressions. You get 3.14 million impressions. What is your approximate bill?

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2 years ago

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

Mkey [24]

Answer:

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

∴ we get

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

6 0

3 years ago