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3 years ago

.What is the probability of two 6 sided diced rolled, adding up to lucky 13

Mathematics

1 answer:

BARSIC [14]3 years ago

5 0

Answer:

Step-by-step explanation:

Assuming the six sided dice are numbered 1 through six

The max each could roll is 6

6+6 = 12

They cannot reach a sum of 13

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love history [14]

Answer:

Step-by-step explanation:

4:2

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2 years ago

PLS HELP ME ASAP FOR 12 (SHOW WORK!!)

alex41 [277]

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3 years ago

I have a bad memory,i learned this but I forgot it (I swear I'm not lying)

mihalych1998 [28]

Answer:

Step-by-step explanation:

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3 years ago

HELP FAST PLEASE!!!!!

lidiya [134]

Alright, lets get started.

The answer Sandra got is correct that is zero but there are some mistakes she made while writing steps.

$2 cot(\frac{-9\pi}{2}) = 2 cot(\frac{9\pi}{2})$ Incorrect step

Because as per even/odd identity,

Cot (-Θ) = - cot (Θ)

Hence

$2 cot (\frac{-9\pi}{2}) = - 2 cot (\frac{9\pi}{2})$

Now,

It could be written as

$-2 cot (\frac{9\pi}{2}) = -2 cot (\frac{\pi}{2})$

$-2 * 0$

Hence the answer is 0. : Answer

Hope it will help :)

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3 years ago

For all real numbers x and y, if x # y = x(x-y), then x # (x # y) =

kati45 [8]

$x\#y=x(x-y)\\\\x\#(x\#y)\\\\x\#y=x(x-y)=x^2-xy\\\\x\#(x\#y)=x\#(x^2-xy)=x[x-(x^2-xy)]=x(x-x^2+xy)\\\\=x^2-x^3+x^2y$

$x^2-x^3+x^2y=8\\\\x^2(1-x+y)=8\\\\x^2(1-x+y)=2^2\cdot4\iff x^2=2^2\ and\ 1-x+y=4\\\\x=2\ and\ y=4-1+x\\\\x=2\ and\ y=3+2\\\\x=2\ and\ y=5$

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4 years ago

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