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Jlenok [28]
3 years ago
8

y^4}{5x^6y^7 })^-2" alt="(\frac{3x^-3y^4}{5x^6y^7 })^-2" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
viktelen [127]3 years ago
7 0

9514 1404 393

Answer:

  (25/9)x^18·y^6

Step-by-step explanation:

Maybe you want this simplified.

The appropriate rules of exponents are ...

  (a^b)/(a^c) = a^(b-c)

  (a^b)^c = a^(bc)

__

  \left(\dfrac{3x^{-3}y^4}{5x^6y^7}\right)^{-2}=\left(\dfrac{3x^{-3-6}y^{4-7}}{5}\right)^{-2}=\dfrac{3^{-2}}{5^{-2}}\cdot x^{(-2)(-9)}y^{(-2)(-3)}\\\\=\boxed{\dfrac{25}{9}x^{18}y^{6}}

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Three identical fair coins are thrown simultaneously until all three show the same face. What is the probability that they are t
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Answer:

The probability that the coins are thrown more than three times to show the same face is 0.3164.

Step-by-step explanation:

The problem is related to Geometric distribution.

The Geometric distribution defines the probability distribution of <em>X</em> failures before the first success.

The probability distribution function is:

P(X=k)=(1-p)^{k}p;\    k = 0, 1, 2, ...

First compute the probability that in the i^{th} throw all the three coins will show the same face.

P (All the 3 coins shows the same face) = P (All the three coins shows Heads) + P (All the three coins shows Tails)

                      =(\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2} )+(\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2} )\\=\frac{1}{8}+\frac{1}{8}\\  =\frac{2}{8} \\=\frac{1}{4}

Now compute the probability that it takes more than 3 throws for the coins to show the same face.

P (<em>X</em> > 3) = 1 - P (<em>X</em> ≤ 3)

=1-[P(X=1)+P(X=2)+P(X=3)]\\=1-[[(1-\frac{1}{4} )^{0}\times\frac{1}{4}]+[(1-\frac{1}{4} )^{1}\times\frac{1}{4}]+ [(1-\frac{1}{4} )^{2}\times\frac{1}{4}]+[(1-\frac{1}{4} )^{3}\times\frac{1}{4}]]\\=1-[0.2500+0.1875+0.1406+0.1055]\\=1-0.6836\\=0.3164

Thus, the probability that it takes more than 3 throws for the coins to show the same face is 0.3164.

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