(2^8 * 3^-5 * 6^0)^-2 * (3^-2/ 2^3) ^4 * 2^28

1 answer:

6 0

$\left(2^8\cdot \:3^{-5}\cdot \:1\right)^{-2}=\left(1\cdot \frac{256}{243}\right)^{-2}=\left(\frac{256}{243}\right)^{-2}=\left(\frac{243}{256}\right)^2$

$\left(2^8\cdot \:3^{-5}\cdot \:6^0\right)^{-2}\left(\frac{3^{-2}}{2^3}\right)^4\cdot \:2^{28}$

$=2^{28}\left(\frac{3^{-2}}{2^3}\right)^4\left(3^{-5}\cdot \:2^8\cdot \:1\right)^{-2}$

Consider

$\left(2^8\cdot \:3^{-5}\cdot \:1\right)^{-2}=\left(1\cdot \frac{256}{243}\right)^{-2}=\left(\frac{243}{256}\right)^2=\frac{243^2}{256^2}$

$=2^{28}\left(\frac{3^{-2}}{2^3}\right)^4\frac{243^2}{256^2}$

$=2^{28}\cdot \frac{1}{2^{12}\cdot \:3^8}\cdot \frac{243^2}{256^2}$

$=\frac{243^2\cdot \:1\cdot \:2^{28}}{256^2\cdot \:3^8\cdot \:2^{12}}$

$=\frac{2^{16}\cdot \:243^2}{3^8\cdot \:256^2}$

$=\frac{2^{16}\cdot \:3^{10}}{2^{16}\cdot \:3^8}$

$=3^2$

$=9$

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Consider rolling two fair dice one 3-sided the other 5-sided

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Since the dice are fair and the rolling are independent, each single outcome has probability 1/15. Every time we choose

$1\leq x\leq 3,\quad 1\leq y \leq 5$

We have $P(X=x)=\frac{1}{3}$ and $P(Y=y)=\frac{1}{5}$ , because the dice are fair.

Now we use the assumption of independence to claim that

$P(X=x, Y=y) = P(X=x)\cdot P(Y=y) =\dfrac{1}{3}\cdot\dfrac{1}{5} = \dfrac{1}{15}$

Now, we simply have to count in how many ways we can obtain every possible outcome for the sum. Consider the attached table: we can see that we can obtain:

2 in a unique way (1+1)
3 in two possible ways (1+2, 2+1)
4 in three possible ways
5 in three possible ways
6 in three possible ways
7 in two possible ways
8 in a unique way

This implies that the probabilities of the outcomes of $W=X+Y$ are the number of possible ways divided by 15: we can obtain 2 and 8 with probability 1/15, 3 and 7 with probability 2/15, and 4, 5 and 6 with probabilities 3/15=1/5