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This is actually mentioned in my module,I am not sure how they have got this result but I think this may not be correct,I tried using wolfarm alpha which gives false or am I missing something?


I was missing approximation,it seems like data interpretation problems keeps relying on this, so,$\frac{44}{9.3} = 4.73118 \text{ and } \frac{5.5}{1.16} = 4.74138 $ Hence, $44:9.3 \approx 5.5:1.16$

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    Jasper, thanks I figured it out, if you post that comment as your answer, I will rather be happy in accepting it. :)2011-01-06

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I eyeballed it this way:

$\frac {44} {9.3} = \frac {11} {2.325}$

while:

$\frac {5.5} {1.16} = \frac {11} {2.32}$

So one expects roughly three significant figures of agreement but not equality.

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A quick computation gives that $ \begin{align} {440 \over 93} = {550 \over 116} &\iff {440 \times 116 \over 93 \times 116} = {550 \times 93 \over 116 \times 93} \\\\ &\iff 440 \times 116 = 550 \times 93 \\\\ &\iff 44 \times 116 = 55 \times 93, \end{align} $ and in the last equality the LHS is even and RHS is odd, so they can't possibly be equal.

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    Okay I figured it out $\frac{44}{9.3} = 4.73118$ and $\frac{5.5}{1.16} = 4.74138$. Hence the approximation.2011-01-06
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Probably the approximation arises from solving $\rm\displaystyle\ \frac{44}{x} = \frac{5.5}{1.16}\ \Rightarrow\;\ x = \frac{1.16\cdot 44}{5.5} = 9.28 \approx 9.3$