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AMC 8 1989 Problem 2

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210+4100+61000=\frac{2}{10}+\frac{4}{100}+\frac{6}{1000}= (A) .012(B) .0246(C) .12(D) .246(E) 246\text{(A)}\ .012 \text{(B)}\ .0246 \text{(C)}\ .12 \text{(D)}\ .246 \text{(E)}\ 246

Dataset text NuminaMath-1.5. Quoted with source.

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0.246

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