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AMC 8 1987 Problem 1

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.4+.02+.006=.4+.02+.006= (A) .012(B) .066(C) .12(D) .24(E).426\text{(A)}\ .012 \text{(B)}\ .066 \text{(C)}\ .12 \text{(D)}\ .24 \text{(E)} .426

Dataset text NuminaMath-1.5. Quoted with source.

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0.426

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