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AMC 8 1989 Problem 5

  • Paper

−15+9×(6÷3)=-15+9\times (6\div 3) = (A) −48(B) −12(C) −3(D) 3(E) 12\text{(A)}\ -48 \text{(B)}\ -12 \text{(C)}\ -3 \text{(D)}\ 3 \text{(E)}\ 12

Dataset text NuminaMath-1.5. Quoted with source.

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