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1979 AMC 12 Question 6

AMC12 · 1979 AMC 12

Question 6

$\frac32+\frac54+\frac98+\frac{17}{16}$$+\frac{33}{32}+\frac{65}{64}-7=$
  1. $-\frac1{64}$
  2. $-\frac1{16}$
  3. $0$
  4. $\frac1{16}$
  5. $\frac1{64}$