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1ES/1L Spé Corrigé du DNS 4

N°59 page 118. Soit V la valeur manquante. L'effectif total est 5+3+6 = 14. La moyenne est donc (5×6+3×10+6×V)÷14 = 12 donc 60 + 6V = 12×14 donc 6V = 168 ? 60 ...

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