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agdgdgdgwngo

  • 4 years ago

Solve this numerical syllogism for me: Eight of ten A's are B's; four of ten A's are C's. What is the minimum number of B's that are also C's?

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  1. osanseviero
    • 4 years ago
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    2 B´s are C´s

  2. genius12
    • 4 years ago
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    2 Bs would be the minimum number of Bs that can be Cs.

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