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Montano1993 [528]
3 years ago
9

At a convention for lawyers it was known that of the 100 present, at least one was honest, yet if you met any two of the lawyers

, you could guarantee that at least one of the two would be crooked. How many honest lawyers were present?
Mathematics
1 answer:
Bogdan [553]3 years ago
4 0
In my opinion there would be fifty one honest lawers
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Step-by-step explanation:

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Width of bigger package =<em><u>2</u></em><em><u>5</u></em><em><u>×</u></em><em><u>1</u></em><em><u>2</u></em><em><u>5</u></em><em><u>/</u></em><em><u>8</u></em><em><u>=</u></em><em><u>3</u></em><em><u>9</u></em><em><u>0</u></em><em><u>.</u></em><em><u>6</u></em><em><u>2</u></em><em><u>5</u></em><em><u>c</u></em><em><u>m</u></em>

Height of bigger package =<em><u>4</u></em><em><u>5</u></em><em><u>×</u></em><em><u>1</u></em><em><u>2</u></em><em><u>5</u></em><em><u>/</u></em><em><u>8</u></em><em><u>=</u></em><em><u>7</u></em><em><u>0</u></em><em><u>3</u></em><em><u>.</u></em><em><u>1</u></em><em><u>2</u></em><em><u>5</u></em><em><u>cm</u></em>

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Answer:

Step-by-step explanation:

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