Question

Media ponderata a numerelor 4,2 si 3,5(3) cu ponderile 7 si 3
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Answer 1

$Mp= \frac{4,2*7+ 3,5(3)*3}{7+3} = \frac{ \frac{42}{10}*7+ \frac{353-35}{90}*3 }{10} = \frac{ \frac{294}{10} + \frac{318}{30} }{10} = \frac{ \frac{882+318}{30} }{10} = \frac{1200}{30} * \frac{1}{10} =$ $\frac{1200}{300} = 4$