$5\cdot \left(4,2+6\frac{1}{3}-2\frac{7}{15}\right)=5\cdot \left(4\frac{2}{10}+6\frac{1}{3}-2\frac{7}{15}\right)=5\cdot \left(\frac{21}{5}+\frac{19}{3}-2\frac{7}{15}\right)-5\cdot \left(\frac{63}{15}+\frac{95}{15}-2\frac{7}{15}\right)=$ $5\cdot \left(4,2+6\frac{1}{3}-2\frac{7}{15}\right)=5\cdot \left(4\frac{2}{10}+6\frac{1}{3}-2\frac{7}{15}\right)=5\cdot \left(\frac{21}{5}+\frac{19}{3}-2\frac{7}{15}\right)-5\cdot \left(\frac{63}{15}+\frac{95}{15}-2\frac{7}{15}\right)=$

$=5\cdot \left(\frac{158}{15}-\frac{37}{15}\right)={\sout{5}}^1\cdot \frac{151}{{\sout{15}}^3}=\frac{151}{3}=40\frac{1}{3}$

Odpowiedź:

B