Math Problem Solver

\(\frac{8}{11}\) rounded to the nearest \(\frac{1}{10}\) is \(0.7\).

Rounding \(\frac{8}{11}\) to the nearest \(\frac{1}{10}\) is equivalent to finding the integer multiple of \(\frac{1}{10}\) nearest to \(\frac{8}{11}\).

The two consecutive multiples of \(\frac{1}{10}\) that \(\frac{8}{11}\) falls between are:\[7 \cdot \frac{1}{10} = 0.7\] and \[(7 + 1) \cdot \frac{1}{10} = 0.8\] So:\[0.7 \le \frac{8}{11} \le 0.8\]

The distance from \(\frac{8}{11}\) to \(0.7\) is \(\left| \frac{8}{11} - 0.7 \right| = 0.0\overline{27}\).

The distance from \(\frac{8}{11}\) to \(0.8\) is \(\left| \frac{8}{11} - 0.8 \right| = 0.0\overline{72}\).

\(\frac{8}{11}\) is closer to \(0.7\) than it is to \(0.8\), so \(\frac{8}{11}\) rounded to the nearest \(\frac{1}{10}\) is \(0.7\).

\(7\) from the above statement about consecutive multiples of \(\frac{1}{10}\) can be computed from the formula for the greatest integer \(n\) such that \(n \cdot \frac{1}{10} \le \frac{8}{11}\): \[ n = \operatorname{floor}\left(\frac{8}{11} \div \frac{1}{10} \right) = \operatorname{floor}\left(7.\overline{27} \right) = 7\]