Math Problem Solver

\(11.1264\) rounded to the nearest \(\frac{1}{10}\) is \(11.1\).

Rounding \(11.1264\) to the nearest \(\frac{1}{10}\) is equivalent to finding the integer multiple of \(\frac{1}{10}\) nearest to \(11.1264\).

The two consecutive multiples of \(\frac{1}{10}\) that \(11.1264\) falls between are:\[111 \cdot \frac{1}{10} = 11.1\] and \[(111 + 1) \cdot \frac{1}{10} = 11.2\] So:\[11.1 \le 11.1264 \le 11.2\]

The distance from \(11.1264\) to \(11.1\) is \(\left| 11.1264 - 11.1 \right| = 0.0264\).

The distance from \(11.1264\) to \(11.2\) is \(\left| 11.1264 - 11.2 \right| = 0.0736\).

\(11.1264\) is closer to \(11.1\) than it is to \(11.2\), so \(11.1264\) rounded to the nearest \(\frac{1}{10}\) is \(11.1\).

\(111\) from the above statement about consecutive multiples of \(\frac{1}{10}\) can be computed from the formula for the greatest integer \(n\) such that \(n \cdot \frac{1}{10} \le 11.1264\): \[ n = \operatorname{floor}\left(11.1264 \div \frac{1}{10} \right) = \operatorname{floor}\left(111.264 \right) = 111\]