Math Problem Solver

\(781.448\) rounded to the nearest \(\frac{1}{10}\) is \(781.4\).

Rounding \(781.448\) to the nearest \(\frac{1}{10}\) is equivalent to finding the integer multiple of \(\frac{1}{10}\) nearest to \(781.448\).

The two consecutive multiples of \(\frac{1}{10}\) that \(781.448\) falls between are:\[7814 \cdot \frac{1}{10} = 781.4\] and \[(7814 + 1) \cdot \frac{1}{10} = 781.5\] So:\[781.4 \le 781.448 \le 781.5\]

The distance from \(781.448\) to \(781.4\) is \(\left| 781.448 - 781.4 \right| = 0.048\).

The distance from \(781.448\) to \(781.5\) is \(\left| 781.448 - 781.5 \right| = 0.052\).

\(781.448\) is closer to \(781.4\) than it is to \(781.5\), so \(781.448\) rounded to the nearest \(\frac{1}{10}\) is \(781.4\).

\(7814\) from the above statement about consecutive multiples of \(\frac{1}{10}\) can be computed from the formula for the greatest integer \(n\) such that \(n \cdot \frac{1}{10} \le 781.448\): \[ n = \operatorname{floor}\left(781.448 \div \frac{1}{10} \right) = \operatorname{floor}\left(7814.48 \right) = 7814\]