Math Problem Solver

\(4.5671\) rounded to the nearest \(\frac{1}{100}\) is \(4.57\).

Rounding \(4.5671\) to the nearest \(\frac{1}{100}\) is equivalent to finding the integer multiple of \(\frac{1}{100}\) nearest to \(4.5671\).

The two consecutive multiples of \(\frac{1}{100}\) that \(4.5671\) falls between are:\[456 \cdot \frac{1}{100} = 4.56\] and \[(456 + 1) \cdot \frac{1}{100} = 4.57\] So:\[4.56 \le 4.5671 \le 4.57\]

The distance from \(4.5671\) to \(4.56\) is \(\left| 4.5671 - 4.56 \right| = 0.0071\).

The distance from \(4.5671\) to \(4.57\) is \(\left| 4.5671 - 4.57 \right| = 0.0029\).

\(4.5671\) is closer to \(4.57\) than it is to \(4.56\), so \(4.5671\) rounded to the nearest \(\frac{1}{100}\) is \(4.57\).

\(456\) from the above statement about consecutive multiples of \(\frac{1}{100}\) can be computed from the formula for the greatest integer \(n\) such that \(n \cdot \frac{1}{100} \le 4.5671\): \[ n = \operatorname{floor}\left(4.5671 \div \frac{1}{100} \right) = \operatorname{floor}\left(456.71 \right) = 456\]