Math Problem Solver

\(4.337704918032787\) rounded to the nearest \(\frac{1}{100}\) is \(4.34\).

Rounding \(4.337704918032787\) to the nearest \(\frac{1}{100}\) is equivalent to finding the integer multiple of \(\frac{1}{100}\) nearest to \(4.337704918032787\).

The two consecutive multiples of \(\frac{1}{100}\) that \(4.337704918032787\) falls between are:\[433 \cdot \frac{1}{100} = 4.33\] and \[(433 + 1) \cdot \frac{1}{100} = 4.34\] So:\[4.33 \le 4.337704918032787 \le 4.34\]

The distance from \(4.337704918032787\) to \(4.33\) is \(\left| 4.337704918032787 - 4.33 \right| = 0.007704918032787\).

The distance from \(4.337704918032787\) to \(4.34\) is \(\left| 4.337704918032787 - 4.34 \right| = 0.002295081967213\).

\(4.337704918032787\) is closer to \(4.34\) than it is to \(4.33\), so \(4.337704918032787\) rounded to the nearest \(\frac{1}{100}\) is \(4.34\).

\(433\) from the above statement about consecutive multiples of \(\frac{1}{100}\) can be computed from the formula for the greatest integer \(n\) such that \(n \cdot \frac{1}{100} \le 4.337704918032787\): \[ n = \operatorname{floor}\left(4.337704918032787 \div \frac{1}{100} \right) = \operatorname{floor}\left(433.7704918032787 \right) = 433\]