Math Problem Solver

\(0.0995\) rounded to the nearest \(\frac{1}{100}\) is \(0.1\).

Rounding \(0.0995\) to the nearest \(\frac{1}{100}\) is equivalent to finding the integer multiple of \(\frac{1}{100}\) nearest to \(0.0995\).

The two consecutive multiples of \(\frac{1}{100}\) that \(0.0995\) falls between are:\[9 \cdot \frac{1}{100} = 0.09\] and \[(9 + 1) \cdot \frac{1}{100} = 0.1\] So:\[0.09 \le 0.0995 \le 0.1\]

The distance from \(0.0995\) to \(0.09\) is \(\left| 0.0995 - 0.09 \right| = 0.0095\).

The distance from \(0.0995\) to \(0.1\) is \(\left| 0.0995 - 0.1 \right| = 0.0005\).

\(0.0995\) is closer to \(0.1\) than it is to \(0.09\), so \(0.0995\) rounded to the nearest \(\frac{1}{100}\) is \(0.1\).

\(9\) from the above statement about consecutive multiples of \(\frac{1}{100}\) can be computed from the formula for the greatest integer \(n\) such that \(n \cdot \frac{1}{100} \le 0.0995\): \[ n = \operatorname{floor}\left(0.0995 \div \frac{1}{100} \right) = \operatorname{floor}\left(9.95 \right) = 9\]