round off to the nearest hundredth: 1.07532
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MathBot Answer:
\(1.07532\) rounded to the nearest \(\frac{1}{100}\) is \(1.08\).
Rounding \(1.07532\) to the nearest \(\frac{1}{100}\) is equivalent to finding the integer multiple of \(\frac{1}{100}\) nearest to \(1.07532\).
The two consecutive multiples of \(\frac{1}{100}\) that \(1.07532\) falls between are:\[107 \cdot \frac{1}{100} = 1.07\] and \[(107 + 1) \cdot \frac{1}{100} = 1.08\] So:\[1.07 \le 1.07532 \le 1.08\]
The distance from \(1.07532\) to \(1.07\) is \(\left| 1.07532 - 1.07 \right| = 0.00532\).
The distance from \(1.07532\) to \(1.08\) is \(\left| 1.07532 - 1.08 \right| = 0.00468\).
\(1.07532\) is closer to \(1.08\) than it is to \(1.07\), so \(1.07532\) rounded to the nearest \(\frac{1}{100}\) is \(1.08\).
\(107\) from the above statement about consecutive multiples of \(\frac{1}{100}\) can be computed from the formula for the greatest integer \(n\) such that \(n \cdot \frac{1}{100} \le 1.07532\): \[ n = \operatorname{floor}\left(1.07532 \div \frac{1}{100} \right) = \operatorname{floor}\left(107.532 \right) = 107\]