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**Topic**

Write the round number of each of the following approximations:

a) -131 298 with precision \(d = 20\)

b) 0.02298 with precision \(d = 0.0006\)

**Solution method – See details**

We say \(a\) is an approximation of \(\overline a \) with precision \(d\) if \({\Delta _a} = \left| {a – \overline a } \right| \le d\)

**Detailed explanation**

a) -131 298 with precision \(d = 20\)

The largest row of precision \(d = 20\) is tens, so we round -131 298 to hundreds we get the rounded number of -131 298 which is -131 300

b) 0.02298 with precision \(d = 0.0006\)

The largest row of precision is \(d = 0.0006\) which is ten thousandths, so we round 0.02298 to thousandths we get 0.02298 rounding number 0.023

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