Justify Round each number to the thousandths place. Describe how you got your answer. \( \begin{array}{ll}\text { c. } 0 . \overline{6} & \text { d. } 5.3 \overline{81}\end{array} \)

The thousandths place is the third digit after the decimal point. We need to look at the digit immediately to the right (the ten-thousandths place) to decide whether to round up. **For \( 0.\overline{6} \):** - Writing out a few digits: \( 0.6666\ldots \) - The digit in the thousandths place is \(6\). - The next digit (ten-thousandths) is also \(6\) (which is \(\geq 5\)), so we round up the thousandths digit. - Rounding \(6\) up gives \(7\). Thus, \( 0.\overline{6} \) rounded to the thousandths place is \( \boxed{0.667} \). **For \( 5.3\overline{81} \):** - Write out the decimal: \( 5.381818\ldots \) - The digits are: \(5.3\) (tenths), \(8\) (hundredths), \(1\) (thousandths), followed by \(8\) (ten-thousandths). - The digit in the thousandths place is \(1\), and the next digit is \(8\) (which is \(\geq 5\)), so we round up \(1\) to \(2\). Hence, \( 5.3\overline{81} \) rounded to the thousandths place is \( \boxed{5.382} \).