7. Ross is comparing \( \sqrt{11} \) and \( 5 . \overline{4} \). He says that \( \sqrt{11}>5 . \overline{4} \) because \( \sqrt{11}=5.5 \) a. What is the correct comparison?

To compare \( \sqrt{11} \) and \( 5 . \overline{4} \), we need to convert \( 5 . \overline{4} \) to a decimal form. Given that \( \sqrt{11} = 5.5 \), we can see that \( \sqrt{11} \) is greater than 5.5. Now, let's convert \( 5 . \overline{4} \) to a decimal form. \( 5 . \overline{4} \) can be written as \( 5 + \frac{4}{9} \) because the repeating decimal \( \overline{4} \) is equivalent to \( \frac{4}{9} \). So, \( 5 . \overline{4} = 5 + \frac{4}{9} = 5 + 0.444... = 5.444... \). Therefore, \( 5 . \overline{4} = 5.444... \). Comparing \( \sqrt{11} \) and \( 5 . \overline{4} \), we have: \( \sqrt{11} = 5.5 \) and \( 5 . \overline{4} = 5.444... \). Since \( 5.5 > 5.444... \), the correct comparison is \( \sqrt{11} > 5 . \overline{4} \).