3. Fill in the missing numbers to form equivalent rates. \( \begin{array}{l}\text { a. } \frac{4 \mathrm{~L}}{10 \mathrm{~m}^{2}}=\frac{10 \mathrm{~L}}{5 \mathrm{~m}^{2}}=\frac{1}{\mathrm{~m}^{2}} \\ \text { b. } \frac{\$ 9}{6 \mathrm{~min}}=\frac{}{2 \mathrm{~min}}=\frac{}{10 \mathrm{~min}}=\frac{-}{1 \text { hour }}\end{array} \)
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To make the first set of equivalent rates, observe the ratios. By simplifying \( \frac{4 \mathrm{~L}}{10 \mathrm{~m}^{2}} \), we can understand that each \( \mathrm{m}^2 \) uses \( \frac{4}{10} = 0.4 \) liters. Thus, for \( 5 \mathrm{~m}^{2} \), it would be \( 0.4 \times 5 = 2 \mathrm{~L} \). Now for the second set, we divide $9 by 6 minutes to find the rate per minute, which is \( \frac{3}{2} \) dollars per minute. So, for 2 minutes, it would be \( \frac{3}{2} \times 2 = 3 \) dollars. Extending this, for 10 minutes, we would use \( 3 \times 5 = 15 \) dollars, and for 1 hour (60 minutes), it would be \( 3 \times 30 = 90 \) dollars. Thus, the filled in missing numbers are: a. \( \frac{2 \mathrm{~L}}{5 \mathrm{~m}^{2}} \) b. \( \frac{3 \$}{2 \mathrm{~min}} \), \( 15 \$ \), and \( 90 \$ \) respectively.