1. The problem asks which table has a constant of proportionality between $y$ and $x$ equal to $\frac{1}{5}$. This means we want to check if $y = kx$ where $k = \frac{1}{5}$. 2. The constant of proportionality $k$ is found by dividing $y$ by $x$: $$k = \frac{y}{x}$$ 3. We check each table's pairs $(x,y)$ to see if $\frac{y}{x} = \frac{1}{5}$. Table A: - For $x=2$, $y=\frac{4}{5}$, $\frac{y}{x} = \frac{\frac{4}{5}}{2} = \frac{4}{5} \times \frac{1}{2} = \frac{4}{10} = \frac{2}{5} \neq \frac{1}{5}$. - For $x=7$, $y=\frac{14}{5}$, $\frac{y}{x} = \frac{\frac{14}{5}}{7} = \frac{14}{5} \times \frac{1}{7} = \frac{14}{35} = \frac{2}{5} \neq \frac{1}{5}$. - For $x=12$, $y=\frac{24}{5}$, $\frac{y}{x} = \frac{\frac{24}{5}}{12} = \frac{24}{5} \times \frac{1}{12} = \frac{24}{60} = \frac{2}{5} \neq \frac{1}{5}$. Table B: - For $x=8$, $y=\frac{8}{5}$, $\frac{y}{x} = \frac{\frac{8}{5}}{8} = \frac{8}{5} \times \frac{1}{8} = \frac{1}{5}$. - For $x=9$, $y=\frac{9}{5}$, $\frac{y}{x} = \frac{\frac{9}{5}}{9} = \frac{9}{5} \times \frac{1}{9} = \frac{1}{5}$. - For $x=10$, $y=2$, $\frac{y}{x} = \frac{2}{10} = \frac{1}{5}$. Table C: - For $x=3$, $y=\frac{16}{5}$, $\frac{y}{x} = \frac{\frac{16}{5}}{3} = \frac{16}{5} \times \frac{1}{3} = \frac{16}{15} \neq \frac{1}{5}$. - For $x=5$, $y=\frac{26}{5}$, $\frac{y}{x} = \frac{\frac{26}{5}}{5} = \frac{26}{5} \times \frac{1}{5} = \frac{26}{25} \neq \frac{1}{5}$. - For $x=7$, $y=\frac{36}{5}$, $\frac{y}{x} = \frac{\frac{36}{5}}{7} = \frac{36}{5} \times \frac{1}{7} = \frac{36}{35} \neq \frac{1}{5}$. 4. Only Table B has $\frac{y}{x} = \frac{1}{5}$ for all pairs, so the constant of proportionality is $\frac{1}{5}$ in Table B. **Final answer:** Table B