1. The problem involves analyzing three tables of values $(x, y)$ and their corresponding constants $k$. 2. Each table suggests a relationship between $x$ and $y$ that can be expressed as $y = kx$. 3. For the first table: $x = \{1.5, 3, 8\}$ and $y = \{1, 2, 4\}$ with $k = 1.5$. 4. Check if $y = kx$ holds: For $x=1.5$, $y = 1.5 \times 1.5 = 2.25$, but given $y=1$, so $k=1.5$ does not fit perfectly. 5. For the second table: $x = \{2, 4, 5, 7\}$ and $y = \{1, 2, 2.5, 3.5\}$ with $k=2$. 6. Check $y = kx$: For $x=2$, $y=2 \times 2=4$, but given $y=1$, so $k=2$ does not fit. 7. For the third table: $x = \{1, 2, 3, 4\}$ and $y = \{2, 4, 6, 8\}$ with $k=0.5$. 8. Check $y = kx$: For $x=1$, $y=0.5 \times 1=0.5$, but given $y=2$, so $k=0.5$ does not fit. 9. Conclusion: The given $k$ values do not match the data in the tables if assuming $y = kx$. 10. To find the correct $k$ for each table, use $k = \frac{y}{x}$ for each pair. 11. For the first table, $k$ values are $\frac{1}{1.5} = 0.666...$, $\frac{2}{3} = 0.666...$, $\frac{4}{8} = 0.5$ (not consistent). 12. For the second table, $k$ values are $\frac{1}{2} = 0.5$, $\frac{2}{4} = 0.5$, $\frac{2.5}{5} = 0.5$, $\frac{3.5}{7} = 0.5$ (consistent). 13. For the third table, $k$ values are $\frac{2}{1} = 2$, $\frac{4}{2} = 2$, $\frac{6}{3} = 2$, $\frac{8}{4} = 2$ (consistent). 14. Therefore, the correct $k$ values are approximately $0.67$ for the first table (not perfectly consistent), $0.5$ for the second table, and $2$ for the third table. 15. The given $k$ values in the problem seem mismatched with the data. Final answer: The constant of proportionality $k$ for each table based on the data is approximately $0.67$, $0.5$, and $2$ respectively, not the given $1.5$, $2$, and $0.5$.