1. The problem is to estimate the square root of 0.036 using a number line. 2. Recall that the square root of a number $x$ is a number $y$ such that $y^2 = x$. 3. Important rule: The square root of a decimal less than 1 will be a decimal greater than the original number but less than 1. 4. Note that $0.036$ is close to $0.04$, and $\sqrt{0.04} = 0.2$ because $0.2^2 = 0.04$. 5. Since $0.036$ is slightly less than $0.04$, its square root will be slightly less than $0.2$. 6. On a number line, locate $0.2$ and estimate a point slightly to the left to represent $\sqrt{0.036}$. 7. To be more precise, calculate $\sqrt{0.036} = \sqrt{36 \times 10^{-3}} = \sqrt{36} \times \sqrt{10^{-3}} = 6 \times 10^{-1.5} = 6 \times 0.031622 = 0.1897$ approximately. 8. Therefore, $\sqrt{0.036} \approx 0.19$ on the number line. 9. This means the point is just before $0.2$ on the number line. 10. Final answer: $\boxed{0.19}$.