The problem asks which sign (greater than > or less than <) makes the statement true:

$-\sqrt{3} \, ? \, -\frac{11}{10}$

Let's evaluate both sides.

1. Evaluate $-\sqrt{3}$:

   • The square root of 3 is approximately $\sqrt{3} \approx 1.732$.
   • Therefore, $-\sqrt{3} \approx -1.732$.

2. Evaluate $-\frac{11}{10}$:

   • $-\frac{11}{10} = -1.1$.

Now, compare the two values:

• $-1.732$ is less than $-1.1$.

Thus, the correct sign to make the statement true is <.

So the complete statement is: $-\sqrt{3} < -\frac{11}{10}$

Would you like any more details or have any other questions?

Here are 5 related questions to consider:

1. What is the value of $\sqrt{2}$ and how does it compare to $-\frac{5}{4}$?
2. How would you express $\frac{11}{10}$ as a decimal?
3. Can you explain why negative square roots are always less than their positive counterparts?
4. How would you determine the correct sign for $-\frac{3}{2}$ and $-1.5$?
5. What is the approximate decimal value of $\sqrt{5}$ and how does it compare to $2$?

Tip: Remember that when comparing negative numbers, the one with the smaller absolute value is actually greater.