The problem asks us to compare $3\frac{2}{3}$ and $\sqrt{11}$. We need to determine if $3\frac{2}{3}$ is less than, greater than, or equal to $\sqrt{11}$.

1. Problem Description

The problem asks us to compare $3\frac{2}{3}$ and $\sqrt{11}$. We need to determine if $3\frac{2}{3}$ is less than, greater than, or equal to $\sqrt{11}$.

2. Solution Steps

First, convert the mixed number $3\frac{2}{3}$ to an improper fraction:

$3\frac{2}{3} = \frac{3 \times 3 + 2}{3} = \frac{9 + 2}{3} = \frac{11}{3}$.

Now, we want to compare $\frac{11}{3}$ with $\sqrt{11}$. To make the comparison easier, we can square both numbers:

$(\frac{11}{3})^2 = \frac{11^2}{3^2} = \frac{121}{9}$

$(\sqrt{11})^2 = 11$

Next, we can express 11 as a fraction with a denominator of 9:

$11 = \frac{11 \times 9}{9} = \frac{99}{9}$

Now we can compare $\frac{121}{9}$ and $\frac{99}{9}$. Since $121 > 99$, we have $\frac{121}{9} > \frac{99}{9}$.

Therefore, $(\frac{11}{3})^2 > (\sqrt{11})^2$.

Since both $\frac{11}{3}$ and $\sqrt{11}$ are positive numbers, if their squares satisfy an inequality, then the numbers themselves also satisfy the same inequality.

Thus, $\frac{11}{3} > \sqrt{11}$, which means $3\frac{2}{3} > \sqrt{11}$.

3. Final Answer

$3\frac{2}{3} > \sqrt{11}$