The problem asks us to order the given numbers from least to greatest. The numbers are $\sqrt{15}$, $3.\overline{4}$, $2\pi - 3$, and $\frac{17}{6}$.

1. Problem Description

The problem asks us to order the given numbers from least to greatest. The numbers are $\sqrt{15}$, $3.\overline{4}$, $2\pi - 3$, and $\frac{17}{6}$.

2. Solution Steps

First, we approximate each number.

$\sqrt{15}$ is between $\sqrt{9} = 3$ and $\sqrt{16} = 4$. Since 15 is closer to 16 than 9, $\sqrt{15}$ is closer to 4 than

3. We can estimate $\sqrt{15} \approx 3.9$.

$3.\overline{4} = 3.444...$.

$2\pi - 3$. We know $\pi \approx 3.14159$, so $2\pi \approx 2(3.14159) = 6.28318$. Therefore, $2\pi - 3 \approx 6.28318 - 3 = 3.28318$.

$\frac{17}{6} = 2\frac{5}{6} = 2 + \frac{5}{6}$. Since $\frac{5}{6} \approx 0.833$, we have $\frac{17}{6} \approx 2.833$.

Now we compare the approximated values:

$\sqrt{15} \approx 3.9$

$3.\overline{4} \approx 3.444...$

$2\pi - 3 \approx 3.28318$

$\frac{17}{6} \approx 2.833$

Ordering from least to greatest gives:

$\frac{17}{6} < 2\pi - 3 < 3.\overline{4} < \sqrt{15}$.

3. Final Answer

The order from least to greatest is: $\frac{17}{6}$, $2\pi - 3$, $3.\overline{4}$, $\sqrt{15}$.