Jenn draws a scalene triangle, and measures the heights from each of the vertices to its opposite side. She discovers that the three height lengths are all roots of the polynomial $x^3 - 3.9 x^2 + 4.4 x - 1.2.$ What is the length of the inradius of the triangle?

$\textbf{(A) }\frac{3}{13}\qquad\textbf{(B) }\frac{3}{11}\qquad\textbf{(C) }\frac{2}{7}\qquad\textbf{(D) }\frac{8}{15}\qquad\textbf{(E) }\frac{9}{14}$
1. Let \( A \) denote the area of the triangle, and let \( h \), \( k \), and \( l \) denote the lengths of the heights from each vertex to the opposite side.
2. The side lengths of the triangle can be expressed in terms of the area and the heights as follows:
   \[
   a = \frac{2A}{h}, \quad b = \frac{2A}{k}, \quad c = \frac{2A}{l}
   \]
3. The semiperimeter \( s \) of the triangle is given by:
   \[
   s = \frac{a + b + c}{2} = \frac{\frac{2A}{h} + \frac{2A}{k} + \frac{2A}{l}}{2} = A \left( \frac{1}{h} + \frac{1}{k} + \frac{1}{l} \right)
   \]
4. The formula for the inradius \( r \) of a triangle is:
   \[
   r = \frac{A}{s}
   \]
5. Substituting the expression for \( s \) into the formula for \( r \), we get:
   \[
   r = \frac{A}{A \left( \frac{1}{h} + \frac{1}{k} + \frac{1}{l} \right)} = \frac{1}{\frac{1}{h} + \frac{1}{k} + \frac{1}{l}}
   \]
6. By Vieta's formulas for the polynomial \( x^3 - 3.9 x^2 + 4.4 x - 1.2 \), the roots \( h \), \( k \), and \( l \) satisfy:
   \[
   h + k + l = 3.9, \quad hk + kl + hl = 4.4, \quad hkl = 1.2
   \]
7. Therefore, the inradius \( r \) can be calculated as:
   \[
   r = \frac{hkl}{hk + kl + hl} = \frac{1.2}{4.4}
   \]
8. Simplifying the fraction:
   \[
   r = \frac{1.2}{4.4} = \frac{12}{44} = \frac{3}{11}
   \]

The final answer is \(\boxed{\frac{3}{11}}\).