The expression in question is:

$\frac{1}{\sqrt{27}} - \frac{5\sqrt{3}}{4\sqrt{24}}$

We need to rewrite this expression in the form:

$a\sqrt{3} - b\sqrt{2}$

where $a$ and $b$ are constants to be determined, and $b > 0$.

Let's break this down step-by-step.

Step 1: Simplify $\frac{1}{\sqrt{27}}$

We can simplify this by recognizing that:

$\sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}$

So:

$\frac{1}{\sqrt{27}} = \frac{1}{3\sqrt{3}} = \frac{\sqrt{3}}{9}$

Step 2: Simplify $\frac{5\sqrt{3}}{4\sqrt{24}}$

Next, we simplify $\frac{5\sqrt{3}}{4\sqrt{24}}$. We first note that:

$\sqrt{24} = \sqrt{4 \times 6} = 2\sqrt{6}$

So:

$\frac{5\sqrt{3}}{4\sqrt{24}} = \frac{5\sqrt{3}}{4 \times 2\sqrt{6}} = \frac{5\sqrt{3}}{8\sqrt{6}}$

Now we can multiply both the numerator and denominator by $\sqrt{6}$ to rationalize the denominator:

$\frac{5\sqrt{3}}{8\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}} = \frac{5\sqrt{18}}{8 \times 6} = \frac{5 \times 3\sqrt{2}}{48} = \frac{15\sqrt{2}}{48} = \frac{5\sqrt{2}}{16}$

Step 3: Combine the two expressions

Now we have:

$\frac{\sqrt{3}}{9} - \frac{5\sqrt{2}}{16}$

This is already in the desired form $a\sqrt{3} - b\sqrt{2}$. From this, we can identify:

$a = \frac{1}{9}, \quad b = \frac{5}{16}$

Step 4: Round to the nearest hundredth

Finally, rounding $b$ to the nearest hundredth:

$b \approx 0.31$

Thus, the value of $b$ is approximately 0.31.

Would you like more details on any of the steps? Let me know if you'd like further clarification!

Here are five related questions:

1. How do you rationalize a denominator with a square root?
2. What are other methods to simplify square roots in fractions?
3. What is the significance of expressing square roots in terms of simpler numbers?
4. How can you check if a simplified expression is correct?
5. Can you give another example of a similar problem?

Tip: Always simplify square roots when you encounter them in fractions to make the calculations easier!