If $$\int_{-0.15}^{0.15} |100x^2 - 1| \ dx = \frac{k}{3000}$$, then $$k$$ is equal to _____.

We need to evaluate $$\int_{-0.15}^{0.15} |100x^2 - 1| \, dx = \frac{k}{3000}$$ and find $$k$$.

$$100x^2 - 1 = 0 \Rightarrow x = \pm \frac{1}{10} = \pm 0.1$$

For $$|x| < 0.1$$: $$100x^2 - 1 < 0$$, so $$|100x^2 - 1| = 1 - 100x^2$$.

For $$0.1 \leq |x| \leq 0.15$$: $$100x^2 - 1 \geq 0$$, so $$|100x^2 - 1| = 100x^2 - 1$$.

Since the integrand is even, the integral equals:

$$2\left[\int_0^{0.1} (1 - 100x^2) \, dx + \int_{0.1}^{0.15} (100x^2 - 1) \, dx\right]$$

$$\int_0^{0.1} (1 - 100x^2) \, dx = \left[x - \frac{100x^3}{3}\right]_0^{0.1} = 0.1 - \frac{100 \times 0.001}{3} = \frac{1}{10} - \frac{1}{30} = \frac{2}{30} = \frac{1}{15}$$

$$\int_{0.1}^{0.15} (100x^2 - 1) \, dx = \left[\frac{100x^3}{3} - x\right]_{0.1}^{0.15}$$

At $$x = 0.15 = \frac{3}{20}$$: $$\frac{100}{3} \cdot \frac{27}{8000} - \frac{3}{20} = \frac{9}{80} - \frac{3}{20} = \frac{9}{80} - \frac{12}{80} = -\frac{3}{80}$$

At $$x = 0.1 = \frac{1}{10}$$: $$\frac{100}{3} \cdot \frac{1}{1000} - \frac{1}{10} = \frac{1}{30} - \frac{1}{10} = -\frac{2}{30} = -\frac{1}{15}$$

Difference: $$-\frac{3}{80} - \left(-\frac{1}{15}\right) = -\frac{3}{80} + \frac{1}{15} = \frac{-45 + 80}{1200} = \frac{35}{1200} = \frac{7}{240}$$

$$\text{Total} = 2\left(\frac{1}{15} + \frac{7}{240}\right) = 2\left(\frac{16}{240} + \frac{7}{240}\right) = 2 \cdot \frac{23}{240} = \frac{23}{120}$$

$$\frac{23}{120} = \frac{k}{3000} \Rightarrow k = \frac{23 \times 3000}{120} = 23 \times 25 = 575$$

The correct answer is 575.