If $4 = 10^{2m}$ and $9 = 10^{2n}$, then $0 \cdot 15$ equals to :

From the first equation:
$4 = (10^m)^2$
Taking the square root of both sides:
$\sqrt{4} = \sqrt{(10^m)^2}$
$2 = 10^m$ (Assuming $10^m$ is positive, which is true for real $m$ as $10^x$ is always positive)

From the second equation:
$9 = (10^n)^2$
Taking the square root of both sides:
$\sqrt{9} = \sqrt{(10^n)^2}$
$3 = 10^n$ (Assuming $10^n$ is positive)

We need to express $0.15$ using $10^m$ and $10^n$.
$0.15 = \frac{15}{100}$
We know $10^n = 3$. Let’s try to express 15 and 100 using 2, 3, and powers of 10.
$0.15 = \frac{3 \times 5}{10^2}$
We have 3 ($10^n$). We need to get 5 and relate it to $10^m$ and $10^n$.
$10 = 10^1$. We know $10^m=2$ and $10^n=3$. $10 = 2 \times 5 = 10^m \times 5$. So, $5 = 10 / 10^m = 10^1 / 10^m = 10^{1-m}$.

Substitute $3 = 10^n$ and $5 = 10^{1-m}$ into the expression for 0.15:
$0.15 = \frac{10^n \times 10^{1-m}}{10^2}$
Using exponent rules ($a^x \times a^y = a^{x+y}$ and $a^x / a^y = a^{x-y}$):
$0.15 = 10^{n + (1-m) – 2}$
$0.15 = 10^{n + 1 – m – 2}$
$0.15 = 10^{n – m – 1}$

Let’s check this against the options. Option C is $10^{n-m-1}$. This matches our result.

Alternatively, express 0.15 in prime factors related to 2 and 3:
$0.15 = \frac{15}{100} = \frac{3 \times 5}{10 \times 10} = \frac{3 \times (10/2)}{10 \times 10} = \frac{3 \times 10 / 2}{10^2} = \frac{3 \times 10}{2 \times 10^2} = \frac{3}{2 \times 10}$
We know $3 = 10^n$ and $2 = 10^m$, and $10 = 10^1$.
$0.15 = \frac{10^n}{10^m \times 10^1} = \frac{10^n}{10^{m+1}} = 10^{n – (m+1)} = 10^{n – m – 1}$.