If the function $\mathrm{f}(x)=x(x+3) \mathrm{e}^{-\frac{x}{2}}$ satisfies all the conditions of Rolle's theorem in $[-3,0]$, then c is

Rolle's Theorem requires finding $c \in (-3,0)$ such that $f'(c) = 0$.

The function $f(x) = x(x+3)e^{-x/2}$ satisfies the theorem's conditions: it is continuous on $[-3,0]$, differentiable on $(-3,0)$, and $f(-3) = f(0) = 0$.

The derivative is computed using the product rule:

$f'(x) = \frac{d}{dx}\left[x(x+3)\right]e^{-x/2} + x(x+3)\frac{d}{dx}\left[e^{-x/2}\right]$

$f'(x) = (2x+3)e^{-x/2} - \frac{1}{2}x(x+3)e^{-x/2}$

Factoring yields:

$f'(x) = e^{-x/2}\left(-\frac{1}{2}x^2 + \frac{1}{2}x + 3\right)$

Setting $f'(c) = 0$ and noting $e^{-c/2} \neq 0$:

$-\frac{1}{2}c^2 + \frac{1}{2}c + 3 = 0$

Multiplying through by $-2$ gives $c^2 - c - 6 = 0$, which factors as $(c-3)(c+2) = 0$.

The solutions are $c = 3$ and $c = -2$. Only $c = -2$ lies in $(-3,0)$.

The value satisfying Rolle's Theorem is $\boxed{-2}$.