Which of the following functions are positive, continuous, and decreasing on an interval $[\mathrm{N}, \infty)$ for some $\mathrm{N}> 0$ ?
$f(x)=\tan ^{-1}(-x)$
$f(x)=\frac{(1-x)}{x}$
$f(x)=\frac{1}{x^{4}}$
$f(x)=\frac{1}{2}$
$f(x)=\sin (x)$
$f(x)=e^{-x}$

Step 1 ：Check each function one by one to see if it is positive, continuous, and decreasing on an interval $[N, \infty)$ for some $N>0$.

Step 2 ：Consider the function $f(x)=\tan^{-1}(-x)$. The function $\tan^{-1}(x)$ is continuous for all real numbers. However, it is not decreasing for $x>0$. Therefore, $f(x)=\tan^{-1}(-x)$ is not decreasing on an interval $[N, \infty)$ for some $N>0$.

Step 3 ：Consider the function $f(x)=\frac{(1-x)}{x}$. This function is continuous for all $x \neq 0$. However, it is not decreasing for $x>1$. Therefore, $f(x)=\frac{(1-x)}{x}$ is not decreasing on an interval $[N, \infty)$ for some $N>0$.

Step 4 ：Consider the function $f(x)=\frac{1}{x^{4}}$. This function is continuous for all $x \neq 0$. It is also decreasing for $x>0$. Therefore, $f(x)=\frac{1}{x^{4}}$ is decreasing on an interval $[N, \infty)$ for some $N>0$.

Step 5 ：Consider the function $f(x)=\frac{1}{2}$. This function is continuous for all real numbers. However, it is not decreasing for any interval. Therefore, $f(x)=\frac{1}{2}$ is not decreasing on an interval $[N, \infty)$ for some $N>0$.

Step 6 ：Consider the function $f(x)=\sin(x)$. This function is continuous for all real numbers. However, it is not decreasing for any interval. Therefore, $f(x)=\sin(x)$ is not decreasing on an interval $[N, \infty)$ for some $N>0$.

Step 7 ：Consider the function $f(x)=e^{-x}$. This function is continuous for all real numbers. It is also decreasing for all $x>0$. Therefore, $f(x)=e^{-x}$ is decreasing on an interval $[N, \infty)$ for some $N>0$.

Step 8 ：The functions that are positive, continuous, and decreasing on an interval $[N, \infty)$ for some $N>0$ are $f(x)=\frac{1}{x^{4}}$ and $f(x)=e^{-x}$. Therefore, the final answer is \(\boxed{f(x)=\frac{1}{x^{4}}, f(x)=e^{-x}}\).