Your digital signal is originally sampled at $380KHz$ and you want to downsample it to sampling rate $44KHz$. Therefore, you will require fractional sampling rate change. You cannot just downsample to $44KHz$ because $\frac{380}{44}$ is not an integer.
First upsample by a factor of $11$ and then downsample by a factor of $95$ to reach your goal. Since you might be knowing upsampling by $L$ is followed by LowPass Filter with cut-off freq of $\omega = \frac{\pi}{L}$ and for downsampling by a factor $M$ we need to first LowPass Filter of cut-off $\omega = \frac{\pi}{M}$. These two Low Pass Filtering operation can be clubbed into one. And you can low-Pass filter with cut-off frequency $Min \{\frac{\pi}{L}, \frac{\pi}{M} \}$.
So, to reach your goal of fractional downsampling by a factor of $\frac{95}{11}$, upsample by a factor of $11$, LowPass Filter with Cut-off frequency $\omega = \frac{\pi}{95}$ and then downsample by a factor of $95$.
Interpolation sampling frequency will be the new $44KHz$. You will retain all frequency component of the original signal till $22KHz$ assuming ideal LPF. With finite order Butterworth LPF, there will be some non-zero transition band which will give a roll-off in the transition band of LPF.