I do this differently.
The transistor config is a common emitter with negative feedback for bias and gain.
Since the Re=0 , it can only be used for small signals like <<100 mVpp
The output impedance of the electret is a open drain buffered affects gain and Q operating point which is not critical for small signals, so 10k is a common value.
If you recognize that negative feedback BJT's lower both the input and output impedance by the amount of negative feedback or forward gain and that the voltage gain is limited by the current gain of the device, it is quite easy to achieve approx voltage gain of 100 with an hFE of 200 +/-50%
I would choose a feedback resistor ratio of Rbc/Rc= 100 for this configuration (with high hFE transistors at low current.)
To be close to mic output impedance for maximum power transfer, Since the feedback R also controls AC gain lets make Rbc= 100 * 10k
where the gain is Av ~= Rbc/Rc (as long as Av < hFE with sufficient margin.
The cap value determines the high pass filter ( HPF ) breakpoint where f =1/(2pi*RC) is -20 dB per decade below cutoff.
Conclusion
Rbc=1M
Rc=10k
Choose C for 100Hz
Gain = Rbc/Rc=100
BJT Zin= Rbc/Av = Rc for Av=100
BJT Zout = Rc/Av = 10k/100 = 100
Since base current is approx 0.5%, OHm's Law determines Q operating point from resistor ratio and using Vbe = 0.6 (i) for low current. or slightly less for determining since Ic=hFE*Ib (ii) and (Vce-Vbe)/Rbc=Ib ...(iii) and (Vbat-Vce)/Rc=Ic ...(iv) you have 4 equations and 4 unknowns so you can solve for to verify what Vc is and that it is close to Vbat/2 for max undistorted switch. Normally you alllow Vce>2V min for linear operation, but with negative feedback, this increases the range to Vce>0.2V.
Let's see if you can figure this out!