Answer: The sound intensity was reduced by -14.77 dB.
Explanation:
The level of the intensity of the sound with respect to the reference value is called Sound Intensity level.
[tex]L_I=10\log\frac{I}{I_o} dB[/tex]
I = Intensity of the sound
[tex]I_o[/tex] = Reference sound intensity
According to question, sound-reflecting windows that reduce the sound intensity level by 30.0 db .
Let the intensity of the sound be [tex]I_o[/tex] and reduced intensity be I.
[tex]I=\frac{I_o}{30} dB[/tex]
[tex]L_I=10\log\frac{I}{I_o} dB[/tex]
[tex]L_I=10\log\frac{\frac{I_o}{30}}{I_o}=10\log\frac{1}{30} dB=-14.77 dB[/tex]
The sound intensity was reduced by -14.77 dB.
