Answer:
[tex] db_1 = 10 log_{10} (\frac{10^{-9}}{10^{-12}})= 30 db[/tex]
And using the same formula we can find the number of decible for the Braylee's music:
[tex] db_2 = 10 log_{10} (\frac{10^{-3}}{10^{-12}})= 90 db[/tex]
And since we want to find how many times louder is Braylee’s music than Jessica’s we can find the following relationship:
[tex]\frac{db_2}{db_1} = \frac{90 dB}{30 dB}= 3[/tex]
So then we can conclude that the Braylee's music is 3 times louder then the Jessica's sound. And the best option would be:
3 times louder
Step-by-step explanation:
For this case we know that the loudness measured in decibels is given by this formula:
[tex] dB = 10 log_{10} (\frac{I}{I_o})[/tex]
Where [tex] I_o = 10^{-12} W/m^2[/tex]. Using this formula we can find the decibels for the sound for Jessica:
[tex] db_1 = 10 log_{10} (\frac{10^{-9}}{10^{-12}})= 30 db[/tex]
And using the same formula we can find the number of decible for the Braylee's music:
[tex] db_2 = 10 log_{10} (\frac{10^{-3}}{10^{-12}})= 90 db[/tex]
And since we want to find how many times louder is Braylee’s music than Jessica’s we can find the following relationship:
[tex]\frac{db_2}{db_1} = \frac{90 dB}{30 dB}= 3[/tex]
So then we can conclude that the Braylee's music is 3 times louder then the Jessica's sound. And the best option would be:
3 times louder
Answer:
B: 3times louder is correct シ
Step-by-step explanation:
good luck
