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(C) The sound intensity level $L$ in decibels is given by $L = 10 \log_{10} \left( \frac{I}{I_0} \right)$,where $I_0 = 10^{-12} \ W/m^2$.For the music

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$60$

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(C) The sound intensity level $L$ in decibels is given by $L = 10 \log_{10} \left( \frac{I}{I_0} ight)$,where $I_0 = 10^{-12} \ W/m^2$.For the music,$L_m = 60 \ dB$,so $I_m = 10^{-12} 	imes 10^{60/10} = 10^{-6} \ W/m^2$.For the spectator's yell,$L_s = 40 \ dB$,so $I_s = 10^{-12} 	imes 10^{40/10} = 10^{-8} \ W/m^2$.The total intensity is $I_{total} = I_m + I_s = 10^{-6} + 10^{-8} = 10^{-6} (1 + 0.01) = 1.01 	imes 10^{-6} \ W/m^2$.The combined sound level is $L_{total} = 10 \log_{10} \left( \frac{1.01 	imes 10^{-6}}{10^{-12}} ight) = 10 \log_{10} (1.01 	imes 10^6)$.$L_{total} = 10 [\log_{10}(1.01) + \log_{10}(10^6)] = 10 [0.0043 + 6] \approx 60.04 \ dB$.Thus,the approximate combined sound level is $60 \ dB$.

$A$ group of people watch an ice show held in a closed area. Skaters perform a music show whose sound level is $60 \ dB$. $A$ spectator can yell at $40.0 \ dB$. Which one of them represents the approximate combined sound level (in $dB$)?

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