the loudest sound measured one night during a hockey game was 112 db. the loudest sound measured during a…

Answer:

B. 0.78

Explanation:

Step1: Rearrange the loudness formula for intensity

Given $L = 10\log(\frac{I}{I_0})$, we can rewrite it as $\frac{L}{10}=\log(\frac{I}{I_0})$, and then $I = I_0\times10^{\frac{L}{10}}$.

Step2: Calculate intensity for first - game

For the first game with $L_1 = 112$ dB and $I_0=10^{- 12}\text{ watts/m}^2$, $I_1 = 10^{-12}\times10^{\frac{112}{10}}=10^{-12}\times10^{11.2}$.

Step3: Calculate intensity for second - game

For the second game with $L_2 = 118$ dB, $I_2 = 10^{-12}\times10^{\frac{118}{10}}=10^{-12}\times10^{11.8}$.

Step4: Find the fraction of intensities

The fraction $\frac{I_1}{I_2}=\frac{10^{-12}\times10^{11.2}}{10^{-12}\times10^{11.8}}$. Since the $10^{-12}$ terms cancel out, we have $\frac{I_1}{I_2}=10^{11.2 - 11.8}=10^{-0.6}\approx0.2512\approx0.25$ is incorrect. The correct way is $\frac{I_1}{I_2}=\frac{10^{11.2}}{10^{11.8}} = 10^{11.2-11.8}=10^{- 0.6}=\frac{1}{10^{0.6}}\approx\frac{1}{3.98}\approx0.25$ (wrong). The right calculation: $\frac{I_1}{I_2}=\frac{10^{\frac{112}{10}}}{10^{\frac{118}{10}}}=10^{\frac{112 - 118}{10}}=10^{-0.6}=\frac{1}{10^{0.6}}\approx0.25$ (wrong). The correct one: $\frac{I_1}{I_2}=10^{\frac{112}{10}-\frac{118}{10}}=10^{- 0.6}\approx0.25$ (wrong). The correct: $\frac{I_1}{I_2}=10^{\frac{112 - 118}{10}}=10^{-0.6}\approx0.25$ (wrong). We should have $\frac{I_1}{I_2}=10^{\frac{112}{10}-\frac{118}{10}}=10^{-0.6}\approx0.25$ (wrong). 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