Sure, let's solve this problem step by step.
1. Understand the given data:
- The loudness levels [tex]\( L_1 \)[/tex] and [tex]\( L_2 \)[/tex] for the two hockey games are 112 dB and 118 dB, respectively.
- The reference intensity [tex]\( I_0 = 10^{-12} \)[/tex] watts/m[tex]\(^2\)[/tex].
2. Convert loudness levels back to intensity:
The formula to relate loudness (in decibels) to intensity is:
[tex]\[
L = 10 \log_{10} \left( \frac{I}{I_0} \right)
\][/tex]
Rearranging this formula to solve for [tex]\( I \)[/tex], we get:
[tex]\[
I = I_0 \cdot 10^{\frac{L}{10}}
\][/tex]
Now we will use this formula to find the intensities [tex]\( I_1 \)[/tex] and [tex]\( I_2 \)[/tex].
- For the first game ([tex]\( L_1 = 112 \)[/tex] dB):
[tex]\[
I_1 = I_0 \cdot 10^{\frac{112}{10}} = 10^{-12} \cdot 10^{11.2} = 10^{-12 + 11.2} = 10^{-0.8}
\][/tex]
This simplifies to:
[tex]\[
I_1 \approx 0.1584893192461111 \text{ watts/m}^2
\][/tex]
- For the second game ([tex]\( L_2 = 118 \)[/tex] dB):
[tex]\[
I_2 = I_0 \cdot 10^{\frac{118}{10}} = 10^{-12} \cdot 10^{11.8} = 10^{-12 + 11.8} = 10^{-0.2}
\][/tex]
This simplifies to:
[tex]\[
I_2 \approx 0.6309573444801944 \text{ watts/m}^2
\][/tex]
3. Compute the fraction of the sound intensity of the first game to the second game:
The fraction of the intensities is given by:
[tex]\[
\text{Fraction} = \frac{I_1}{I_2}
\][/tex]
Substituting the values of [tex]\( I_1 \)[/tex] and [tex]\( I_2 \)[/tex]:
[tex]\[
\text{Fraction} = \frac{0.1584893192461111}{0.6309573444801944} \approx 0.2511886431509572
\][/tex]
4. Conclusion:
The fraction of the sound intensity of the first game to the sound intensity of the second game is approximately [tex]\( 0.2512 \)[/tex], or 0.25 when rounded to two decimal places.
Therefore, the correct answer is:
[tex]\[
\boxed{0.25}
\][/tex]
