A singer is performing to a crowd. How many times more intense is her singing at 70 dB, compared to a bird whose call is 34 dB? Show ALL work and its conclusion sentence. Round to one decimal place, if necessary.
To determine how many times more intense the singer's singing is compared to the bird's call, we need to calculate the difference in sound intensity levels and then convert it to a ratio.
The formula to convert dB to intensity ratio is:
Intensity ratio = 10^((dB2 - dB1) / 10)
Where dB1 and dB2 are the sound intensity levels of the bird's call and the singer's performance, respectively.
Given that the bird's call is 34 dB and the singer's performance is 70 dB, we can plug these values into the formula:
Intensity ratio = 10^((70 - 34) / 10)
= 10^(36 / 10)
= 10^(3.6)
= 3981.071705534973
Therefore, the singer's singing is approximately 3981 times more intense compared to the bird's call.
Conclusion: The singer's singing is approximately 3981 times more intense compared to the bird's call at 70 dB.