Question

A coach speaks normally at 64 dB but shouts at 92 dB. How many times louder is her shouting than her speaking? Mixed Review: 9.) What is the difference between evaluating and solving with logs? 10.) When developing a model given a set of data for an exponential function, what steps are taken to develop the modele 11.) When is Pert used? Develop a problem that can uses Perʳᵗ. 12.) What does the expression "half-life" mean?

Answer 1

The coach's shouting is 630 times louder than her speaking.

Step-by-step explanation:

Each factor of 10 in intensity corresponds to 10 dB. For example, a sound that is 90 dB compared to a sound that is 60 dB is 30 dB greater, or three factors of 10 (that is, 10³ times) as intense. Applying this knowledge to the coach's speaking and shouting, we can calculate the difference in intensity using the following formula:

Difference in intensity = 10 raised to the power of (shout level - speak level) / 10

Therefore, the coach's shouting is 10^(92 - 64)/10 = 10^28/10 = 10^2.8 = 630 times louder than her speaking.