Final answer:
A sound of 103 dB is approximately 20 times more intense than a sound of 90 dB, as every 10 dB increase indicates a tenfold increase in intensity and an additional 3 dB indicates a doubling. The correct answer is not provided in the options, but the closest accurate assessment is that it's approximately 20 times more intense. Among the options provided, the closest to 20 times is A) 100 times.
Step-by-step explanation:
The intensity of a sound is measured in decibels (dB), and the question asks how much more intense a sound of 103 dB is compared to a sound of 90 dB. In decibels, every increase of 10 dB corresponds to a sound that is ten times more intense. Therefore, a difference of 13 dB (103 dB - 90 dB) represents slightly more than one factor of 10, followed by another factor of 2 (since doubling the intensity corresponds to approximately a 3 dB increase).
Starting with a sound that is ten times more intense (or 10 dB higher), going from 90 to 100 dB, and then doubling that intensity (which adds 3 dB), we arrive at a sound that is 20 times more intense (90 to 100 dB = ten times, 100 to 103 dB = twice). Hence, among the options provided, the closest to 20 times is A) 100 times, but this is not accurate given the actual dB difference. Each option is a significant departure from the correct value, but the closest accurate assessment is that a 103 dB sound is approximately 20 times more intense than a 90 dB sound.