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The Athletic Department at your school hosts a celebrity football game for which no tickets are sold but cash donations are welcome and expected. Since tickets are not collected, they are looking for a creative manner to get a good estimate of the number of fans attending the event. Before the day of the game, you stand in the middle of the field and record the sound intensity level of a friend midway up in the stands as 43.2 dB. Before the game starts, the announcer asks for a big round of applause and appreciation for participating celebrities and you record the sound intensity level as 97 dB. What is your estimate of the number of people attending the event

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okpalawalter8

Answer:

The number of people attending the events [tex]n = 239883.292 \ \approx 2.4*10 ^5[/tex] people

Explanation:

Generally total sound intensity is given as

            [tex]I_t =I_s + 10log(n)[/tex]

Where

         [tex]I_t[/tex] is the total sound intensity

        [tex]I_s[/tex] is the signal level from each single source

        n is the number of source

        Substituting the given value

                  97 = 43.2 + 10log(n)

                  53.8 = 10log(n)

             log(n)  = 5.38

Raising both sides to the power of 10

               [tex]10^{log(n)} = 10^{5.38}[/tex]        Note [   [tex]10^{log(n)} = n[/tex]]

                => [tex]n = 239883.292 \ \approx 2.4*10 ^5[/tex]