In 2005, Consumers recived, on average, 3253 spam messages.The volume of spam me

Question

In 2005, Consumers recived, on average, 3253 spam messages.The volume of spam messages per con-sumer is decreasingexponentially with an exponential decay rate of 13.7% peryear. A. Find the exponential decay function that can be used topredict the average number of spam messages, m(t),tyearsafter2005. B. predict the number of spam messages received per consumerin 2010. C. inwhat year, theoretically, will the average consumerreceive 100spam messages. In 2005, Consumers recived, on average, 3253 spam messages.The volume of spam messages per con-sumer is decreasingexponentially with an exponential decay rate of 13.7% peryear. A. Find the exponential decay function that can be used topredict the average number of spam messages, m(t),tyearsafter2005. B. predict the number of spam messages received per consumerin 2010. C. inwhat year, theoretically, will the average consumerreceive 100spam messages.

Explanation / Answer

exponential growth modeled by N(t) =N0*ekt were t = timeperiods k = growth rate per period t = 2010 - 1005 = 5 k = -.137 N(2010) = 3253*e(-.137)(5) = the number of spam messagesin 2010 N(t) = 100 = 3253*e(-.137)(t) divide both sides by 3253 100/3253 = e(-.137)(t) take ln of both sides t = (ln(100/3253))/(-.137) = the time that will elapse 2005 + the time that elapses in years = the year the spam decreasesto 100 per consumer

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