In 2005, consumers recived 3253 spam messages. Thevalume of spam per consumer is

Question

In 2005, consumers recived 3253 spam messages. Thevalume of spam per consumer is decreasing exponentially with anexponential decay rate of 13.7% per year. A. find the exponential decay function that can be used topredict the average number of spam messages, M(t) t yearsafter 2005 B. predict the number of spam messages recived per consumer in2010. C. in what year, theoretically, will the average consumerreceive 100 spam messages In 2005, consumers recived 3253 spam messages. Thevalume of spam per consumer is decreasing exponentially with anexponential decay rate of 13.7% per year. A. find the exponential decay function that can be used topredict the average number of spam messages, M(t) t yearsafter 2005 B. predict the number of spam messages recived per consumer in2010. C. in what year, theoretically, will the average consumerreceive 100 spam messages A. find the exponential decay function that can be used topredict the average number of spam messages, M(t) t yearsafter 2005 B. predict the number of spam messages recived per consumer in2010. C. in what year, theoretically, will the average consumerreceive 100 spam messages

Explanation / Answer

A) P = Poe-.0137t   B) the year 2010 will be 5 years after 2005 so t=5 P = 3253 e-.0137(5) P = 3253 e-.0685 P = 3253 (.9338) P = 3038 spam in 2010 C) 100 = 3253 e-.0137t   divide bothsides by 3253 .03074 = e-.0137t    to solve for a variable, t, in an exponent of e, take the LN ofboth sides of the equal LN .03074 = LN e-.0137t  do the left side on your calculator; on the right side, LN isbase e so it is asking "what exponent do I put on e to gete-.0137t ?" answer: -.0137t -3.4822 = -.0137t    divide bothsides by -.0137 254.17 = t 254 years after 2005 would be 2259 In the year 2259, the spam isexpected to be 100

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