the function f(x)=60e^(-0.5x)+40 describes the percentage of information, f(x), that a particular person…

the function f(x)=60e^(-0.5x)+40 describes the percentage of information, f(x), that a particular person remembers x weeks after learning the information. a. substitute 0 for x and, without using a calculator, find the percentage of information remembered at the moment it is first learned. b. substitute 1 for x and find the percentage of information that is remembered after 1 week. c. find the percentage of information that is remembered after 12 weeks. d. find the percentage of information that is remembered after one year (52 weeks). a. at the moment it is first learned, 100 % of the information is remembered. (round to one decimal place as needed.) b. after one week, 76.4 % of the information is remembered. (round to one decimal place as needed.) c. after twelve weeks, 40.1 % of the information is remembered. (round to one decimal place as needed.) d. after one year, % of the information is remembered. (round to one decimal place as needed.)

the function f(x)=60e^(-0.5x)+40 describes the percentage of information, f(x), that a particular person remembers x weeks after learning the information. a. substitute 0 for x and, without using a calculator, find the percentage of information remembered at the moment it is first learned. b. substitute 1 for x and find the percentage of information that is remembered after 1 week. c. find the percentage of information that is remembered after 12 weeks. d. find the percentage of information that is remembered after one year (52 weeks). a. at the moment it is first learned, 100 % of the information is remembered. (round to one decimal place as needed.) b. after one week, 76.4 % of the information is remembered. (round to one decimal place as needed.) c. after twelve weeks, 40.1 % of the information is remembered. (round to one decimal place as needed.) d. after one year, % of the information is remembered. (round to one decimal place as needed.)

Answer

Explanation:

Step1: Recall the function

The function is $f(x)=60e^{- 0.5x}+40$.

Step2: Substitute $x = 52$

We need to find $f(52)$, so we substitute $x = 52$ into the function: $f(52)=60e^{-0.5\times52}+40$. First, calculate the exponent: $-0.5\times52=-26$. Then we have $f(52)=60e^{-26}+40$. Since $e^{-26}=\frac{1}{e^{26}}$, and $e\approx2.718$, $e^{26}$ is a very large number. So $60e^{-26}\approx0$. Then $f(52)\approx0 + 40=40.0$.

Answer:

$40.0$