Question 1100283
First, find t(15)
t(15)=0+(160-0)e^-15k
47=160e^-15k
0.29375=e^-15k
ln 0.29375=ln e^-15k=-15k ln e=-15k
k=0.081668
So, for t(20), we have:
t(20)=0+(160-0)e^-(20*.081668)
t(20)=31.2433 deg. F as the temperature after 20 mins.
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