Using an exponential function, the expected mass of the sample in the year 2001 would be of 16 mg.
The function is:
[tex]A(t) = A(0)e^{-kt}[/tex].
In which:
The information given is as follows:
A(0) = 440, A(8) = 40.
Hence:
[tex]A(t) = A(0)e^{-kt}[/tex].
[tex]40 = 440e^{-8k}[/tex].
[tex]e^{-8k} = 0.09090909[/tex]
[tex]\ln{e^{-8k}} = \ln{0.09090909}[/tex]
[tex]-8k = \ln{0.09090909}[/tex]
[tex]k = -\frac{\ln{0.09090909}{8}[/tex]
k = 0.29973691
Then the function is:
[tex]A(t) = 440e^{-0.29973691t}[/tex]
2001 is 11 years after 1990, hence the amount is:
[tex]A(11) = 440e^{-0.29973691 \times 11} = 16[/tex]
More can be learned about exponential functions at https://brainly.com/question/25537936
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