Question

a radioactive material decays exponentially at a rate of 1.68% per year. the amount of the radioactive material, y, left after t years can be found by the formula ( y = y_0e^{-0.0168t} ), where ( y_0 ) is the original amount of radioactive material present. if there are originally 8 grams of the radioactive material, find the amount of the radioactive material after 50 years. the amount of the radioactive material left after 50 years is grams (round to the nearest thousandth)

Explanation:

Step1: Substitute the values into the formula

Given \(y_0 = 8\), \(t = 50\), and the formula \(y=y_0e^{- 0.0168t}\). Substitute these values: \(y = 8e^{-0.0168\times50}\).
First, calculate the exponent: \(-0.0168\times50=- 0.84\). So the formula becomes \(y = 8e^{-0.84}\).

Step2: Calculate the value of \(e^{-0.84}\)

We know that \(e^{-x}=\frac{1}{e^{x}}\). Using a calculator, \(e^{0.84}\approx2.316367\), so \(e^{-0.84}=\frac{1}{e^{0.84}}\approx\frac{1}{2.316367}\approx0.4317\).

Step3: Calculate the value of \(y\)

Now, \(y = 8\times0.4317\). \(y=3.4536\approx3.454\)

Answer:

\(3.454\)