the radioactive substance cesium - 137 has a half - life of 30 years. the amount a(t) (in grams) of a sample…

the radioactive substance cesium - 137 has a half - life of 30 years. the amount a(t) (in grams) of a sample of cesium - 137 remaining after t years is given by the following exponential function. a(t)=458(1/2)^(t/30) find the amount of the sample remaining after 40 years and after 50 years. round your answers to the nearest gram as necessary. amount after 40 years: grams amount after 50 years: grams

the radioactive substance cesium - 137 has a half - life of 30 years. the amount a(t) (in grams) of a sample of cesium - 137 remaining after t years is given by the following exponential function. a(t)=458(1/2)^(t/30) find the amount of the sample remaining after 40 years and after 50 years. round your answers to the nearest gram as necessary. amount after 40 years: grams amount after 50 years: grams

Answer

Explanation:

Step1: Calculate amount after 40 years

Substitute $t = 40$ into $A(t)=458\left(\frac{1}{2}\right)^{\frac{t}{30}}$. So $A(40)=458\left(\frac{1}{2}\right)^{\frac{40}{30}}=458\left(\frac{1}{2}\right)^{\frac{4}{3}}$. First, calculate $\left(\frac{1}{2}\right)^{\frac{4}{3}}=\sqrt[3]{\left(\frac{1}{2}\right)^4}=\sqrt[3]{\frac{1}{16}}\approx0.397$. Then $A(40)=458\times0.397\approx182$.

Step2: Calculate amount after 50 years

Substitute $t = 50$ into $A(t)=458\left(\frac{1}{2}\right)^{\frac{t}{30}}$. So $A(50)=458\left(\frac{1}{2}\right)^{\frac{50}{30}}=458\left(\frac{1}{2}\right)^{\frac{5}{3}}$. First, calculate $\left(\frac{1}{2}\right)^{\frac{5}{3}}=\sqrt[3]{\left(\frac{1}{2}\right)^5}=\sqrt[3]{\frac{1}{32}}\approx0.315$. Then $A(50)=458\times0.315\approx144$.

Answer:

Amount after 40 years: 182 grams Amount after 50 years: 144 grams