A scientist has 100 milligrams of a radioactive element. The amount of radioactive element remaining after t days can be determined using the equation f(t)=100(1/2)^t/10. After three days, the scientist receives a second shipment of 100 milligrams of the same element. The equation used to represent the amount of shipment 2 remaining after t days is f(t)=100(1/2)^t-3/10. After any time, t, the mass of the element remaining in shipment 1 is what percentage of the mass of the element remaining in shipment 2?A.78.1%B.81.2%C.123.1%D.128.0%

The mass of the first shipment at time t is
[tex]m_{1}=100( \frac{1}{2} )^{ \frac{t}{10}} [/tex]

The mass of the second shipment at time t is
[tex]m_{2}=100( \frac{1}{2} )^{ \frac{t-3}{10} }[/tex]

At time t, the ratio of m₁ to m₂ is
[tex] \frac{m_{1}}{m_{2}} = \frac{100}{100}. \frac{(1/2)^{t/10}}{(1/2)^{(t-3)/10}} \\ = \frac{(1/2)^{t/10}}{(1/2)^{t/10}}. \frac{1}{(1/2)^{-3/10}} \\ = (1/2)^{3/10} \\ = 0.8123[/tex]

Therefore as a percentage,
[tex] \frac{m_{1}}{m_{2}} =100*0.8123 = 81.23 \%[/tex]

Answer: B. 81.2%