The amount of polonium-210 remaining, p(t), after t days in a sample can be modeled by the exponential function p(t) = 100e−0.005t, where 100 represents the initial number of grams in the sample. what is an equivalent expression, written as a percentage rate of polonium-210 lost, and how much polonium-210 remains (rounded to the nearest whole number) after 63 days?

Answer:% Po lost = 100[1 - e^(-0.005t)]  %; 73.0 g Step-by-step explanation:p(t) = 100e^(-0.005t) Initial amount:        p(0) = 100 Amount remaining: p(t) = 100e^(-0.005t) Amount lost: p(0) – p(t) = 100 - 100e^(-0.005t) = 100[1 - e^(-0.005t)] % of Po lost  = amount lost/initial amount × 100 %= [1 - e^(-0.005t)]  × 100 % = 100[1 - e^(-0.005t)]  % p(63) = 100e^(-0.005 × 63) = 100e^(-0.315) = 100 × 0.730 = 73 g The mass of polonium remaining after 63 days is 73 g.