+109 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?
Hint: Find the value of e^−0.005 on calculator.
a.) P(t) = 100e0.005t, 73 grams remain
b.) P(t) = 100(0.995)t, 73 grams remain
c.)P(t) = 100e0.995t, 2 grams remain
d.)P(t) = 100(0.995)−t, 137 grams remain
please help :-(
Hint: Find the value of e^−0.005 on calculator.
a.) P(t) = 100e0.005t, 73 grams remain
b.) P(t) = 100(0.995)t, 73 grams remain
c.)P(t) = 100e0.995t, 2 grams remain
d.)P(t) = 100(0.995)−t, 137 grams remain
Find the value of e^−0.005 on calculator=.995
b.) P(t)=100(.995)^63=100 X 0.7292 =~73 grams remain, which is the right answer.
You could tell your teacher that there is simpler way of doing the same thing, as long as you know the half-life of the radioactive element. In this case the half-life of Polonium-210 is: 138.376 days. Then you simply find out the number of half-lives expired as follows: 63/138.376=0.45528..........
Then the remaining % is: 2^-.45528=~73% of the original element left.
