SOLUTION: Lead shielding is used to contain radiation. The percentage of a certain radiation that can penetrate x millimeters of lead shielding is given by I(x) = 100e^−1.5x. (a) Wh

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Question 1081003: Lead shielding is used to contain radiation. The percentage of a certain radiation that can penetrate x millimeters of lead shielding is given by
I(x) = 100e^−1.5x.
(a) What percentage of radiation, to the nearest tenth of a percent, will penetrate a lead shield that is 1 millimeter thick?
%
(b) How many millimeters of lead shielding are required so that less than 0.02% of the radiation penetrates the shielding? Round to the nearest millimeter.
mm
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
I(x)=100e^(-1.5)=22.3%
0.02=100e^(-1.5x)
ln both sides
-3.9120=ln 100-1.5x=4.605-1.5x
-8.517=-1.5x
x=5.68 or 6 mm