SOLUTION: Lead shielding is used to contain radiation. The percent of a certain radiation that can penetrate x millimeters of lead shielding is given by r(x) = 100e^-1.5x. What percent of ra

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Question 92001: Lead shielding is used to contain radiation. The percent of a certain radiation that can penetrate x millimeters of lead shielding is given by r(x) = 100e^-1.5x. What percent of radiation, to the nearest tenth of a percent, will penetrate a lead shield that is 3 millimeters thick?
Found 2 solutions by mathispowerful, stanbon:
Answer by mathispowerful(115) About Me  (Show Source):
You can put this solution on YOUR website!
all you need to do is replace x by 3
so r%283%29+=+100e%5E%28-1.5%2A3%29
=100e%5E-4.5
= 100%2A0.0111
= 1.1
so the answer is 1.1%

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
The percent of a certain radiation that can penetrate x millimeters of lead shielding is given by r(x) = 100e^-1.5x. What percent of radiation, to the nearest tenth of a percent, will penetrate a lead shield that is 3 millimeters thick?
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Let x = 3 and solve for r:
r(3) = 100e^(-1.5*3)
r(3) = 100e^(-4.5)
r(3) = 100*0.011089965
r(3) = 11.09 %
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Cheers,
Stan H.