Question

Lead shielding is used to contain radiation. The percentage of a certain radiation that can penetrate $x$ millimeters of lead shielding is given by $I(x)=100 e^{-1.5 x}$ 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 $0.05 \%$ of the radiation penetrates the shielding? Round to the nearest millimeter.

          Lead shielding is used to contain radiation. The percentage of a certain radiation that can penetrate $x$ millimeters of lead shielding is given by $I(x)=100 e^{-1.5 x}$
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 $0.05 \%$ of the radiation penetrates the shielding? Round to the nearest millimeter.

Added by Joseph C.

Grade 6 Mathematics: Open Up Resources, Common Core State Standards Edition

N/A 6th Edition

Instant Answer

Solved by Expert Cheryl Glor

07/14/2023

Step 1

$I(1) = 100 e^{-1.5(1)} \approx 33.2$ So, approximately 33.2% of the radiation will penetrate a lead shield that is 1 millimeter thick. b. Show more…

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Lead shielding is used to contain radiation. The percentage of a certain radiation that can penetrate $x$ millimeters of lead shielding is given by $I(x)=100 e^{-1.5 x}$ 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 $0.05 \%$ of the radiation penetrates the shielding? Round to the nearest millimeter.