Answered

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The half-life of krypton-91 (91Kr) is 10 s. At time t = 0 a heavy canister contains 3 g of this radioactive gas. Find a function m(t) = m02−t/h that models the amount of 91Kr remaining in the canister after t seconds.

Sagot :

The answer is m(t) = 3.2 -t/10
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