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Cesium 137 (Cs137)is a Short-Lived Radioactive Isotope e(ln230)te^{-\left(\frac{\ln 2}{30}\right) t} )As a Result of Its Operations, a Nuclear Power Plant

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Cesium 137 (Cs137) is a short-lived radioactive isotope.It decays at a rate proportional to the amount of itself present and has a half-life of 30 years (i.e., the amount of Cs137 remaining t years after A0 mg of the radioactive isotope is released is given by  Cesium 137 (Cs<sup>137</sup>) is a short-lived radioactive isotope.It decays at a rate proportional to the amount of itself present and has a half-life of 30 years (i.e., the amount of Cs<sup>137</sup> remaining t years after A<sub>0</sub> mg of the radioactive isotope is released is given by   e^{-\left(\frac{\ln 2}{30}\right)  t}  ) .As a result of its operations, a nuclear power plant releases Cs<sup>137</sup> at a rate of 0.12 mg per year.The plant began its operations in 1990, which we will designate as t = 0.Assume there is no other source of this particular isotope.Which of the following integrals give the total amount of Cs<sup>137</sup> T years after the plant began operations? A)   \int_{0}^{T} e^{\left(-\frac{0.12 \ln 2}{30}\right)  t} d t  B)   \int_{0}^{T} 0.12 e^{\left(-\frac{\ln 2}{30}\right)  t} d t  C)   \int_{0}^{0.12} \ t e^{\left(-\frac{\ln 2}{30}\right)  t} d t  D)   \left.\int_{0}^{T}\left(e^{\left(-\frac{\ln 2}{30}\right.}\right)  t+0.12\right)  d t e(ln230) te^{-\left(\frac{\ln 2}{30}\right) t} ) .As a result of its operations, a nuclear power plant releases Cs137 at a rate of 0.12 mg per year.The plant began its operations in 1990, which we will designate as t = 0.Assume there is no other source of this particular isotope.Which of the following integrals give the total amount of Cs137 T years after the plant began operations?


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