How do I enter this into the calculator for either problem to arrive at the same answer? **Transcription for Educational Website****Radioactive Decay and Half-Life Calculations**(d) What is the half-life of strontium 90? (Round to the nearest tenth)**Note:** The *half-life* is the time it takes an initial amount of a radioactive substance to decay to one half of it.\[\frac{1}{2}A_0 = A_0 e^{-0.0244t}\]\[\frac{1}{2} = e^{-0.0244t}\]\[\ln\left(\frac{1}{2}\right) = \ln\left(e^{-0.0244t}\right)\]\[t = \frac{\ln\left(\frac{1}{2}\right)}{-0.0244} \approx 28.4 \text{ years}\]---9. The half-life of radium is 1690 years. If 30 grams are present now, how much will be present in 280 years? (Do not round until the final answer. Then round to the nearest tenth.)\[\frac{1}{2}A_0 = A_0 e^{1690k}\]\[\frac{1}{2} = e^{1690k}\]\[\ln\left(\frac{1}{2}\right) = 1690k\]\[k = \frac{\ln\left(\frac{1}{2}\right)}{1690} \approx -0.000410146\]\[A = 30 \cdot e^{\left(\frac{\ln\left(\frac{1}{2}\right)}{1690} \cdot 280\right)} \approx 26.7 \text{ grams}\]---**Answers:**1. $451.002. 2.13 years3. 3.94 years; 3.85 years4. 9.576%5. Bank B6. $172.20

Newton's law of Decay: The Half life of radium Problems:Given personality constant k=1690 Now A=A0ekt Half life radium is 1690 12A0=A0e1690k12=e1690k1690k =ln(12)k=ln(12)1690IF 30 grms radium present now, how much radium present in after 290 years A=A0 ektsince AA0=30, t=280 and k=ln0.51690A=30eln0.51690280≈30.087849A=30.09 grams