1. The problem asks us to determine the value of $a$ in the exponential decay model $$A(t) = A_0 a^t$$ where $A_0$ is the initial amount of radium-223 and $t$ is time in days. 2. This formula models exponential decay or growth, where $a$ is the base that determines how the amount changes each day. 3. Since radium-223 is radioactive and decays over time, $a$ must be a number between 0 and 1. 4. To find $a$, we need additional information such as the half-life or the amount remaining after a certain number of days. 5. The half-life of radium-223 is approximately 11.4 days, meaning after 11.4 days, half of the original amount remains. 6. Using the half-life, we set $$A(11.4) = \frac{1}{2} A_0$$ and substitute into the formula: $$\frac{1}{2} A_0 = A_0 a^{11.4}$$ 7. Dividing both sides by $A_0$ gives: $$\frac{1}{2} = a^{11.4}$$ 8. Taking the 11.4th root of both sides to solve for $a$: $$a = \left(\frac{1}{2}\right)^{\frac{1}{11.4}}$$ 9. Calculating this value: $$a \approx 0.940618$$ 10. Therefore, the value of $a$ rounded to six decimal places is $0.940618$. This means each day, about 94.0618% of the radium-223 remains compared to the previous day.