1. **State the problem:** We are given a table of values for $t$ (days) and $D(t)$, and we want to find which function among the given options models the data correctly. 2. **Given data:** $$ \begin{array}{c|ccccc} t & 0 & 1 & 2 & 3 & 4 \\ \hline D(t) & 40.5 & 27 & 18 & 12 & 8 \end{array} $$ 3. **Check each function option by evaluating at $t=0$ and $t=1$ to see if it matches the data:** - Option 1: $y = 40.5\left(\frac{1}{3}\right)^t$ - At $t=0$: $40.5 \times 1 = 40.5$ (matches) - At $t=1$: $40.5 \times \frac{1}{3} = 13.5$ (does not match 27) - Option 2: $y = 40.5 - 27t$ - At $t=0$: $40.5 - 0 = 40.5$ (matches) - At $t=1$: $40.5 - 27 = 13.5$ (does not match 27) - Option 3: $y = 40.5 - \left(\frac{2}{3}\right)^t$ - At $t=0$: $40.5 - 1 = 39.5$ (does not match 40.5) - Option 4: $y = 40.5\left(\frac{2}{3}\right)^t$ - At $t=0$: $40.5 \times 1 = 40.5$ (matches) - At $t=1$: $40.5 \times \frac{2}{3} = 27$ (matches) - At $t=2$: $40.5 \times \left(\frac{2}{3}\right)^2 = 40.5 \times \frac{4}{9} = 18$ (matches) - At $t=3$: $40.5 \times \left(\frac{2}{3}\right)^3 = 40.5 \times \frac{8}{27} = 12$ (matches) - At $t=4$: $40.5 \times \left(\frac{2}{3}\right)^4 = 40.5 \times \frac{16}{81} = 8$ (matches) 4. **Conclusion:** The function that fits the data is $$ y = 40.5\left(\frac{2}{3}\right)^t $$ This is an exponential decay function where the quantity decreases by a factor of $\frac{2}{3}$ each day. Final answer: $y = 40.5\left(\frac{2}{3}\right)^t$