1. **State the problem:** We want to find an exponential function $w=f(t)$ that models the number of whooping cough cases $w$ over time $t$ years since 1980, given $w(1)=1248$ in 1981 and $w(24)=25827$ in 2004. 2. **Formula and explanation:** The general form of an exponential function is $$w = w_0 \cdot b^t$$ where $w_0$ is the initial amount at $t=0$ and $b$ is the growth factor per year. 3. **Identify initial value:** Since $t$ is years since 1980, at $t=1$ (1981), $w=1248$. We can rewrite the function as $$w = w_1 \cdot b^{t-1}$$ where $w_1=1248$. 4. **Use data point at $t=24$ (2004):** $$25827 = 1248 \cdot b^{24-1} = 1248 \cdot b^{23}$$ 5. **Solve for $b$: $$b^{23} = \frac{25827}{1248} = 20.7$$ 6. **Take the 23rd root:** $$b = 20.7^{\frac{1}{23}}$$ 7. **Calculate $b$:** Using a calculator, $$b \approx 1.139$$ (rounded to three decimals) 8. **Write the exponential function:** $$w = 1248 \cdot 1.139^{t-1}$$ 9. **Find the average annual percent growth rate:** The growth rate $r$ is related to $b$ by $$b = 1 + r$$ so $$r = b - 1 = 1.139 - 1 = 0.139 = 13.9\%$$ Rounded to nearest percent, the annual growth rate is 14%. 10. **Check if cases more than doubled between 2000 and 2004:** - At $t=20$ (year 2000): $$w(20) = 1248 \cdot 1.139^{19}$$ - At $t=24$ (year 2004): $$w(24) = 1248 \cdot 1.139^{23} = 25827$$ Calculate $w(20)$: $$w(20) = 1248 \cdot 1.139^{19} \approx 1248 \cdot 9.04 = 11283$$ Double $w(20)$ is $$2 \times 11283 = 22566$$ Since $w(24) = 25827 > 22566$, the number of cases more than doubled between 2000 and 2004, confirming the report. **Final answers:** - (a) $$w = 1248 \cdot 1.139^{t-1}$$ - (b) Annual percent growth rate is 14%. - (c) Yes, the model confirms the number of cases more than doubled between 2000 and 2004.