1. **State the problem:** We have the probability function $$P(t) = e^{2.1.394t}$$ representing the percent chance of an automobile accident based on blood alcohol level $$t$$. (a) We need to graph $$P(t)$$ for $$0 \leq t \leq 0.2$$ and $$0 \leq P(t) \leq 100$$. (b) We want to find the blood alcohol level $$t$$ where the probability $$P(t)$$ is at least 30%. 2. **Formula and explanation:** The function is exponential: $$P(t) = e^{2.1.394t}$$. - Exponential functions grow rapidly. - To find $$t$$ for a given $$P(t)$$, use the natural logarithm: $$t = \frac{\ln(P(t))}{2.1.394}$$. 3. **Part (a) Graph:** - At $$t=0$$, $$P(0) = e^0 = 1$$. - At $$t=0.2$$, $$P(0.2) = e^{2.1.394 \times 0.2} = e^{0.2788} \approx 1.321$$. Since the problem states the curve rises sharply toward 100 by $$t=0.2$$, the exponent likely is $$2 \times 1.394 t = 2.788 t$$ (assuming a typo in the problem). Using $$P(t) = e^{2.788 t}$$: - At $$t=0.2$$, $$P(0.2) = e^{2.788 \times 0.2} = e^{0.5576} \approx 1.746$$, still less than 100. Given the problem's description, the graph labeled C matches the exponential growth shape. 4. **Part (b) Find $$t$$ for $$P(t) \geq 30$$:** Set $$P(t) = 30$$: $$30 = e^{2.1.394 t}$$ Take natural log on both sides: $$\ln(30) = 2.1.394 t$$ Solve for $$t$$: $$t = \frac{\ln(30)}{2.1.394}$$ Calculate: $$\ln(30) \approx 3.401$$ $$t = \frac{3.401}{2.1.394} \approx 1.622$$ Since $$t$$ must be between 0 and 0.2, this suggests the exponent might be miswritten. Assuming the exponent is $$2 \times 1.394 t = 2.788 t$$: $$t = \frac{\ln(30)}{2.788} = \frac{3.401}{2.788} \approx 1.22$$ still outside the range. If the problem means $$P(t) = 100 e^{2.1.394 t}$$ or similar, the exact value depends on the correct formula. **Assuming the problem's original formula is $$P(t) = e^{2.1.394 t}$$ and the probability is in percent, the blood alcohol level for at least 30% probability is approximately:** $$t \approx 1.622$$ (rounded to three decimals). **Summary:** - (a) Graph C matches the function. - (b) Blood alcohol level for $$P(t) \geq 30\%$$ is approximately $$t = 1.622$$. Note: The problem's exponent seems unusual; verify the exact formula for precise answers.