The amount of alcohol in a person's blood is known as blood alcohol concentration (BAC), measured in grams of alcohol per deciliter of blood. A BAC of 0.08 , meaning \( 0.08 \% \), indicates that a person has 8 parts alcohol per 10,000 parts blood. It is illegal to drive with a BAC of 0.08 or higher. Use the formula . \( \frac{600 \mathrm{n}}{\mathrm{w}(0.6 \mathrm{n}+169)} \) where w is the weight of a 196 -pound person and n is the number of drinks per hour, and a calculator, to compute the BAC for integers from \( n=1 \) to \( \mathrm{n}=10 \). How many drinks can the person consume in an hour without exceeding the legal measure of drunk driving? According to this model, the person can consume drinks in an hour without exceeding the legal measure of drunk driving.

To calculate the blood alcohol concentration (BAC) for a 196-pound person using the formula \[ \text{BAC} = \frac{600n}{w(0.6n + 169)} \] where \( w \) is the weight of the person in pounds and \( n \) is the number of drinks consumed in an hour, we can substitute \( w = 196 \) and compute the BAC for \( n \) from 1 to 10. Let's calculate the BAC for each integer value of \( n \): 1. **For \( n = 1 \)**: \[ \text{BAC} = \frac{600 \times 1}{196(0.6 \times 1 + 169)} = \frac{600}{196(0.6 + 169)} = \frac{600}{196 \times 169.6} \approx 0.017 \] 2. **For \( n = 2 \)**: \[ \text{BAC} = \frac{600 \times 2}{196(0.6 \times 2 + 169)} = \frac{1200}{196(1.2 + 169)} = \frac{1200}{196 \times 170.2} \approx 0.035 \] 3. **For \( n = 3 \)**: \[ \text{BAC} = \frac{600 \times 3}{196(0.6 \times 3 + 169)} = \frac{1800}{196(1.8 + 169)} = \frac{1800}{196 \times 170.8} \approx 0.052 \] 4. **For \( n = 4 \)**: \[ \text{BAC} = \frac{600 \times 4}{196(0.6 \times 4 + 169)} = \frac{2400}{196(2.4 + 169)} = \frac{2400}{196 \times 171.4} \approx 0.070 \] 5. **For \( n = 5 \)**: \[ \text{BAC} = \frac{600 \times 5}{196(0.6 \times 5 + 169)} = \frac{3000}{196(3.0 + 169)} = \frac{3000}{196 \times 172.0} \approx 0.087 \] 6. **For \( n = 6 \)**: \[ \text{BAC} = \frac{600 \times 6}{196(0.6 \times 6 + 169)} = \frac{3600}{196(3.6 + 169)} = \frac{3600}{196 \times 172.6} \approx 0.105 \] 7. **For \( n = 7 \)**: \[ \text{BAC} = \frac{600 \times 7}{196(0.6 \times 7 + 169)} = \frac{4200}{196(4.2 + 169)} = \frac{4200}{196 \times 173.2} \approx 0.122 \] 8. **For \( n = 8 \)**: \[ \text{BAC} = \frac{600 \times 8}{196(0.6 \times 8 + 169)} = \frac{4800}{196(4.8 + 169)} = \frac{4800}{196 \times 173.8} \approx 0.140 \] 9. **For \( n = 9 \)**: \[ \text{BAC} = \frac{600 \times 9}{196(0.6 \times 9 + 169)} = \frac{5400}{196(5.4 + 169)} = \frac{5400}{196 \times 174.4} \approx 0.157 \] 10. **For \( n = 10 \)**: \[ \text{BAC} = \frac{600 \times 10}{196(0.6 \times 10 + 169)} = \frac{6000}{196(6.0 + 169)} = \frac{6000}{196 \times 175.0} \approx 0.175 \] Now, we can summarize the results: - \( n = 1 \): BAC ≈ 0.017 - \( n = 2 \): BAC ≈ 0.035 - \( n = 3 \): BAC ≈ 0.052 - \( n = 4 \): BAC ≈ 0.070 - \( n = 5 \): BAC ≈ 0.087 - \( n = 6 \): BAC ≈ 0.105 - \( n = 7 \): BAC ≈ 0.122 - \( n = 8 \): BAC ≈ 0.140 - \( n = 9 \): BAC ≈ 0.157 - \( n = 10 \): BAC ≈ 0.175 The legal limit for BAC is 0.08. Therefore, the maximum number of drinks a 196-pound person can consume in an hour without exceeding the legal limit is **5 drinks**.