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 \( \frac{600 n}{w(0.6 n+169)} \) where \( w \) is the weight of a 133 -pound person and \( n \) is the number of drinks per hour, and a calculator, to compute the BAC for integers from \( n=1 \) to \( n=10 \). How many drinks can the person consume in an hour without exceeding the legal measure of drunk driving?

To calculate the blood alcohol concentration (BAC) for a 133-pound person using the formula \[ \text{BAC} = \frac{600 n}{w(0.6 n + 169)} \] we first need to convert the weight of the person from pounds to grams, since the formula uses weight in grams. 1 pound is approximately 453.592 grams, so: \[ w = 133 \text{ pounds} \times 453.592 \text{ grams/pound} \approx 60,330.5 \text{ grams} \] Now we can substitute \( w \) into the formula and calculate the BAC for \( n \) from 1 to 10. The formula becomes: \[ \text{BAC} = \frac{600 n}{60330.5(0.6 n + 169)} \] Next, we will compute the BAC for each integer value of \( n \) from 1 to 10 and check if it exceeds 0.08. ### Calculations: 1. **For \( n = 1 \)**: \[ \text{BAC} = \frac{600 \times 1}{60330.5(0.6 \times 1 + 169)} = \frac{600}{60330.5(169.6)} \approx 0.0053 \] 2. **For \( n = 2 \)**: \[ \text{BAC} = \frac{600 \times 2}{60330.5(0.6 \times 2 + 169)} = \frac{1200}{60330.5(170.2)} \approx 0.0106 \] 3. **For \( n = 3 \)**: \[ \text{BAC} = \frac{600 \times 3}{60330.5(0.6 \times 3 + 169)} = \frac{1800}{60330.5(170.8)} \approx 0.0159 \] 4. **For \( n = 4 \)**: \[ \text{BAC} = \frac{600 \times 4}{60330.5(0.6 \times 4 + 169)} = \frac{2400}{60330.5(171.4)} \approx 0.0212 \] 5. **For \( n = 5 \)**: \[ \text{BAC} = \frac{600 \times 5}{60330.5(0.6 \times 5 + 169)} = \frac{3000}{60330.5(172)} \approx 0.0265 \] 6. **For \( n = 6 \)**: \[ \text{BAC} = \frac{600 \times 6}{60330.5(0.6 \times 6 + 169)} = \frac{3600}{60330.5(172.6)} \approx 0.0318 \] 7. **For \( n = 7 \)**: \[ \text{BAC} = \frac{600 \times 7}{60330.5(0.6 \times 7 + 169)} = \frac{4200}{60330.5(173.2)} \approx 0.0371 \] 8. **For \( n = 8 \)**: \[ \text{BAC} = \frac{600 \times 8}{60330.5(0.6 \times 8 + 169)} = \frac{4800}{60330.5(173.8)} \approx 0.0424 \] 9. **For \( n = 9 \)**: \[ \text{BAC} = \frac{600 \times 9}{60330.5(0.6 \times 9 + 169)} = \frac{5400}{60330.5(174.4)} \approx 0.0477 \] 10. **For \( n = 10 \)**: \[ \text{BAC} = \frac{600 \times 10}{60330.5(0.6 \times 10 + 169)} = \frac{6000}{60330.5(175)} \approx 0.0530 \] ### Conclusion: The BAC values for \( n = 1 \) to \( n = 10 \) are all below 0.08. Therefore, the person can consume **up to 10 drinks in an hour** without exceeding the legal limit for driving.