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 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 \( n=10 \). 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 \( \square \) drinks in an hour without exceeding the legal measure of drunk driving.

To solve this problem, we need to use the given formula to calculate the blood alcohol concentration (BAC) for each integer value of \( n \) from 1 to 10. The formula is: \[ \text{BAC} = \frac{600n}{w(0.6n + 169)} \] where \( w \) is the weight of a 196-pound person, which we will use as a constant in our calculations. Since the weight \( w \) does not change for each value of \( n \), we can calculate it once and use it for all values of \( n \). First, let's calculate the weight \( w \) in grams: \[ w = 196 \text{ pounds} \times 453.592 \text{ grams/pound} \] \[ w = 88623.312 \text{ grams} \] Now we can plug this value into the formula and calculate the BAC for each \( n \) from 1 to 10. We will then compare each BAC to the legal limit of 0.08 to determine the maximum number of drinks that can be consumed without exceeding the legal measure of drunk driving. Let's calculate the BAC for each \( n \): For \( n = 1 \): \[ \text{BAC} = \frac{600 \times 1}{88623.312 \times (0.6 \times 1 + 169)} \] \[ \text{BAC} = \frac{600}{88623.312 \times (0.6 + 169)} \] \[ \text{BAC} = \frac{600}{88623.312 \times 169.6} \] \[ \text{BAC} = \frac{600}{15051000.81952} \] \[ \text{BAC} \approx 0.00399 \] For \( n = 2 \): \[ \text{BAC} = \frac{600 \times 2}{88623.312 \times (0.6 \times 2 + 169)} \] \[ \text{BAC} = \frac{1200}{88623.312 \times (1.2 + 169)} \] \[ \text{BAC} = \frac{1200}{88623.312 \times 170.2} \] \[ \text{BAC} = \frac{1200}{15051000.81952} \] \[ \text{BAC} \approx 0.00800 \] For \( n = 3 \): \[ \text{BAC} = \frac{600 \times 3}{88623.312 \times (0.6 \times 3 + 169)} \] \[ \text{BAC} = \frac{1800}{88623.312 \times (1.8 + 169)} \] \[ \text{BAC} = \frac{1800}{88623.312 \times 170.8} \] \[ \text{BAC} = \frac{1800}{15051000.81952} \] \[ \text{BAC} \approx 0.01200 \] For \( n = 4 \): \[ \text{BAC} = \frac{600 \times 4}{88623.312 \times (0.6 \times 4 + 169)} \] \[ \text{BAC} = \frac{2400}{88623.312 \times (2.4 + 169)} \] \[ \text{BAC} = \frac{2400}{88623.312 \times 171.4} \] \[ \text{BAC} = \frac{2400}{15051000.81952} \] \[ \text{BAC} \approx 0.01600 \] For \( n = 5 \): \[ \text{BAC} = \frac{600 \times 5}{88623.312 \times (0.6 \times 5 + 169)} \] \[ \text{BAC} = \frac{3000}{88623.312 \times (3 + 169)} \] \[ \text{BAC} = \frac{3000}{88623.312 \times 172} \] \[ \text{BAC} = \frac{3000}{15301736.0624} \] \[ \text{BAC} \approx 0.01977 \] For \( n = 6 \): \[ \text{BAC} = \frac{600