The Blood Alcohol Concentration (BAC) of a person who has been drinking is given by the formula BAC \( =\frac{\text { (ounces } \times \text { percent alcohol } \times 0.075 \text { ) }}{\text { body weight in } \mathrm{lb}}-( \) hours of drinking \( \times 0.015) \). In two hours, a \( 200-\mathrm{lb} \) person has drunk 516 -oz beers, each having a \( 3.6 \% \) alcohol content. The formula would then read BAC \( =\frac{[(5 \times 16) \times 3.6 \times 0.075]}{200}-(2 \times 0.015) \). Answer parts (a) and (b) below. (a) The person's BAC is \( \square \). (Round to the nearest thousandth as needed.)

1. Write the formula for BAC: \[ \text{BAC} = \frac{(\text{ounces} \times \text{percent alcohol} \times 0.075)}{\text{body weight in lb}} - (\text{hours of drinking} \times 0.015) \] 2. Substitute the given values into the formula. The person drinks \(5\) beers, each of \(16\) oz, so the total amount of beer is: \[ 5 \times 16 = 80 \text{ oz}. \] The percent alcohol is \(3.6\), the body weight is \(200\) lb, and the drinking period is \(2\) hours. Thus, the expression becomes: \[ \text{BAC} = \frac{[(5 \times 16) \times 3.6 \times 0.075]}{200} - (2 \times 0.015) \] 3. Compute the total ounces: \[ 5 \times 16 = 80 \text{ oz}. \] 4. Calculate the numerator: \[ 80 \times 3.6 \times 0.075 = 80 \times 0.27 = 21.6. \] (Note: \(3.6 \times 0.075 = 0.27\)) 5. Divide by the body weight: \[ \frac{21.6}{200} = 0.108. \] 6. Subtract the alcohol elimination over the drinking period: \[ 2 \times 0.015 = 0.03. \] So, \[ 0.108 - 0.03 = 0.078. \] 7. Therefore, the person's BAC is: \[ 0.078. \] Thus, the answer for part (a) is \( \boxed{0.078} \).