Chapter R - Section R.2 - Real Numbers and Their Properties - R.2 Exercises - Page 19: 119

$\displaystyle \begin{array}{|r|c|} \hline \mathrm{Man's\ BAC} & \mathrm{0.024}\\ \hline \mathrm{Woman's\ BAC} & \mathrm{0.023}\\ \hline \end{array}$

$\displaystyle \begin{array}{ l } \text{The BAC Formula for a person who has been drinking alcohol is} :\\ \#\ \text{oz} \ \times \ \%\ \text{alcohol} \ \times \dfrac{75}{1000} \times \dfrac{1}{\text{lbs}} - \text{hrs} \times \dfrac{15}{1000}\\ \\\\ \begin{array}{|l|l|} \hline a) & \text{Calculate the man's BAC concentration level} :\\ \hline & \begin{array}{{>{\displaystyle}l}} \text{If the man weighed 25 lbs heavier than his current weight, he would}\\ \text{weigh approximately 215 lbs. Considering this while all other variables}\\ \text{remain unchanged, his BAC concentration level is found by} :\\ \\ 48\ \times 3.2\ \times \dfrac{75}{1000} \ \times \dfrac{1}{215} -2\ \times \dfrac{15}{1000}\\ \\ \text{Applying the order of operations as follows} :\\ \\ \begin{array}{ l } \begin{array}{{>{\displaystyle}l}} \text{Man's BAC}\\ =\left( 48\ \times 3.2\ \times \dfrac{75}{1000} \ \times \dfrac{1}{215}\right) -\left( 2\ \times \dfrac{15}{1000}\right)\\ \end{array}\\ \begin{array}{{>{\displaystyle}l}} \approx ( 0.053581395348838) \ -\ ( 0.03)\\ \approx 0.024 \end{array} \end{array} \end{array}\\ & \\ \hline b.) & \text{Calculate the woman's BAC concentration level} :\\ \hline & \begin{array}{{>{\displaystyle}l}} \text{If the woman weighed 25 lbs heavier than her current weight, she would}\\ \text{weigh approximately 160 lbs. Considering this while all other variables}\\ \text{remain unchanged, her BAC concentration level is found by:}\\ \\ 36\ \times 4\ \times \dfrac{75}{1000} \ \times \dfrac{1}{160} -3\ \times \dfrac{15}{1000}\\ \\ \text{Applying the order of operations as follows:}\\ \begin{array}{ l } \begin{array}{{>{\displaystyle}l}} \text{Woman's BAC}\\ =\left( 36\ \times 4\ \times \dfrac{75}{1000} \ \times \dfrac{1}{160}\right) -\left( 3\ \times \dfrac{15}{1000}\right)\\ \end{array}\\ \begin{array}{{>{\displaystyle}l}} =\left(\dfrac{27}{400}\right) \ -\ \left(\dfrac{45}{1000}\right)\\ =\dfrac{27}{400} -\dfrac{9}{200}\\ =\dfrac{27}{400} -\dfrac{18}{400}\\ =\dfrac{9}{400}\\ \approx 0.023 \end{array} \end{array} \end{array}\\ & \\ \hline & \begin{array}{{>{\displaystyle}l}} \text{These calculations seem to suggest that if all other variables,}\\ \text{remain the same, an increase in weight will result in a decrease in}\\ \text{BAC output.} \end{array}\\ \hline \end{array} \end{array}$