3.6: Balancing Blood Alcohol
- Page ID
- 148751
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)INTRODUCTION
As you keep taking more advanced math courses in the future, you will keep seeing more operations, procedures, or functions that cancel each other out. Many operations, procedures, or functions come in pairs. Whatever one operation does, the other one will undo.
In this collaboration, we will be focusing on two pairs of such operations:
- Addition and subtraction
- Multiplication and division
Let’s illustrate the operation of addition and subtraction with a simple example:
- Pick a number. You can pick any number you want: positive, negative, a fraction, or a decimal. In this example, we’ll use 25.
- Next, add 10 to the number that you picked. In the example, that would be 25+10=35.
- Next, subtract 10 from the number you found in the previous step. In this example, that would be 35-10=25.
- From this example, and your own work, you can see we just got the same number that we started with! This shows how addition and subtraction cancel each other out.
This is not a coincidence. It works for any numbers and variables. For example,
- 50 + 2 - 2 = 50
- -20 - 76 + 76 = -20
- x + 2 - 2 = x
- -60R - 91 + 91 = -60R
The same applies to multiplication and division. To illustrate, let’s look at another example, but instead of adding and subtracting, we will be multiplying and dividing.
- Pick a number. You can pick any number you want - positive, negative, a fraction, or a decimal. This example will use 80.
- Next, multiply the number that you picked by 20. In this example, 80\(\times\)20=1,600.
- Next, divide the number that you found in the previous step by 20. In this example, 1,600\(\div\)20=80.
- From this example, and your own work, you can see we just got the same number that we started with! This shows how multiplication and division cancel each other out.
This is not a coincidence. It works for any numbers and variables. For example,
- 50 × 10 ÷ 10 = 50
- T × 25.81 ÷ 25.81 = T
- -7 ÷ 8 × 8 = -7
- 25 × y ÷ 25 = y
SPECIFIC OBJECTIVES
By the end of this collaboration, you should understand that
- addition and subtraction are inverse operations.
- multiplication and division are inverse operations.
- solving for a variable includes isolating it by “undoing” the actions to it.
By the end of this collaboration, you should be able to
- solve for a variable in a linear equation.
- explicitly write out the order of operations to evaluate a given equation.
PROBLEM SITUATION: CALCULATING BLOOD ALCOHOL CONTENT
Blood alcohol content (BAC) is a measurement of how much alcohol is in someone’s blood. It is usually measured as a percentage. So, a BAC of 0.3% is three-tenths of 1%. That is, there are 3 grams of alcohol for every 1,000 grams of blood. A BAC of 0.05% impairs reasoning and the ability to concentrate. A BAC of 0.30% can lead to shortness of breath, a blackout, and loss of bladder control. In most states, the legal limit for driving is a BAC of 0.08%.21
BAC is usually determined by a Breathalyzer, urinalysis, or blood test. However, the Swedish physician E.M.P. Widmark developed the following equation for estimating an individual’s BAC.22 This formula is widely used by forensic scientists:
\[B = -0.015t + \left( \dfrac{2.84\cdot N}{W\cdot r}\right) \nonumber \]
B = percentage of BAC
N = number of “standard drinks” (A standard drink is one 12-ounce beer, one 5-ounce glass of wine, or one 1.5-ounce shot of liquor.) N should be at least one.
W = weight in pounds
r = the distribution rate for alcohol through the body (this value is a constant), 0.68 for males
and 0.55 for females
t = number of hours since the first drink
(1) Looking at the equation, discuss why the items (t, N, W, and r) on the right of the equation make sense in calculating BAC.
(2) Consider the case of a male student who has three beers and weighs 120 pounds. Simplify the equation as much as possible for this case. What variables are still unknown in the equation? Round values where necessary to the nearest thousandth.
(3) (a) Using your simplified equation from Question 2, find the estimated BAC for this student
one hour, three hours, and five hours after his first drink.
1 hour:
3 hours:
5 hours:
(b) What patterns do you notice in the data from 3(a)?
(4) Discuss with your group how you arrived at the BAC values mathematically. For example, did you multiply, add, subtract, etc., and what did you do first? Outline the steps that you took to get the answers for Question 3(a).
(5) How long will it take for this student’s BAC to be 0.08, the legal limit? How long will it take for the alcohol to be completely metabolized resulting in a BAC of 0.0? Round to one decimal place.
When BAC = 0.08, t =
When BAC = 0.0, t =
(6) A female student, weighing 110 pounds, plans on going home in two hours. The simplified Widmark formula is:
\[B = -0.03 + \dfrac{2.84\cdot N}{60.5}\nonumber \]
(a) Compare the student’s BAC for one glass of wine versus three glasses of wine at the time she will leave. Round to three decimal places.
When N = 1, B =
When N = 3, B =
(b) In this scenario, determine how many drinks she can have so that her BAC remains less than 0.08. Round to the nearest drink.
FURTHER APPLICATIONS
(7) Solve the following equation for the values given in (a) and (b). In each case, outline the steps you took as you did in Question 4.
y = −4x – 2
(a) Solve for y if x = −3. Outline your steps.
(b) Solve for x if y = −3. Outline your steps.
MAKING CONNECTIONS
Record the important mathematical ideas from the discussion.
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