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This unit was moved from 7th grade, so this is a new unit for 6th grade.
The new Common Core State Standards for Mathematics (CCSSM) describe quite specific objectives in algebra for grade six. While some introduction to algebra is done in Prime Time, Let&rsquos be Rational, Covering and Surrounding, and Decimal Ops, the most natural way for CMP3 to respond to those enhanced expectations was to move a revised version of the Variables and Patterns unit from the beginning of grade seven to the end of grade six. The revision for CMP3includes some streamlining of the first three investigations (in part responding to CCSSM requirements that objectives like coordinate graphing be treated in earlier grades and units) and development of new material on order of operations, equivalent expressions, and solving equations that also responds to CCSSM requirements. The study of patterns of change between two variables and how these patterns are captured in verbal contexts, tables, graphs and equations are maintained from CMP2 and are also part of the CCSSM for 6 th grade.
Investigation 1: This set of 4 problems begins the development of students&rsquo ability and disposition to find relationships between variables and the representation of those relationships in tables and graphs and &lsquostories.&rsquo It focuses almost exclusively on change over time relationships, especially the fundamental connections between distance, rate, and time.
The major changes for CMP3 are:
 Delay of the introduction of the terms &lsquoindependent&rsquo and &lsquodependent&rsquo for describing variables until Investigation 2. The rationale for this is that in the typical time and distance situations it doesn&rsquot make much sense to think of distance as a variable that depends on time. The more natural settings for thinking about dependence occur in Investigation 2, where changes in the one variable really do cause changes in another.
 More explicit development of the important relationship between distance, rate, and time in Problem 1.4. This is really only a modest enhancement of issues that have been in this unit all along. It seems a desirable focus of the investigation, since those variables are so important in many mathematical and applied contexts.
Investigation 2: This set of 4 problems continues the development of students&rsquo ability and disposition to find relationships between variables and the representation of those relationships in tables and graphs and &lsquostories.&rsquo The focus in this investigation is on cause and effect relationships between dependent and independent variables.
The only significant change in CMP3 is introduction of problem 2.3 on fourquadrant graphing. This is included to satisfy CCSSM objective 6.NS.8.
Investigation 3: This investigation focuses attention on writing symbolic equations for relationships between two variables. It responds directly to the heavy CCSSM emphasis on expressing relationships in symbolic form, and it also deals explicitly with proportionality relationships of the y = mx type involving rates, another CCSSM requirement.
The major changes proposed for this investigation are:
 Introduction of Problem 3.2 to deal explicitly with CCSS 6.RP.2 and 6.RP.3 on proportionality.
 Use tables, graphs and numeric reasoning to find values of the variables.
 Introduction of Problem 3.4 to develop the order of operation conventions that are needed to express quantitative relationships accurately. It addresses CCSSM objectives 6.EE.1, 6.EE.2a, 6.EE.2c, 6.EE.3, and 6.EE.4.
Investigation 4: This investigation is a substantial replacement of the material in CMP2, reworked to address the heavy agenda of CCSSM objectives for equivalent expressions and solving equations and inequalities. It has five problems. While we still encourage students to use calculator or computergenerated tables and graphs to study relationships and solve problems, we have deleted the explicit directions on how to use those tools. We are assuming that teachers can provide the &lsquohow to&rsquo instruction without our including it in the student text.
Ch 4.1 Discrete Random Variable
Random Variable: is a variable (X) that has a single numerical value, determined by chance, for each outcome of a procedure.
Probability distribution: is a table, formula or graph that gives the probability of each value of the random variable.
A Discrete Random Variable has a collection of values that is finite or countable similar to discrete data.
A Continuous Random Variable as infinitely many values and the collection of values is not countable.
Ex : X = the number of times &ldquofour&rdquo shows up after tossing a die 10 times is a discrete random variable.
X = weight of a student randomly selected from a class. X is a continuous random variable
X = the method a friend contacts you online. X is not a random variable. ( X is not numerical)
A Probability Distribution (PDF) for a Discrete Random Variable is a table, graph or formula that gives Probability of each value of X.
A Probability Distribution Function (PDF or PD) satisfies the following requirements:
1. The value X is numerical, not categorical and each P(x) is associated with the corresponding probability.
2 &SigmaP(X) = 1 . A &SigmaP(X) = 0.999 or 1.01 is acceptable as a result of rounding.
3. 0 &leP(x) &le 1 for all P(x) in the PDF.
Ex1: PDF for number of heads in a twocoin toss are given as a table and a graph.
Both are valid PDF because &Sigma P(X) = 1
and each value of P(X) is between 0 and 1.
Ex2: The number of medical tests a patient receives after entering a hospital is given by the PDF below
a) Is the table a valid PDF?
The table is not a probability distribution because &Sigma P(x) = 0.02+0.18+0.3+0.4 = 0.9 is not 1
b) Define the random variable x.
x = no. of medical tests a patient receives after entering a hospital.
c) Explain why the x = 0 is not in the PDF?
A patient always receives at least one medical test in the hospital.
Parameters of a Probability distribution:
Mean &mu for a probability distribution:
Variance &sigma 2 for a probability distribution:
Standard deviation for a probability distribution:
To calculate Mean, variance and standard deviation of a probability distribution by technology:
Use Libretext statistics calculator:
Enter Number of outcomes, each X and P(X), calculate.
round off rule: one more decimal place than for E(x)
Two decimal places for &sigma and &sigma 2 .
Expected value = the longterm outcome of average of x when the procedure is repeated infinitely many times. Round to one decimal place.
Nonsignificant values of X.
1. The range of X from ( mu  2cdotsigma ext< to >mu +2cdotsigma ) are nonsignificant. (Range of rule of Thumb)
2. X that are outside of ( mu  2cdotsigma ext < to >mu +2cdotsigma ) are significant that is unlikely to occur.
Ex1: X = no. of year a new hire will stay with the company. P(x) = Prob. that a new hire with stay for x year.
a) Find mean, variance, st. deviation and determine the Expected number of years a new hire will stay.
Use easycalculation.com statistics discrete random variable calculator,
Enter number of outcomes = 7. Mean = 2.4, &sigma 2 = 2.73, &sigma =1.65
The Expected no. of year a new hire will stay is 2.4 years.
b) Find probability that a new hire will stay for 4 years or more.
Add P(4), P(5) and P(6) = P( 4 or more) = 0.1 + 0.1 + 0.05 = 0.25
c) Find probability that a new hire will stay for between 3 or 5 years inclusive.
P( 3 to 5 inclusive) = 0.15 + 0.1+0.1 = 0.35
d) Find the probability that a new hire will stay for 2 years or fewer.
P(2 or fewer) = 0.12 + 0.18 + 0.30 = 0.6
e) Find the range of nonsignificant year of stay.
2.4 2(1.65) to 2.4 + 2(1.65) is 0.9 to 5.7
Ex2: Given x = of number of textbooks a student buy per semester. What is the expected number of textbooks?
a) Find mean, variance and standard deviation.
Use easycalculation.com statistics discrete random variable calculator, Enter number of outcomes = 6
Expected number of textbook is 3.5 books.
b) Find Probability that a student buys at least 5 textbook.
P( at least 5) = P(5 or more) = 0.03 + 0.02 = 0.05,
c) Find probability that x is at most 2.
P(at most 2 ) = P( 2 or fewer) = 0.02 + 0.03 = 0.05
c) Find the range nonsignificant.
Range of nonsignificant is 3.5 &ndash 2(0.78) to 2.5 + 2(0.78) is 1.94 to 5.06.
4.1: What is a Variable?  Mathematics
In any technical course, math plays a vital role in obtaining correct solutions for the problems. Statics is no exception. The student must have a solid base in several mathematical disciplines to setup and solve the equations of statics. Algebra, trigonemetry, geometry, and calculus are all very important in the study of statics and beyond. This course is intended for the first year technology student, and will therefore not require calculus. This section is intended to be a review of the basic mathematical principles that are used exetensively in this course. It is an overview, and is not intended to be all inclusive. The principles involved will be presented primarily in the form of examples.
Algebra : Most of the problems in this course rely heavily on algebra to solve the equations which are used. The student will be required to solve equations with one variable and sets of equations with several variables. In addition, the student should be familiar with the quadratic formula, and with natural logarithms.
Equations With One Variable : A very common problem is an equation with just one variable. The variable may appear once or several times throughout the equation. The equation may be linear or nonlinear, and may contain trig functions or logarithmic functions.
The basic approach used to solve an equation with one variable is to manipulate the equation until the variable is isolated on one side of the equation, and everything esle is on the other side. In the case of a linear equation, this can be done by adding or subtracting equals to each side, multiplying both sides by equals, or dividing both sides by equals. Often it will require a series of operations to arrive at the final solution. In the case of nonlinear equations, other operations may have to be performed, such as taking the square root of each side or the tangent of each side. Examples 21 thru 24 are all examples taken from statics problems, showing step by step manipulations to determine the final solution. For several reasons, it is recommended that the student does not skip steps while solving a problem. Skipping steps can lead to errors, and will always making checking the work more difficult.
Example 21  Linear Equation With 1 Variable Solve for X in the equation: Multiply both sides by 3: Multiply thru by 2: Subtract 24 from both sides: Add 18X to both sides: Do the arithmetic on the left side: Divide both sides by 18: Substitute 2 into the original equation to check:
Example 22  NonLinear Equation With 1 Variable Using the Quadratic Formula Solve for X in the equation: X (5X + 1.5)  2.6 = 1.6X (2.5X+.5) Multiply thru by X on the left and 1.6X on the right: 5X 2 + 1.5X  2.6 = 4X 2 +.8X Subtract 4X 2 +.8X from both sides: X 2 + .7X  2.6 = 0 Apply the quadratic formula: Substitute values into the quadratic formula: Solve for X: X 1 = 1.3
X 2 = 2
Notice that there are 2 solutions for X in this equation. There will always be 2 solutions to a quadratic equation, however, usually only one solution will satisfy the conditions of the problem. After the 2 solutions have been calculated, then they must be evaluated in the context of the problem to determine which solution is the appropriate one for the situation.
The 2 solutions should be substituted back into the original equation to check the work. The checks will not be shown here, but should be done.


Equations With Several Variables : In statics, as in most technical courses, problems arise which involve several variables. If a system of equations has N variables, then there must be N independent equations in those variables to be able to find a solution. There are several methods which can be used to solve such a system of equations. The examples below show 3 ways to solve the same system of equations. The equations used are:
X + 2Y  3Z = 1 (1)
2X + Z = 0 (2)
3X  4Y + 4Z = 2 (3)
One method used to solve simultaneous equations is to solve one of the equations for one of the variables, then substituting that into one of the other equations, thereby eliminating that variable from the second equation. That process continues until there is only one variable remaining in an equation. The lone variable is solved for, then back substituted into one of the other equations, until all of the solutions have been found.
Example 25  Systems of Equations Write down all of the equations:
X + 2Y  3Z = 1 (1)
2X + Z = 0 (2)
3X  4Y + 4Z = 2 (3)
Z = 2X Substitute the result into one of the other equations. For example, substite Z=2X into equation 1: X + 2Y  3Z = 1
X + 2Y  3(2X) = 1
X + 2Y + 6X = 1
5X + 2Y = 1 Solve this new equation for one of the remaining variables: 5X + 2Y = 1
5X  1 = 2Y
1  5X = 2Y
Y = (1  5X) / 2 Substitute this result, and the result for Z into the unused equation, in this case equation 3: 3X  4Y + 4Z = 2
3X  4[(1  5X) / 2] + 4(2X) = 2
3X  2(1  5X) + 4(2X) = 2
3X  2 + 10X  8X = 2 Solve the resulting equation for the remaining variable: 3X  2 + 10X  8X = 2
5X = 4
X = .8 Back substitute this result into equation 2: 2X + Z = 0
2(.8) + Z = 0
1.6 + Z = 0
Z = 1.6 Back substitute this result into equation 3 (or 1): 3X  4Y + 4Z = 2
3(.8)  4Y + 4(1.6)= 2
Y = 1.5 So, the final solutions are: X = .8
Y = 1.5
Z = 1.6
The same system of equations can be solved by using another method of eliminating variables in the equations. In order to take full advantage of this second method, it is important to understand the concept of equivalent systems of equations. Two systems of equations are equivalent if they have precisely the same solution set. There are three operations that can be performed on any system of equations which will produce equivalent systems:
1.) Interchange two equations.
2.) Multiply an equation by a nonzero constant.
3.) Add a multiple of an equation to another equation.
The first of these is important in other methods of solving a system of equations, but will not be used here. The method that will be used in the next example involves eliminating variables by repeatedly using the second and third operations above.
Example 25  Systems of Equations Write down all of the equations:
X + 2Y  3Z = 1 (1)
2X + Z = 0 (2)
3X  4Y + 4Z = 2 (3)
2X + 4Y  6Z = 2 (1)
2X + Z = 0 (2)
3X  4Y + 4Z = 2 (3)
Multiply equation 2 by 2: 4 X + 2Z = 0
Add this to the previous result: 5 X = 4 Solve the resulting equation: X = .8 Back substitute this result into equation 2: 2X + Z = 0
2(.8) + Z = 0
1.6 + Z = 0
Z = 1.6 Back substitute this result into equation 3 (or 1): 3X  4Y + 4Z = 2
3(.8)  4Y + 4(1.6)= 2
Y = 1.5 So, the final solutions are: X = .8
Y = 1.5
Z = 1.6
The same system of equations can be solved by using a third method known as Cramer's Rule. This method is efficient for systems of equations involving three variables, but is very tedious for larger systems. However, this method can be programmed fairly easily, so it lends itself well to computer solutions of larger systems. Cramer's rule involves finding the determinant of matrices. Therefore, a brief review will be given for finding the determinant of matrices of order 2 and order 3. Refer to any linear algebra text for other orders.
A matrix is an array of numbers arranged in row and column format. For applying Cramer's rule, the matrices will be square, meaning that the number of rows and the number of columns is the same. A matix is represented by the array shown inside a set of brackets:
The determinant of a matrix is represented by the same array shown inside a set of vertical line:
The determinant of a matrix is a number, which is evaluated by manipulating the numbers in the array.
Determinant of a matrix of order 2 :
To evaluate the determinant of a matrix of order 2, use:
A = a 11 ( C 11 ) + a 12 (C 12 ) + a 13 (C 13 )
With this background, it is now possible to discuss Cramer's Rule. This will be done by way of an example using a matrix of order 3. Refer to a linear algebra text for applications of Cramer's Rule to other size matrices.
Example 28  Cramer's Rule for a Systems of Equations
A =
Trigonometry : Trigonometry is used extensively in statics. Trigonometry is the study of relationships in triangles. These relationships include functions of angles such as sine and cosine, and relationships involving the lengths of the sides, such as the Pythagorean theorem.
Basic Trigonometric Functions: The three basic trig functions used extensively in statics are sine, cosine, and tangent. Consider the triangle below:
It is important to recognize that these relationships only hold for right triangles. In statics, right triangles appear quite regularly, so these relationships will be used often, however, there is also a need for relationships between angles and sides on triangles which do not have a right angle. Those relationships are called the Law of Sines and the Law of Cosines. Consider the triangle below:
Example 29  Trig. Functions
For the triangle on the left, find a and b.
Now that a is known, the tangent function can be used again to find b:
Example 210  Law of Cosines & Law of Sines
For the triangle on the left, find the length of the unknown side and the angle .
Geometry : There are several geometric relationships which will be needed in this course.
Other examples showing the identity property of multiplication.
Multiplicative inverse property
If you multiply two numbers and the product is 1, we call the two numbers multiplicative inverses or reciprocals of each other.
For example, 4 is the multiplicative inverse of 1/4 because 4 × 1/4 = 1.
1/4 is also the multiplicative inverse of 4 because 1/4 × 4 = 1.
Notice that the multiplicative inverse of 1 is 1 since 1 × 1 is 1 . By the same token, the multiplicative inverse of 1 is 1 since 1 × 1 is 1. In fact, 1 and 1 are the only two numbers that can be their own multiplicative inverse.
Notice also that any number divided by 1 return the same number. We call this the identity property of division.
Contingency Tables
A contingency table relates two categories of data. In the example above, the relationship is between the gender of the student and his/her response to the question.
A marginal distribution of a variable is a frequency or relative frequency distribution of either the row or column variable in the contingency table.
If we consider the previous example:
The entire table is referred to as the contingency table.
The marginal distribution for gender removes the effect of whether or not the student enjoys math:
Determinant of a matrix of order 3 : Cofactors will always be either +1 or 1 times the minor, and can be determined from the following matrix: This shows that the cofactor for a 11 = C 11 = (+1)M 11 . Having defined these two terms, it is now possible to define a procedure for evaluating the determinant of a matrix of order 3: A = a 11 ( C 11 ) + a 12 (C 12 ) + a 13 (C 13 ) Example 27  Determinant of a Matrix of Order 3
 
Find the determinant of the following matrix:  A =  
Write down all of the equations: 
X + 2Y  3Z = 1 (1)  
Set up a "coefficient matrix". This is a matrix with the coefficients of X in column 1, coefficients of Y in column 2, and coefficients of Z in column 3:  
Calculate the determinant of the coefficent matrix using the method shown in Example 27:  A = 10  
Set up 3 more matrices by replacing individual columns with the values of the constants on the right side of the equations:  
Calculate the determinants of these matrices using the method shown in Example 27:  A 1  = 8
A 3  = 16  
Apply Cramer's Rule:  
So, the final solutions are:  X = .8 Y = 1.5 Z = 1.6  
This is a right triangle, and both a & b can be determined using basic trig. functions:  
Since two sides and the angle between them are given, this is an example of a problem suited to the Law of Cosines. Call the side opposite the given angle C, and apply the Law of Cosines:  
C 2 = A 2 + B 2 + 2AB cos c
C 2 = 6 2 + 4 2 + 2(6)(4) cos 20 The Law of Sines can now me used to find the angle .  
Strongly Agree  Agree  Neutral  Disagree  Strongly Disagree  Total  
Men  9  13  5  2  1  30 
Women  12  18  11  6  5  52 
Whereas, the marginal distribution for whether or not the student enjoys math removes the effect of gender:
Strongly Agree  Agree  Neutral  Disagree  Strongly Disagree  
Men  9  13  5  2  1 
Women  12  18  11  6  5 
Total  21  31  16  8  6 
We can also create a relative frequency marginal distribution, which, as expected, is simply relative frequencies rather than frequencies.
The combined relative frequency marginal distributions would look like this:
SA  A  N  D  SD  Total  
Men  9  13  5  2  1  30/82 &asymp 0.37 
Women  12  18  11  6  5  52/82 &asymp 0.63 
Total  21/82 &asymp 0.26  31/82 &asymp 0.39  16/82 &asymp 0.20  8/82 &asymp 0.10  6/82 = 0.07  1 
Let's consider the frequency marginal distributions from Example 2.
We might now be interested in comparing the two variables. For example:
 What proportion of women strongly agreed with the statement "I enjoy math"?
 What proportion of women disagreed?
 What proportion of men were neutral?
 What proportion of men strongly agreed?
 There were 12 women who strongly agreed, and 52 women in all, so 12/52 &asymp 0.23
 Similarly, there were 6 women who disagreed, and 52 overall, so 6/52 &asymp 0.12
 5/30 &asymp 0.17
 9/30 &asymp 0.30
If we completed the table in this fashion, we get something called a conditional distribution.
A conditional distribution lists the relative frequency of each category of variable, given a specific value of the other variable in the contingency table.
For another explanation of marginal and conditional distributions, watch this YouTube video:
The conditional distribution of how the students feel about math by gender would be as follows:
SA  A  N  D  SD  Total  
Men  9/30 &asymp 0.30  13/30 &asymp 0.43  5/30 &asymp 0.17  2/30 &asymp 0.07  1/30 &asymp 0.03  30/30 = 1 
Women  12/52 &asymp 0.23  18/52 &asymp 0.35  11/52 &asymp 0.21  6/52 &asymp 0.12  5/52 &asymp 0.10  52/52 = 1 
Note: The row totals sometimes do not add up to 1 due to rounding.
Another way to think of this distribution is that it's the distribution of how students feel for each gender. That's what the "by gender" indicates.
The conditional distribution of gender by how the student feels would be:
SA  A  N  D  SD  
Men  9/21 &asymp 0.43  13/31 &asymp 0.42  5/16 &asymp 0.31  2/8 = 0.25  1/6 &asymp 0.17 
Women  12/21 &asymp 0.57  18/31 &asymp 0.58  11/16 &asymp 0.69  6/8 = 0.75  5/6 &asymp 0.83 
Total  21/21 = 1  31/31 = 1  16/16 = 1  8/8 = 1  6/6 = 1 
4.1: What is a Variable?  Mathematics
Finite Mathematics Lesson 4
You just learned a method for solving linear programming problems using a graphical approach. This method is not practical if there are more than 2 variables in the the problem. Many business or economics problems may involve thousands or millions of variables. Chapter 4 introduces a new method to handle these problems more efficiently. The simplex method is an algorithmic approach and is the principle method used today in solving complex linear programming problems. Computer programs are written to handle these large problems using the simplex method.
Just a little history on the simplex method
George Dantzig 'invented' the simplex method while looking for methods for solving optimization problems. He used a primitive computer in 1947 to achieve his success in developing the simplex method . Dantzig is currently a professor of operations research and computer science at Stanford.
In 1984, Narenda Karmarker, a research mathematician at Bell Laboratories, invented a powerful new linear programming algorithm that is faster and more efficient than the simplex method . I think this may be "owned" by AT&T and is said to have "a direct impact on the efficiency and profitability of of numerous industry".
Section 4.1 and 4.2 Slack Variables and the Simplex Tableau and The Simplex Method
Read both section sections first. What are we trying to accomplish by using the simplex method ? This method gives the maximum or minimum value of a system that has many variables. We will limit our discussion to chapters 4.1 and 4.2 (maximizing problems).
 Pivoting around a selected element means to make all the entries above and below it 0
 Verify the solutions in the original inequalities and objective function is not easy when dealing with 3 or more values. Although we can actually verify solutions for 2 variable problems, we will just accept that the theory works for higher dimensional problems.
 Objective function : This is the function you are trying to minimize or maximize.
 Slack variables: These are the 'extra' variables put into the table (tableau). They will form a diagonal of 1's
 Lets go through an entire problem from start to finish.
 These are the constraints. Note: Each variable has to be positive.
 60x + 90y + 300z is the objective function to be maximized
 The top row identifies the variables.
 u,v,w, and M are slack variables
 The numbers in bold are from the original constraints.
 The bottom row comes from setting the equation
M = 60x + 90y + 300z to 0
60x  90y  300z +M = 0
 Determine if the left part of the bottom row contains negative entries.
If none, problem solved.  If yes, the pivot column is the column with the most negative entry in the last row. In this case it is column z
 The left part is columns x, yor z
 Divide the last column by the entries in each pivot column.
 The pivot row is the row with the least nonnegative ratio. negative values or undefined values are ignored
 The pivot row is the row in bold (600 is least)
 The pivot element is the entry where the pivot column and pivot row intersect.
 The pivot element is the number in bold (1)
 Make all the numbers above or below the pivot element 0.
 The entry directly below pivot element is already 0
 We need to make the other entries 0
 Multiply 1 times ROW 1 and add it to ROW 3
 ROW 1 is the first row containing numbers
 Multiply 300 times ROW 1 and add it to ROW 4
 ROW 1 is the first row containing numbers
 Pivoting this element is complete
 Determine if the left part of the bottom row contains negative entries.
If none, problem solved.  If yes, the pivot column is the column with the most negative entry in the last row. In this case it is column z
 The left part is columns x, yor z
The solution.
x = 0, y = 0, z = 600, u = 0
v = 600, w = 300, M = 180,000
 The only columns of interest to us are x,y,z and M
 Columns z,v,w,and M are the only columns pivoted
 Columns not pivoted are set equal to 0
 0 + 0 + 600 <= 600 true
 0 +3(0) <= 600 true
 2(0) + 600 <= 900 true
 180,000 = 60(0) + 90(0) + 300(600)
A store sells three brands of of stereo systems, brands A,B, and C. It can sell a total of 100 stereo systems per month.
First constraint a + b + c <= 100 note: <= means less than or equal to.
Brands A, B, and C take up, respectively, 5, 4, and 4 cubic feet of warehouse space and a maximum of 480 cubic feet of warehouse space is available.
Second constraint 5a + 4b + 4c <= 480
Brands A, B, and C generate sales commissions of $40, $20, and $30, respectively, and $3200 is available to pay the sale commissions.
Third constraint 40a + 20b + 30c <= 3200
 Lets go through an entire problem from start to finish.
 These are the constraints. Note: Each variable has to be positive.
 70a + 210b + 140c is the objective function to be maximized
 The top row identifies the variables.
 u,v,w, and M are slack variables
 The numbers in bold are from the original constraints.
 The bottom row comes from setting the equation
M = 70a + 210b + 140c to 0
70a  210b  140c +M = 0
 Determine if the left part of the bottom row contains negative entries.
If none, problem solved.  If yes, the pivot column is the column with the most negative entry in the last row. In this case it is column b
 The left part is columns a,bor c
 Divide the last column by the entries in each pivot column.
 The pivot row is the row with the least nonnegative ratio. negative values or undefined values are ignored
 The pivot row is the row in bold (100 is least)
 The pivot element is the entry where the pivot column and pivot row intersect.
 The pivot element is the number in bold (1)
 Make all the numbers above or below the pivot element 0.
 The entry directly below pivot element is already 0
 We need to make the other entries 0
 Multiply 20 times ROW 1 and add it to ROW 3
 ROW 1 is the first row containing numbers
 Multiply 210 times ROW 1 and add it to ROW 4
 ROW 1 is the first row containing numbers
 Pivoting this element is complete
 Determine if the left part of the bottom row contains negative entries.
If none, problem solved.  If yes, the pivot column is the column with the most negative entry in the last row. In this case it is column z
 The left part is columns a, bor c
The solution.
a = 0, b = 100, c = 0, u = 0
v = 80, w = 1200, M = 21,000
4.1 Probability Distribution Function (PDF) for a Discrete Random Variable
There are two types of random variables , discrete random variables and continuous random variables. The values of a discrete random variable are countable, which means the values are obtained by counting. All random variables we discussed in previous examples are discrete random variables. We counted the number of red balls, the number of heads, or the number of female children to get the corresponding random variable values. The values of a continuous random variable are uncountable, which means the values are not obtained by counting. Instead, they are obtained by measuring. For example, let X = temperature of a randomly selected day in June in a city. The value of X can be 68°, 71.5°, 80.6°, or 90.32°. These values are obtained by measuring by a thermometer. Another example of a continuous random variable is the height of a randomly selected high school student. The value of this random variable can be 5'2", 6'1", or 5'8". Those values are obtained by measuring by a ruler.
A discrete probability distribution function has two characteristics:
 Each probability is between zero and one, inclusive.
 The sum of the probabilities is one.
Example 4.1
A child psychologist is interested in the number of times a newborn baby's crying wakes its mother after midnight. For a random sample of 50 mothers, the following information was obtained. Let X = the number of times per week a newborn baby's crying wakes its mother after midnight. For this example, x = 0, 1, 2, 3, 4, 5.
P(x) = probability that X takes on a value x.
X takes on the values 0, 1, 2, 3, 4, 5. This is a discrete PDF because we can count the number of values of x and also because of the following two reasons:
 Each P(x) is between zero and one, therefore inclusive
 The sum of the probabilities is one, that is,
A hospital researcher is interested in the number of times the average postop patient will ring the nurse during a 12hour shift. For a random sample of 50 patients, the following information was obtained. Let X = the number of times a patient rings the nurse during a 12hour shift. For this exercise, x = 0, 1, 2, 3, 4, 5. P(x) = the probability that X takes on value x. Why is this a discrete probability distribution function (two reasons)?
Example 4.2
Suppose Nancy has classes three days a week. She attends classes three days a week 80 percent of the time, two days 15 percent of the time, one day 4 percent of the time, and no days 1 percent of the time. Suppose one week is randomly selected.
Describe the random variable in words. Let X = the number of days Nancy ________.
4.1: What is a Variable?  Mathematics
Random variable is basically a function which maps from the set of sample space to set of real numbers. The purpose is to get an idea about result of a particular situation where we are given probabilities of different outcomes. See below example for more clarity.
Formal definition :
X: S > R
X = random variable (It is usually denoted using capital letter)
S = set of sample space
R = set of real numbers
 0 p1 + p2 + p3 = 1
==> p1 + 0.3 + 0.5 = 1
==> p1 = 0.2
Continuous Random Variable:
A random variable X is said to be continuous if it takes on infinite number of values. The probability function associated with it is said to be PDF = Probability density function
PDF: If X is continuous random variable.
P (x K*[x^4]/4 = 1 [Note that [x^4]/4 is integral of x^3]
==> K*[3^4 – 0^4]/4 = 1
==> K = 4/81
The value of P (1
Next Topic :
Linearity of Expectation
The Article is contributed by Anil Saikrishna Devarasetty
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Solving Literal Equations (Transforming Formulas)
In this next example, we’ll actually solve for the variable (x) “in terms of” some other variables (meaning that we still solve for (x), but may not get a number) notice that we treat variables just like numbers!
This is called solving Literal Equations, since sometimes variables are called literals. These are also called Transforming Formulas, or solving Multivariable Equations.
Let’s use the formula that converts Celsius temperatures (the type of temperatures they use in Paris!) to Fahrenheit (the type of temperatures they use here in the United States) to get the formula that converts Fahrenheit to Celsius. elsius is the “metric system” way to figure temperature, and we use Fahrenheit. ee, this stuff can really be useful!
Now that we know what (C) is in terms of (F), we can plug in the 86 degrees for (F), and get (displaystyle C=left( <8632> ight) imes frac<5><9>=30) degrees back. So again, when it’s 86 degrees Fahrenheit in New York, it’s 30 degrees Celsius in Paris.
See – algebra is pretty cool (no pun intended)!
You’ll see later that these two equations are called inverses, since one is sort of the opposite of the other.
Note that when we have something like “(5+x)” in the numerator, it’s like it has parentheses around it.
Note that we “pushed” the 4 through (distributive property) to get rid of the parentheses. When we pushed it through to a fraction, we put the 4 in the numerator, since (displaystyle 4=frac<4><1>)).
Let’s do the literal equation problem from above again, this time using the multiplythrough fraction approach:
We have to make sure that we multiply the (4s) through every term (push it through).
Remember that we must constantly thrive to get the (x) on one side by itself.
It looks like this wasn’t too much easier than doing this the original way this is a really tough problem!
Learn these rules and practice, practice, practice!
Click on Submit (the arrow to the right of the problem) to solve this problem. You can also type in more problems, or click on the 3 dots in the upper right hand corner to drill down for example problems.
If you click on “Tap to view steps”, you will go to the Mathway site, where you can register for the full version (steps included) of the software. You can even get math worksheets.
You can also go to the Mathway site here, where you can register, or just use the software for free without the detailed solutions. There is even a Mathway App for your mobile device. Enjoy!