The media equation is a theory developed by two professors of communication, Byron Reeves and Clifford Nass, at Stanford University. The theory is simple. They state that people treat the media as if they were real, hence the equation: media = real life. Basically Reeves and Nass are saying that people on an unconscious level perceive the media as real. People view objects of the media are talking to them personally. Reeves and Nass view things such as computers, televisions, radios, and other media's as inanimate objects. They don't believe that these objects are about to get up and move as if they were alive, but that the objects that relay the media are reacted to as though they were alive.

"Reeves and Nass credit the slow pace of evolution as the reason that the human race responds socially and naturally to the media: "The Human brain evolved in a world in which only humans exhibited rich social behaviors, and a world in which all perceived objects were real physical objects. Anything that seemed to be a real person or place was real." So we haven't yet adapted to the twentieth century media that only depict images, but which themselves personify the characteristics of human actors." (Griffin, pages 375-376)

To prove their theory Reeves and Nass held experiments. One is an experiment that they did with television. They gathered a group of students to participate. "The goal of the study was to show that responses to television content could be changed when the television sets were assigned particular roles." (Reeves and Nass, page 122). They took two groups of students and gave them specific tasks. The first group was to watch two separate televisions, called specialist TVs. One TV was identified as "News" and the other one was identified as "Entertainment". For each TV the participants wee in different chairs. The other group was to watch one TV with both news and entertainment, called generalist TVs. The TV was labeled "News and...

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My Mathematical Equations
By Kathleen Rossi
MAT 222 Intermediate Algebra
Instructor: Mohamed Elseifieen
May 12, 2013
My Mathematical Equations
This paper will show two mathematical problems, the first is “To estimate the size of the bear population on the Keweenaw Peninsula, conservationists captured, tagged, and released 50 bears. One year later, a random sample of 100 bears included only 2 tagged bears. What is the conservationist's estimate of the size of the bear population?” (Dugopolski, 2013, pp. 437, probem 56). The second will be to complete problem 10 on page 444 of Elementary and Intermediate Algebra. Here all steps in solving the problem will be explained step by step.
The first problem is to estimate the size of the bear population located on the Keweenaw Peninsula conservation. In reading over the “Bear Population” method #56 on page 437you will notice we are to assume that the ratio of originally tagged bears to the whole population is equal to the ratio of recaptured bears to the size of the sample.
The ratio of the originally tagged bears to the whole population is 2100
The ration of the recaptured tagged bears to the sample size is 50x
2100=50x Since x is on the right-hand side of the equation, we need to switch the sides so it is on the left-hand side.
50x=2100 This is the proportion set up and ready to solve. I will cross multiply setting the extremes equal to the means.
100(50) = 2x Here 100...

...communication research includes media institutions and processes such as diffusion of information, and media effects such as persuasion or manipulation of public opinion. In the United States, for instance, several university departments were remodeled into schools or colleges of mass communication or "journalism and mass communication". In addition to studying practical skills of journalism, public relations or advertising, they offer programs on "mass communication" or "mass communication research." The latter is often the title given to doctoral studies in such schools, whether the focus of the student's research is journalism practice, history, law or media effects. Departmental structures within such colleges may separate research and instruction in professional or technical aspects of mass communication. With the increased role of the Internet in delivering news and information, mass communication studies and media organizations tend to focus on the convergence of publishing, broadcasting and digital communication. The academic mass communication discipline historically differs from media studies and communication studies programs with roots in departments of theatre, film or speech, and with more interest in "qualitative," interpretive theory, critical or cultural approaches to communication study. In contrast, many mass communication programs historically lean toward empirical analysis and...

...Quadratic Equation:
Quadratic equations have many applications in the arts and sciences, business, economics, medicine and engineering. Quadratic Equation is a second-order polynomial equation in a single variable x.
A general quadratic equation is:
ax2 + bx + c = 0,
Where,
x is an unknown variable
a, b, and c are constants (Not equal to zero)
Special Forms:
* x² = n if n < 0, then x has no real value
* x² = n if n > 0, then x = ± n
* ax² + bx = 0 x = 0, x = -b/a
WAYS TO SOLVE QUADRATIC EQUATION
The ways through which quadratic equation can be solved are:
* Factorizing
* Completing the square
* Derivation of the quadratic formula
* Graphing for real roots
Quadratic Formula:
Completing the square can be used to derive a general formula for solving quadratic equations, the quadratic formula. The quadratic formula is in these two forms separately:
Steps to derive the quadratic formula:
All Quadratic Equations have the general form, aX² + bX + c = 0
The steps to derive quadratic formula are as follows:
Quadratic equations and functions are very important in business mathematics. Questions related to quadratic equations and functions cover a wide range of business concepts that includes COST-REVENUE, BREAKEVEN ANALYSIS, SUPPLY/DEMAND & MARKET EQUILIBRIUM....

...Quadratic equation
In elementary algebra, a quadratic equation (from the Latin quadratus for "square") is any equation having the form
where x represents an unknown, and a, b, and c represent known numbers such that a is not equal to 0. If a = 0, then the equation is linear, not quadratic. The numbers a, b, and c are the coefficients of the equation, and may be distinguished by calling them, the quadratic coefficient, the linear coefficient and the constant or free term.
Solving the quadratic equation
A quadratic equation with real or complex coefficients has two solutions, called roots. These two solutions may or may not be distinct, and they may or may not be real.
Factoring by inspection
It may be possible to express a quadratic equation ax2 + bx + c = 0 as a product (px + q)(rx + s) = 0. In some cases, it is possible, by simple inspection, to determine values of p, q, r, and s that make the two forms equivalent to one another. If the quadratic equation is written in the second form, then the "Zero Factor Property" states that the quadratic equation is satisfied if px + q = 0 or rx + s = 0. Solving these two linear equations provides the roots of the quadratic.
Completing the square
The process of completing the square makes use of the algebraic identity...

...There are now two separate equations: 60 = 6b - 6c and 60 = 3b + 3c
Solve both equations for b: b = 10 + c b = 10 - c
Now make both equations equal each other and solve for c: 10 + c = 10 - c 2c = 0 c = 0
The speed of the current was 0 mph Now, plug the numbers into one of either the original equations to find the speed of the boat in still water.
I chose the first equation: b = 10 + c or b = 10 + 0 b = 10
The speed of the boat in still water must remain a consistent 10 mph or more in order for Wayne and his daughter to make it home in time or dinner.
My Solution: c = current of river b = rate of boat d = s(t) will represent (distance = speed X time) Upstream: 60 = 6(b-c)
Downstream: 60 = 3(b+c)
There are now two separate equations: 60 = 6b - 6c and 60 = 3b + 3c
Solve both equations for b: b = 10 + c b = 10 - c
Now make both equations equal each other and solve for c: 10 + c = 10 - c 2c = 0 c = 0
The speed of the current was 0 mph Now, plug the numbers into one of either the original equations to find the speed of the boat in still water.
I chose the first equation: b = 10 + c or b = 10 + 0 b = 10
The speed of the boat in still water must remain a consistent 10 mph or more in order for Wayne and his daughter to make it home in time or dinner.
My Solution: c = current of river b = rate of boat d = s(t) will...

...
The short story Cold Equations by Tom Godwin takes place on a ship called EDS. The space cruiser is piloted by a man named Barton. He has an order of killing the stowaway who snuck onto the ship because the weight on the EDS is too much for the ship to handle. In the process of hunting down the stowaway, he realizes it was a young innocent girl named Marilyn. Once Barton understands what kind of person Marilyn is, he doesn’t kill her immediately because he knows her reasons were pure. Marilyn only wanted to see her brother, Gerry, again after ten years of being apart and was ignorant to the fact that her life can end with the decision of sneaking onto the ship. Barton begins to feel compassion after being with her and tries to comfort her, but knows what her fate is. He lets Marilyn live long enough to let her speak with Gerry once more before he follows through with the command. After Gerry and Marilyn speak he ejects her out into space. The ending was logical and no other endings would be possible because one the equation that was calibrated delicately, and two Barton could not throw the out the fever serums because that is the main reason for going on the trip to Woden.
A theoretical ending of Cold Equations could have been that Barton sacrifices himself for Marilyn, but since she is lighter than him, the fragile calibrated equation would be disrupted due to the change in weight. On EDS everything on ship is...

...Summer 2010-3 CLASS NOTES CHAPTER 1
Section 1.1: Linear Equations
Learning Objectives:
1. Solve a linear equation
2. Solve equations that lead to linear equations
3. Solve applied problems involving linear equations
Examples:
1. [pic]
[pic]
3. A total of $51,000 is to be invested, some in bonds and some in certificates of deposit (CDs). If the amount invested in bonds is to exceed that in CDs by $3,000, how much will be invested in each type of investment?
4. Shannon, who is paid time-and-a-half for hours worked in excess of 40 hours, had gross weekly wages of $608 for 56 hours worked. What is her regular hourly wage?
Answers: 1. [pic]
2. [pic]
3. $24,000 in CDs, $27,000 in bonds 4. $9.50/hour
Section 1.2: Quadratic Equations
Learning Objectives:
1. Solve a quadratic equation by (a) factoring, (b) completing the square, (c) the
quadratic formula
2. Solve applied problems involving quadratic equations
Examples:
1. Find the real solutions by factoring: [pic]
2. Find the real solutions by using the square root method: [pic]
3. Find the real solutions by completing the square: [pic]
4. Find the real solutions by using the quadratic formula: [pic]
5. A ball is thrown vertically upward from the top of a...

...Balancing Equations
Balancing equations is a fundamental skill in Chemistry. Solving a system of linear equations is a fundamental skill in Algebra. Remarkably, these two field specialties are intrinsically and inherently linked.
2 + O2 ----> H2OA. This is not a difficult task and can easily be accomplished using some basic problem solving skills. In fact, what follows is a chemistry text's explanation of the situation:
Taken from: Chemistry
Wilberham, Staley, Simpson, Matta
Addison Wesley
1. Determine the correct formulas for all the reactants and products in the reaction.
2. Write the formulas for the reactants on the left and the formulas for the products on the right with an arrow in between. If two or more reactants or products are involved, separate their formulas with plus signs.
3. Count the number of atoms of each element in the reactants a products. A polyatomic ion appearing unchanged on both sides of the equation is counted as a single unit.
4. Balance the elements one at a time by using coefficients. A is a small whole number that appears in front of a formula a an equation. When no coefficient is written, it is assumed to be 1. It is best to begin with an element other than hydrogen or oxygen. These two elements often occur more than twice in an equation.
5. Check each atom or polyatomic ion to be sure that the equation is balanced....