Negative Slope and Negative Correlation

Negative Slope and Negative Correlation In mathematics, the slope of a line (m) describes how rapidly or slowly change is occurring and in which direction, whether positive or negative. Linear functions- those whose graph is a straight line- have four possible types of slope: positive, negative, zero, and undefined.  A function with a positive slope is represented by a line that goes up from left to right, while a function with a negative slope is represented by a line that goes down from left to right. A function with zero slope is represented by a horizontal line, and a function with an undefined slope is represented by a vertical line. Slope is usually expressed as an absolute value. A positive value indicates a positive slope, while a negative value indicates a negative slope. In the function y 3x, for example, the slope is positive 3, the coefficient of x. In statistics, a graph with a negative slope represents a negative correlation between two variables. This means that as one variable increases, the other decreases- and vice versa. Negative correlation represents a significant relationship between the variables x and y, which, depending on what they are modeling, can be understood as input and output, or cause and effect. How to Find Slope Negative slope is calculated just like any other type of slope. You can find it by dividing the rise of two points (the difference along the vertical or y-axis) by the run (the difference along the x-axis). Just remember that the rise is really a fall, so the resulting number will be negative. m (y2 - y1) / (x2 - x1) Once the line is graphed, youll see that the slope is negative because the line will go down from left to right. Even without drawing a graph, you will be able to see that the slope is negative simply by calculating m using the values given for the two points. For example, the slope of a line that contains the two points (2,-1) and (1,1) is: m [1 - (-1)] / (1 - 2)m (1 1) / -1m 2 / -1m -2 A slope of -2 means that for every positive change in x, there will be twice as much negative change in y. Negative Slope = Negative Correlation A negative slope demonstrates a negative correlation between the following: variables x and yinput and outputindependent variable and dependent variablecause and effect Negative correlation occurs when the two variables of a function move in opposite directions. As the value of x increases, the value of y decreases. Likewise, as the value of x decreases, the value of y increases. Negative correlation, then, indicates a clear relationship between the variables, meaning one affects the other in a meaningful way. In a scientific experiment, a negative correlation would show that an increase in the independent variable (the one manipulated by the researcher) would cause a decrease in the dependent variable (the one measured by the researcher). For example, a scientist might find that as predators are introduced into an environment, the number of prey gets smaller. In other words, there is a negative correlation between number of predators and number of prey. Reals A simple example of negative slope in the real world is going down a hill. The further you travel, the further down you drop. This can be represented as a mathematical function where x equals distance traveled and y equals elevation. Other examples of negative slope demonstrate the relationship between two variables: Mr. Nguyen drinks caffeinated coffee two hours before his bedtime. The more cups of coffee he drinks (input), the fewer hours he will sleep (output).Aisha is purchasing a plane ticket. The fewer days between the purchase date and the departure date (input), the more money Aisha will have to spend on airfare (output).John is spending some of the money from his last paycheck on presents for his children. The more money John spends (input), the less money John will have in his bank account (output).Mike has an exam at the end of the week. Unfortunately, he would rather spend his time watching sports on TV than studying for the test. The more time Mike spends watching TV (input), the lower Mikes score will be on the exam (output). (In contrast, the relationship between time spent studying and exam score would be represented by a positive correlation, since an increase in studying would lead to a higher score.)

Browns Chemistry The Central Science,15.8 Exercise 1

Brown's Chemistry The Central Science,15.8 Exercise 1 SAT / ACT Prep Online Guides and Tips This posts contains aTeaching Explanation. You can buyChemistry: The Central Sciencehere. Why You Should Trust Me:I’m Dr. Fred Zhang, and I have a bachelor’s degree in math from Harvard. I’ve racked up hundreds and hundreds of hours of experienceworking withstudents from 5thgradethroughgraduate school, and I’m passionate about teaching. I’ve read the whole chapter of the text beforehand and spent a good amount of time thinking about what the best explanation is and what sort of solutions I would have wanted to see in the problem sets I assigned myself when I taught. Exercise: 15.8 Practice Exercise 1 Question: †¦ When 9.2g of frozen $N_2O_4$ is added to a .50L reaction vessel †¦ [What is the value of $K_c$] Part 1: Approaching the Problem The question is asking for an equilibrium constant ($K_c$). We want to know$K_c$. Generally, we can know the equilibrium constant ONLY IF we can figure out the equilibrium concentrations of the species (nitrous oxide and dinitrogen tetraoxide): $$K_c = [NO_2]^2/[N_2O_4]$$ Thus, the entire game to figuring out the equilibrium constant here is to figure out the equilibrium concentrations. We are already given that in equilibrium, the concentration of $[N_2O_4]$=.057 molar. So we have half the puzzle: $$K_c = [NO_2]^2/.057$$ The other half of the puzzle if figuring out the equilibrium concentration $[NO_2]$. Sadly, the question doesn’tjust give us this. But we have a piece of information nearly as good, which is the starting (initial) amount of$[N_2O_4]$. Because we know the reaction equation, thekey now is to go from initial amount of$[N_2O_4]$ to the final (equilibrium) concentration $[NO_2]$. Part 2: Converting Grams to Molar We are given that the reaction started out with 9.2g of $N_2O_4$ in a 0.50L reaction vessel. For equilibrium calculations, we generally want to know concentrations of types molecules, instead of actual mass or volume. We apply stoichiometry here and convert grams per liter to molarity using molar mass. We use the periodic table to look up the molar mass of$N_2O_4$ is 92.01 grams per mole. We get that: $$(9.2g N_2O_4)/(0.50L) *(1 mol)/(92.01 g N_2O_4) = (0.100mol)/L = 0.200 molar$$ Thus the initial concentration of$N_2O_4$is 0.200 molar, and written as [$N_2O_4$]=.200 Part 3: Running the Reaction Now that we know the starting concentration, we want to get to final concentrations. The algebraic equation that links the two is the equation of reaction: $$N_2O_4 (g) ↔ 2 NO_2 (g)$$ This means that for every molecule of$N_2O_4$ we get two molecules of $NO_2$. As the reaction goes forward, when$N_2O_4$ decreases by $x$ molar,$NO_2$ increases by $2x$ molar. The concentration table is then: $N_2O_4 (g)$ $2 NO_2 (g)$ Initial Concentration (M) 0.200 0 Change in Concentration (M) -x +2x Equilibrium Concentration (M) 0.200-x 2x Part 4: Calculating the Equilibrium We are given that the equilibrium concentration of[$N_2O_4$]=.057 molar. The concentration table above gives the equilibrium concentration of[$N_2O_4$]=0.200-x, so we just equate the two and solve for x. 0.200-x = 0.057 x = .143 Now that we know x, 2x = .268 Or that in equilibrium, $[NO_2]=.268$ To calculate the equilibrium constant Kc, we plug in the information above: $$K_c = [NO_2]^2/[N_2O_4]=.268^2/.057= 1.43$$ Therefore, the right answer is d) 1.4 Video Solution Get full textbook solutions for just $5/month. PrepScholar Solutions has step-by-step solutions that teach you critical concepts and help you ace your tests. With 1000+ top texts for math, science, physics, engineering, economics, and more, we cover all popular courses in the country, including Stewart's Calculus. Try a 7-day free trial to check it out.