If the correlation coefficient is positive then the slope must also be positive. Is this statement T/F and why
450656
The statement is true.  The formula for the slope of the regression line is\n\nr(SDy)/SDx\n\nSince SDy and SDx are always positive, then the only thing that determines the sign of the slope is r, the correlation coefficient.  So if r is positive, then the slope is positive; if r is negative, then the slope is negative.
914585
f(x) satisfies the mean value theorem because it is continuous and differentiable on [-2, 1].\nSecant slope = (1 - 4) / (1 + 2) = -1\nDerivative f'(x) = 2x\n2x = -1\nx=-1/2 (this is what c equals to)
746423
If from left to right gets bigger or more positive, and from down to up gets bigger or more positive regarding numeric values, then it makes sense to label the quadrants in that manner.  You start with all positive values in the first quadrant.  Then y values are positive and x values are negative in the second.  Then all values are negative in the third.  Last, y values are negative and x values are positive.\nThe assumption is that x is on the horizontal axis and y is on the vertical axis.
820527
criterion is that changing sign of x changes the sign of y\n\nso a & b are symmetric, and other's aren't, since even powers and adding a constant both ruin the symmetry
668722
If the graph of Volume vs Pressure is a diagonal line with a positive slope, you should be able to conclude that the graph of Pressure vs Volume is also a diagonal line with a positive slope!\n\nSince we know that the gas has expanded, the work done by the gas is positive. Hence, W= corresponding area under Pressure vs Volume graph. If you make a sketch of the graph and indicate the intial and final pressures and volumes of the gas, finding W should not be too much of a problem!
1126609
Actually, no.\n\nIf dy/dx = 2, then I'd say yes, in moving from one coordinate pair to another, the y-value rises twice as fast as the x-value.  (That would be the case for a line with a slope of 2.)\n\nBut when dy/dx = 2x, as it does with a curve like y = x^2 + C, the ratio (of movement in the y-direction to movement in the x-direction) isn't constant -- it varies depending on where you are on the curve.  At the point (1, 2), for instance, that "slope," a.k.a. the slope of the tangent line at that point, is 2 -- the ratio of y-movement to x-movement is 2:1.  But at the point (3, 9), the slope of the tangent line is 6.\n\nHope that helps clear it up.  The important thing is that the slope of the tangent line is changing depending on where you are on the curve.
925132
TimGNO is correct, although a pedant would point out that he hasn't shown this point is a maximum.  But  it is common sense.  See my response to the second time you asked the question, I got the same answer (foolishly mucking around with tangents), but make a couple of other observations you may wish to look at.
339574
If both the dependent variable (Y) and the independent variable (X) are standardized, then both their means are zero and standard deviations are one.  The correlation coefficient is not effected by a change in scale, meaning that the correlation doesn't change if you add or subtract the same number from each X or Y or if you multiply X or Y by the same positive number.  An example of this is if the correlation between height and weight when height is in inches and weight is in pounds is 0.6, then the correlation if height is in centimeters and weight is in kilograms is also 0.6.  We would hope that the relationship between the two variables doesn't change if there is a change in how the variables were measured.\n\nThis is the case when standardizing, since standardizing has you subtract the mean from each value and then divide by the standard deviation.  So the correlation coefficient for the standardized values of X and Y are the same as the correlation coefficient for the original values of X and Y.\n\nNow, the formula for the slope coefficient is r(sy)/(sx), where r is the correlation coefficient and sx and sy are the standard deviations of X and Y respectively.  Since both X and Y are standardized, then both of their standard deviations are equal to one, so putting that back into the expression above, the slope coefficient is equal to r.\n\nTo summarize, if X and Y are standardized, then the slope coefficient is the same as the correlation coefficient.
662293
You might be looking for Regression analysis.\nThis is the statistical technique used to look for correlations between variables.
748261
if we assume that the following reaction occured\n\nA(l)----->A(g)\n\n[that means no A is not a polymer in liquid phase, the only thing occured  is the evaporation],\n\ntry plot ln(P) against 1/T, most likely you will have a straight line and the slope is - delta H/R, please notice the sign.
856807
Yes that will be correct because after you isolate the (y) in other words put it in slope inercect form it would be \ny=2x+4 \nm=2\nb=4
139195
A correlation in statistics represents a relationship between two variables.  A positive correlation indicates that the two variables move together, and the higher the correlation the more related they are.  A negative correlation shows that the two variables move in opposite directions. (For example, if you are looking at the correlation between the number of days it was sunny and the number of umbrellas sold, that would be an inverse correlation because if there are lots of sunny days, very few people will want umbrellas.)  The more negative the correlation, the stronger the relationship between two variables.  Correlation is on a scale from -1 to 1.  The closer to -1 or 1, the stronger the relationship.  A correlation of 0 indicates no relationship.  (So, if your correlation coefficicent is -1, the variables are very inversely related, but if it is just -.02 they are not very related at all)
109708
if i am not wrong..presence of one component( i forget the name)..makes its positive...n negative..no need to worry..cheers
507403
Behind all the nomenclature, this is just a simple minimization problem. You want to find local minima for A(q). Thus, take the derivative of A(q) to look for critical points (places where the graph is locally flat). These will be places where the tangent line has slope 0; so set the derivative equal to 0 and solve for q. You will get two possible choices. Use the second derivative to see what the behavior of the first derivative is at each point (increasing or decreasing). An increasing first derivative around a critical point (where the derivative is zero) implies that the slope locally changed from negative values to positive values. Visualize this to see that it is a minimum point for the graph of A(q). The q that gives you the positive second derivative and has 0 first derivative is thus q of the Wilson lot size formula.\nFor b, simply replace k with k + bq in A(q) and do a similar analysis as above.\n\nPS. The answerer above has forgotten the negative value of q that also satisfies the first equation. Although a negative q does not correspond to the problem, it is unrigorous to say that there is only one answer to the equation. It should at least be explained as being unphysical.
961818
If x and y have a strong negative linear correlation, r is close to -1.  An r value of exactly -1 indicates a perfect negative fit.  Negative values indicate a relationship between x and y such that as values for x increase, values for y decrease. \n\nSo I would say that x and y have a strong negative linear correlation.  \n\nTaking r^2 = 0.6724 = 67.24% of the total variation in y can be explained by the linear relationship between x and y.  The other 32.76% remains unexplained.
592597
It's actually D.  A isn't correct because we aren't given the covariance between the two securities.  The portfolio risk is not the simple weighted average of the individual risks. \n\nThe formula is\n\nsquare root of the sum of the weights multiplied with the covariances.\n\nIf the securities are negatively correlated, the total risk can be less than either of the individual securities.  If they're positively correlated, the portfolio risk goes up.
866301
because y- x > 0\n\n|y-x | = y- x (positive value)\n\nx-y < 0\n\nso | x-y| = y-x (as absolute value positive)