A researcher analyzed the results of her data and calculated the standard deviation of scores at 3.5. what is the variance?
Solution: We are given the standard deviation of scores = 3.5
We know that the variance is the square of the standard deviation.
Therefore, the variance is:



Therefore, the variance of the scores is 12.25
F(X) = 6/X
F(X) = 6 • 1/X
F(X) = 6 • x^-1
F(X) = 6x^-1
F'(X) = 6 • d(x^-1)/dx
F'(X) = 6 • -1x^-1-1
F'(X) = 6 • -1x^-2
F'(X) = -6x^-2
F'(X) = -6/x^2
F'(-2) = -6/(-2)^2
F'(-2) = -6/4
F'(-2) = -3/2
The solution would be C. -3/2.
We are asked to determine the value of a such that the function f(x) = ax^2 + 5 is fit for the point given (-1,2). In this case, we substitute 2 to y and -1 to x. The result is then 2 = a*(-1)^2 + 5 ; 2 = a + 5; a is then equal to -3
Here you go enjoy
Have a good rest of your night