Using the definition of covariance, cov(x, y) = e[(x – my)(y – my)], prove the followings. a. cov(x, y) = e(xy) - mx hy b. cov(x, y) = cov(y, x) c. cov(x, x) = var(x) d. cov(x + z, y) = cov(x, y) + cov(z, y) e. cov(ex, y)= xcov(x, y) f. var(x + y) = var(x) + var(y) + 2cov(x, y) g. var(ex) = var(x) + ej cov(x, x; )

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step-by-step explanation:

region b will make the inequalities y < 2x + 1 and y > -x - 2 true ⇒ answer a

step-by-step explanation:

* lets study the graph and the inequality to solve the question

- the inequality y > -x - 2

∵ the form of the inequality is y > mx + c , m is the slope of the line

  and c is the y-intercept (the line intersect y-axis at point (0 , c)

∴ c = -2 that means the line intersect the y-axis at point (0 , -2)

∵ the sign of the inequality is >

∴ we shad the part over the line

  region a and b

- the inequality y < 2x + 1

∵ the form of the inequality is y > mx + c , m is the slope of the line

  and c is the y-intercept (the line intersect y-axis at point (0 , c)

∴ c = 1 that means the line intersect the y-axis at point (0 , 1)

∵ the sign of the inequality is <

∴ we shad the part under the line

  region b and c

∵ the common region between the two lines is region b

∴ region b will make the inequalities y < 2x + 1 and y > -x - 2 true