Suppose, X|0~ Binom(m,0)and 0~ Beta(a,ß), then Var (Y) is equals to, тав a) (a+B)°(a+ ß+1) maß(a+B+m) (a+ B)* maß(a+B+m) (a+B)´(a+B+1) тав (а + В+т) (&+ B)(g+ß+1) b) c) d)

Show that (z3 z4) -1 = (z3-1. z4-1)

Find A and B are independent, positively associated or negatively associated, in each of the following cases : (i) N= 1000, (A)= 470, (B) = 620 and(AB)= 320 (ii) (A)= 490, (AB)= 294,:(a)= 570 and (aB)= 380 (iii) (AB)= 250, (aB)= 768, (AB)= 48 and (aß)= 144.

(V4z-x2 dydxdz -4 (2vz Evaluate * SvS

Let Use the Gram-Schmidt procedure to produce an orthogonal set with the same span. (Hint: If you have unlimited submissions, it might be useful to submit to check your answer for each vector u, before continuing.) Note: The u, must be given in the same order as provided by the standard procedure. U₁ = (U₁, V₂) = U₂== (u₁, Vs): and vs = =

Why the 1st row and 1st column of the JACOBIAN MATRX is always Zero ?

Which of the following is the general solution to the equation?

Which of the following is the general solution of the equation?

Students at certain university have to pass of one these three courses (MAT102, STA105, and PHY110). The pass rates for the courses MAT 102, STA105, and PHY110 are 0.54, 0.79, and 0.86, respectively. Suppose that 30 candidates take the MAT102 course, 25 take the STA105 course and 20 take the PHY110 coruse. A student takes one of these courses and passes. What are the probabilities that she took MAT102, STA105 or PHY110? Hint: Denote by Q the event a student to pass their course. Now denote by R1, R2 and R3 the events to take courses MAT102, STA105, and PHY110. The pass reates then represent the conditional probabilites: P(Q| R1), P(Q| R2) and P(Q| R3). The number of students taking the courses can be used to estimate the probabilities for R1, R2 and R3. For example, since there are total of 30 + 25 + 20 = 75 students that take the courses, for R1 we have: P(R1) = 30/75 = 2/5 What you are asked is: given that a student passes (event Q) estimate the probabilities that she took MAT102…

A. G(3, -4); H(8, 1)   B. G(6, -8); H(16, 8)   C. G(8, -6); H(8, 16)   D. G(4, -3); H(1, 8)