Given the existence of the first three moments of the cycle duration Y of an ordinary renewal process, verify the formulas: 12 + o? E(S) = %3D 43 E(S?) = Зд where o2 = V (Y) |3D and 13 = E(Y)

Q: (-14) + (–7)

A:

Q: (42x² – 33) ÷ (7.x + 7)

A: We have to solve                           42x2-33÷7x+7

Q: 1.25a +3> 0.5a-6 and 2.5a 129-1.5a

A:

Q: xp xペ-4xt13

A:

Q: +–3m+–9m+4+5+–10m

A: The expression is given as 1+–3m+–9m+4+5+–10m. Simplifying the given expression. 1+–3m+–9m+4+5+–10m…

Q: 18) 2x y

A:

Q: Evaluate. (3- 5)² + 4(1 – 5)

A: Given, (3-5)2 + 4(1-5)

Q: Find the LU factorization of A = A = 1 1 -3 0 9 3 -17 5 2

A:

Q: 2070 = (16a + 10)15

A: 2070 = (16a+10)15 And (16e+11)17 =3723 To find  the values of a and e

Q: 36-9z

A: Given that:36-9z

Q: 1 + 2n < n° for all integers n² 2

A:

Q: Evalvate S (4sin co) -8cos co) )do

A: Given that: ∫0π(4sinθ-8cosθ)dθ

Q: –4(–6m – 7).

A:

Q: Z4×Z6 has an element of order 24. true or false ?

A:

Q: -96 6 -7 A = -3 d 4A.

A:

Q: (2n)! 323. 2n

A: