A3. Let X and Y be independent random variables with probability density functions fx(x) %3D e, >0 and 0 otherwise; fr(4) 2e-", y >0 and 0 otherwise. %3D (a) Derive the probability density function of Z, = min(X, Y); (b) Derive the probability density function of Zz = y.

Q: 2 3 (25-x2) 2

A: Given:The integral is,

Q: V(3278)3 + 6230 – 12e90i

A:

Q: (4) y=x +4 -5 -4 -3 6.

A: The given equation is: y=x+4

Q: 1/3+2/9

A: Given 1/3 + 2/9

Q: (1-2=) dr

A:

Q: 4n – 1

A:

Q: Find m2C and mZD. B. (3c - 17) (c+ 1) A

A:

Q: 5 1 1 + 2r dr

A:

Q: - (2x – 1)(4x² + 2x + 1)

A:

Q: 196. 2(n+1)

A:

Q: 1 7. 27 2.

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: b) S 9+x

A: ∫x39+x8dx

Q: 2(3+d-1)

A: Solve the given expression  given expression is 2(3+d-1)

Q: 7 Find mLGBF. (4x-4)* (6x + 4)*

A:

Q: -2(5a)

A: To evaluate-2(5a)

Q: 9. (x² – 2)3

A:

Q: Identify the coordinates of the point in polar form based upon the given conditions. Use pi for T. r…

A: We will evaluate the required value.

Q: (n + 4)3 =

A: Given query is to expand (n+4)3

Q: スメ3X5

A: 2×3×5  Means multiply numbers =2×3×5=30

Q: 1 + 2n < n° for all integers n² 2

A:

Q: 1) 6dx 3/3

A: ∫33361+x2dx=6∫33311+x2dx=6tan-1x333=6tan-13-tan-133=6tan-13-tan-113=6π3-π6=2π-π=π

Q: ydy (1+ у)*

A: We have to find the integral.

Q: (9, -7) (3, 1). Find 吕 ロ师仁 ?

A:

Q: ) 2x+4 16

A:

Q: いーx)dx

A:

Q: 3 5.) (x+ 7x-44) ÷ (x+ 9)

A:

Q: 2. (1 + 3/2x − 1)² = 3 + √3/2x − 1

A:

Q: 4 -3 4m n 16) 22 4m n

A:

Q: 3(m – 5) + m

A: 3(m-5)+m=0 3m-15+m=0 add both side 15 we get 2m=15

Q: -5 4 -2 -7 -8 2.

A:

Q: 6 1/6 + 6

A: