Two components of a minicomputer have the following joint pdf for their useful lifetimes X and Y: f(x, y) = xe−x(1 + y) x ≥ 0 and y ≥ 0 0 otherwise (a) What is the probability that the lifetime X of the first component exceeds 5? (Round your answer to three decimal places.)

Step-by-step explanation:

Given that,

f(x, y) = xe^−x(1 + y) x ≥ 0 and y ≥ 0

It is f(x, y)= 0 otherwise

To find probability of life time, we will take the double integral of the function with respect to X and Y

Y ranges from 0 to ∞

X exceed 5, so it ranges from 5 to ∞

∫ ∫ f(x, y) dxdy

∫ ∫ xe^−x(1 + y) dxdy

∫ ∫ x•e^(−x -xy) dxdy

Separating the exponential

∫ ∫ x•e^(−x) • e^(-xy) dxdy

Integrating with respect to y and keeping x as a constant.

∫ x•e^(−x) • e^(-xy) / -x dx

∫ - e^(−x) • e^(-xy) dx y = 0 to y=∞

∫ - e^(−x) • [e^(-∞) - e^(0) ]dx

∫ - e^(−x) • [ 0 - 1] dx

∫ - e^(−x) • -1 dx

∫ e^(−x) dx

Now integrating this

e^(-x) /-1 from x=5 to x=∞

-e^(-x) from x=5 to x=∞

- [ e^(-∞) - e^(-5)]

- (0-e^(-5))

- - e^(-5)

e^(-5) = 0.006738

To 3d.p =0.007

Then probability that the life time X exceed 5 is 0.007

an = 22 + (n-1) *6

110 seats

step-by-step explanation:

an=a1+(n-1)(d)

a1 = 22   this is the first item in the sequence

d =6   this is how the sequence goes up

a)   an = 22 + (n-1) *6

b) a12 = 22 + (12-1) *6

    a12 = 22+11*6

  a12 = 22 +88

  a12 = 110

cluster sampling.

step-by-step explanation: