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

abde or 1,2,4, 5 is the answer

step-by-step explanation:

80π/3 or about 83.775

step-by-step explanation:

since a cone is 1/3 of a cylinder, we need to find the base's area. since area = πr^2, we get 16π. then, multiplying this by 5, we get 80π. dividing by 3, since the cone's volume is 1/3 of that of a cylinder's, we get 80π/3.

hope this !