You have a ruler of length 2 and you choose a place to break it using a uniform probability distribution. let random variable x represent the length of the left piece of the ruler. x is distributed uniformly in [0,2] . you take the left piece of the ruler and once again choose a place to break it using a uniform probability distribution. let random variable y be the length of the left piece from the second break. draw a picture of the region in the x-y plane for which the joint density of x and y is non-zero.

Hi!

When you break the rule for the first time, de probability distribution of X is constant in [0, 2], and is zero outside. If the the value obtained is X = a, then the probability distribution of Y (conditioned on X = a) is constaint in interval [0, a], and zero outside. Then the region in the X-Y plane where the joint density is non-zero is the filled triangle  in the figure.

step-by-step explanation:

u = -(6/7)w

step-by-step explanation:

place all values with 'u' on one side of the equation.

4u + 8w = - 3u + 2w

4u + 3u = 2w - 8w

further simplify the equation by solving the addition and subtraction.

7u = - 6w

make 'u' the independent variable by placing its co-efficient on the right-hand side.

u = - (6/7)w