F nine executives in a business firm, four are married, three have never married, and two are divorced. Three of the executives are to be selected for promotion. Let Y1 denote the number of married executives and Y2 denote the number of never-married executives among the three selected for promotion. Assume that the three are randomly selected from the nine available. We determined that the joint probability distribution of Y1, the number of married executives, and Y2, the number of never-married executives, is given by p(y1, y2) = 4 y1 3 y2 2 3 − y1 − y2 9 3 where y1 and y2 are integers, 0 ≤ y1 ≤ 3, 0 ≤ y2 ≤ 3, and 1 ≤ y1 + y2 ≤ 3. (a) Find the marginal probability distribution of Y1, the number of married executives among the three selected for promotion. (Enter your probabilities as fractions.)

mr. flanders’ class sold 47 doughnuts and   mr. rodriquez’s class sold 38 cartons of milk

step-by-step explanation:

let x be the number of doughnuts sold by mr. flanders’ class and y be the number of cartons sold by mr. rodriquez’s class.

1. together, the classes sold 85 items, then

x+y=85.

2. both classes earn earned $104.49 for their school. if doughnuts were sold for $1.35 each, then x doughnuts costed $1.35x. if cartons of milk were sold for $1.08 each, then y cartons costed $1.08y. hence,

1.35x+1.08y=104.49.

3. solve the system of two equations:

false

step-by-step explanation:

with btw u said 100pts but it clearly says

i answerd and just got 10pts

no hate

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