Abag contains 5 yellow, 6 red, and 4 green marbles. two marbles are drawn. the first is replaced before the second is drawn. a random variable assigns the number of green marbles to each outcome. calculate the expected value of the random variable.

The expected value of the random variable : 0, 1 and 2

Step-by-step explanation:

Number of yellow balls in the bag = 5

Number of red balls in the bag = 6

Number of green balls in the bag = 4

No. of balls needed to be drawn from the bag = 2

A random variable assigns the number of green marbles to each outcome.

So, Expected value of random variable is given by :

No green ball in two draws, so random variable is assigned 0

One green ball in two draws, so random variable is assigned 1

Two green ball in two draws, so random variable is assigned 2

Hence, The expected value of the random variable : 0, 1 and 2

a_n = a_1 + (n - 1)d

the first term is 6.

the common difference d = ?

when you find d, replace n with 7 and d with the common difference to find your answer.

delaney would like to make a 5lb nut mixture that is 60% peanuts and 40% almonds.

she has several pounds of peanuts and several pounds of a mixture that is 20% peanuts and 80% almonds.

let 'p' represent the number of pounds of peanuts needed to make the new mixture, and  

let 'm' represent the number of pounds of the 80% almond-20% peanut mixture.

:

p + m = 5

therefore

m = (5-p)

write an equation based on the percent of peanuts

1p + .2m = .6(5)

replace m with (5-p)

p + .2(5-p) = .6(5)

p + 1 - .2p = 3

p - .2p = 3 - 1

.8p = 2

p = 2/.8

p = 2.5 lbs of peanuts required

then

5 - 2.5 = 2.5 lbs of the peanut/almond mixture required

:

a) what is the system that models this situation?

p + m = 5

1p + .2m = .6(5)