Let X be a random variable that has value 1 on "heads" (H) and 2 on "tails" (T), for the toss of a fair coin. Suppose we have a second coin that is biased with 'H' having a probability of ¾ and 'T' having a probability ¼. Let Y be a random variable that has value 1 on a 'H' with this biased coin, and value 2 on a 'T' with this biased coin. What is the covariance of X and Y? Hint: Compute the E(X) and E(Y). Also, you will need the joint probability distribution that was computed above. Then use those values in the formula for covariance.

answer: i think it is (1,-2)

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

now, we can solve for w

subtract both sides by 12

divide both sides by 2

and we get

so,

the smallest possible measurement of the width is 14 feet

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)

answer: (x+2) (x+3) are the binomials for your question : )