And 3y - 1 is also 0.5.

To see why this is the case, let's use the formula for the correlation coefficient:

r = (covariance of x and y) / (standard deviation of x * standard deviation of y)

We can rewrite this formula in terms of the variables 4x^2 and 3y-1:

r' = (covariance of 4x^2 and 3y-1) / (standard deviation of 4x^2 * standard deviation of 3y-1)

Now we just need to plug in some values. First, we need to find the covariance of 4x^2 and 3y-1:

cov(4x^2, 3y-1) = E[(4x^2 - E[4x^2]) * (3y-1 - E[3y-1])]

Since we don't know the means of 4x^2 and 3y-1, we'll assume they're both zero (this won't affect the answer, since we're only interested in the correlation coefficient).

cov(4x^2, 3y-1) = E[4x^2 * 3y-1]

= 12 * E[x^2 * y]

Now we need to use the fact that the correlation coefficient between x and y is 0.5:

r = cov(x,y) / (std(x) * std(y))

0.5 = cov(x,y) / (std(x) * std(y))

cov(x,y) = 0.5 * std(x) * std(y)

So we can rewrite the covariance of 4x^2 and 3y-1 as:

cov(4x^2, 3y-1) = 12 * E[x^2 * y]

= 12 * E[(x * sqrt(x))^2 * (y * sqrt(y))^2]

= 12 * E[(x * sqrt(x)) * (y * sqrt(y))]^2

= 12 * (sqrt(E[x^3 * y]) * sqrt(E[x * y^3]))

= 12 * (sqrt(cov(x^3, y^3)) + sqrt(E[x^4] * E[y^4]))

Again, we'll assume that the means of x^3, y^3, x^4, and y^4 are all zero.

cov(4x^2, 3y-1) = 12 * (sqrt(cov(x^3, y^3)) + sqrt(E[x^4] * E[y^4]))

= 12 * (sqrt(E[x^3 * y^3]) + sqrt(E[x^4] * E[y^4]))

Now we need to find the standard deviations of 4x^2 and 3y-1.

std(4x^2) = sqrt(E[(4x^2 - E[4x^2])^2])

= sqrt(16 * E[x^4])

= 4 * sqrt(E[x^4])

Similarly,

std(3y-1) = sqrt(E[(3y-1 - E[3y-1])^2])

= sqrt(9 * E[y^2])

=

2.2