Q1: The equilibrium data of component A in the two phases B and C are given below.

The estimate of Y for X = 4 by fitting a quadratic expression of a form Y = mX² for the above data is

ANS:C - 16.5

To estimate the value of 𝑌Y for 𝑋=4X=4 using a quadratic expression of the form 𝑌=𝑚𝑋2Y=mX2, we need to fit the given equilibrium data into this quadratic equation and solve for the coefficient 𝑚m. Given that 𝑋=4X=4, let's denote the equilibrium values of 𝑌Y for phases B and C as 𝑌𝐵YB​ and 𝑌𝐶YC​ respectively. The equilibrium data is not provided, so let's assume hypothetical values for illustration purposes: Y_B &= 4 & Y_B &= 9 & Y_B &= 16 & Y_B &= 25 & Y_B &= 36 \\ Y_C &= 10 & Y_C &= 20 & Y_C &= 32 & Y_C &= 48 & Y_C &= 66 \\ \end{align*} \] Now, let's calculate the value of \( m \) using the quadratic expression \( Y = mX^2 \). We'll choose two points (X, Y) from the provided data and solve for \( m \) using these points. We'll choose points where \( X = 3 \) and \( X = 5 \), as these are closer to \( X = 4 \). Using the point \( (X = 3, Y = 9) \): \[ Y = mX^2 \] \[ 9 = m \times 3^2 \] \[ m = \frac{9}{9} = 1 \] Using the point \( (X = 5, Y = 32) \): \[ Y = mX^2 \] \[ 32 = m \times 5^2 \] \[ m = \frac{32}{25} = 1.28 \] Now, we'll take the average of these two values of \( m \): \[ m = \frac{1 + 1.28}{2} = 1.14 \] Finally, we'll substitute \( X = 4 \) into the equation \( Y = mX^2 \) and solve for \( Y \): \[ Y = 1.14 \times 4^2 = 1.14 \times 16 = 18.24 \] Therefore, the estimate of \( Y \) for \( X = 4 \) using the quadratic expression is approximately \( \boxed{18.24} \).