Answer:
The viscosity at 140°C is predicted to be 7.2
Step-by-step explanation:
The function that model the relationship between viscosity and temperature = Quadratic model
The general form of a quadratic equation is y = a·x² + b·x + c
Therefore, we have;
When y = 10.8, x = 110, which gives;
10.8 = a·110² + b·110 + c = 12100·a + 110·b + c
10.8 = 12100·a + 110·b + c
When y =8.2, x = 130
8.2 = a· 130² + b· 130 + c = 16900·a + 130·b + c
8.2 = 16900·a + 130·b + c
When y = 160, x = 5.8
5.8 = a·160² + b·160 + c = 25600·a + 160·b + c
5.8 = 25600·a + 160·b + c
The three equations above can be listed as follows;
10.8 = 12100·a + 110·b + c
8.2 = 16900·a + 130·b + c
5.8 = 25600·a + 160·b + c
Solving using matrices gives;
![\begin{bmatrix}12100 & 110 & 1\\ 16900 & 130 & 1\\ 25600 & 160 & 1\end{bmatrix} \begin{bmatrix}a\\ b\\ c\end{bmatrix} = \begin{bmatrix}10.8\\ 8.2\\ 5.8\end{bmatrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bbmatrix%7D12100%20%26%20110%20%26%201%5C%5C%2016900%20%26%20130%20%26%201%5C%5C%2025600%20%26%20160%20%26%201%5Cend%7Bbmatrix%7D%20%5Cbegin%7Bbmatrix%7Da%5C%5C%20b%5C%5C%20c%5Cend%7Bbmatrix%7D%20%3D%20%5Cbegin%7Bbmatrix%7D10.8%5C%5C%208.2%5C%5C%205.8%5Cend%7Bbmatrix%7D)
![\begin{bmatrix}a\\ b\\ c\end{bmatrix} = -\dfrac{1}{3000}\begin{bmatrix}3 & -5 & 2\\ -867 & 1350 & -480\\ 62400 & -88000 & 28600\end{vmatrix} \begin{bmatrix}10.8\\ 8.2\\ 5.8\end{bmatrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bbmatrix%7Da%5C%5C%20b%5C%5C%20c%5Cend%7Bbmatrix%7D%20%3D%20-%5Cdfrac%7B1%7D%7B3000%7D%5Cbegin%7Bbmatrix%7D3%20%26%20-5%20%26%202%5C%5C%20-867%20%26%201350%20%26%20-480%5C%5C%2062400%20%26%20-88000%20%26%2028600%5Cend%7Bvmatrix%7D%20%5Cbegin%7Bbmatrix%7D10.8%5C%5C%208.2%5C%5C%205.8%5Cend%7Bbmatrix%7D)
From which we have;
a = 0.001, b = -0.37, c = 39.4
Substituting gives;
y = 0.001·x² - 0.37·x + 39.4
When x = 140
y = 0.962·140² - 0.37·140 + 39.4= 7.2
The viscosity at 140°C = 7.2.