PLEASE DO NOT COPY THE PREVIOUS ANSWER FOR THIS QUESTION WHICH IS FOR 2.5 BAR AND AND GIVES ERROR. PLEASE SOLVE IT IN YOUR OWN. MAKE SURE THAT IS WORKING . THANK YOU.

Create the model by using least square method and find a 2nd degree equation. after that find the root when x_initial= 0 ( where interceps 0) by Newton Raphson method. then a code should be created to calculate predicted volume when pressure is 1.05 bar from the equation. Lagrange interpolation is applicated for 1.05 bar.

At last, absolute difference should be calculated between the volume calculated using interpolation and the volume calculated using the model.

here is the question. please solve it making sure it is true. thank. AND DO NOT COPY PASTE

Use the following set of pressure-volume data to develop a script m-file to find a₀ and a₁ in the following exponential model.

After the model development;

Calculate the root of developed model equation using Newton-Raphson method (x_initial=0). Then,

the code should automatically calculate the predicted volume @ 1.05 bar pressure using the regression model. Also, in the same script m-file this value must be calculated using Lagrange Interpolation and the absolute difference between two finding must be displayed as well.

Data

Pressure (bar)	Volume (m3)
0.644	32
0.985	25
1.108	22.2
1.363	18
1.631	15
1.934	12
2.356	9
3.1	7.2

Note: The script m-file should work itself . All inputs must be defined in the beginning of the code.

Warnings:

The results should be presented with fprintf command.

The following results should be displayed at the end of the code in this order (just these) a- the model, with a₀ and a₁ written on it

b- root of the model equation

c- result of the predicted volume at 1.05bar using regression model

d- interpolation result of volume at 1.05bar using lagrange method,

e- the absolute difference between the volume values calculated from model and interpolation.

When script executed above text with correct result should be displayed on command window: answers:

a) The Model: y=()+()*x+()*x^2

b) root value :

c) From model: V= when P=1.05bar

d) According to LAGRANGE Method

f interpolated value for P = 1.05 -> V=

e) absolute difference between calculated value and interpolation

difference: