You have two kinds of data $ and %!
Align them properly and perform cross multiplication.
16% corresponds to $38 as
100% corresponds to $x
Following this, x = (38*100)/16 = $237.5, which is the original price before the discount.
The initial step is to analyze the graph's behavior.
It doesn’t follow a linear pattern, as the rate of change is inconsistent.
It also isn’t exponential since the change rate doesn't fluctuate significantly.
For this scenario, the most suitable regression model would be of the type:

Where,
a <0: This indicates that the parabola opens downwards.
Hence, the most appropriate regression model is quadratic.
Answer:
From visual analysis, the most fitting regression model for the data observed is:
A. Quadratic
The question lacks details. Below is the complete version provided.
Let M =
![\left[\begin{array}{cc}6&5\\-1&-4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%265%5C%5C-1%26-4%5Cend%7Barray%7D%5Cright%5D)
. Find

and

such that

, where

is the identity 2x2 matrix and 0 is the zero matrix of the appropriate dimension.
Answer: 

Step-by-step explanation: The identity matrix is a square matrix with 1's along its main diagonal and 0's elsewhere. For example, a 2x2 identity matrix is:
![\left[\begin{array}{cc}1&0\\0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%260%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D)
![M^{2} = \left[\begin{array}{cc}6&5\\-1&-4\end{array}\right]\left[\begin{array}{cc}6&5\\-1&-4\end{array}\right]](https://tex.z-dn.net/?f=M%5E%7B2%7D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%265%5C%5C-1%26-4%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%265%5C%5C-1%26-4%5Cend%7Barray%7D%5Cright%5D)
![M^{2}=\left[\begin{array}{cc}31&10\\-2&15\end{array}\right]](https://tex.z-dn.net/?f=M%5E%7B2%7D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D31%2610%5C%5C-2%2615%5Cend%7Barray%7D%5Cright%5D)
When solving the equation:
![\left[\begin{array}{cc}31&10\\-2&15\end{array}\right]+c_{1}\left[\begin{array}{cc}6&5\\-1&-4\end{array}\right] +c_{2}\left[\begin{array}{cc}1&0\\0&1\end{array}\right] =\left[\begin{array}{cc}0&0\\0&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D31%2610%5C%5C-2%2615%5Cend%7Barray%7D%5Cright%5D%2Bc_%7B1%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%265%5C%5C-1%26-4%5Cend%7Barray%7D%5Cright%5D%20%2Bc_%7B2%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%260%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0%260%5C%5C0%260%5Cend%7Barray%7D%5Cright%5D)
The multiplication of a matrix by a scalar results in each term being scaled by that scalar. Matrices of different sizes cannot be combined.
So, we structure the equation as follows:
![\left[\begin{array}{cc}31&10\\-2&15\end{array}\right]+\left[\begin{array}{cc}6c_{1}&5c_{1}\\-1c_{1}&-4c_{1}\end{array}\right] +\left[\begin{array}{cc}c_{2}&0\\0&c_{2}\end{array}\right] =\left[\begin{array}{cc}0&0\\0&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D31%2610%5C%5C-2%2615%5Cend%7Barray%7D%5Cright%5D%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6c_%7B1%7D%265c_%7B1%7D%5C%5C-1c_%7B1%7D%26-4c_%7B1%7D%5Cend%7Barray%7D%5Cright%5D%20%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dc_%7B2%7D%260%5C%5C0%26c_%7B2%7D%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0%260%5C%5C0%260%5Cend%7Barray%7D%5Cright%5D)
And the system of equations can be written as:

Various methods are available to solve this system. One way is to multiply the second equation by -1 and then add the equations together:



Following this, substitute variables back into one of the equations to find
:





For the equation,
and 