The table lists the values for two parameters, [tex]\( x \)[/tex] and [tex]\( y \)[/tex], of an experiment.

[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
2.5 & 6.25 \\
\hline
9.4 & 88.36 \\
\hline
15.6 & 243.63 \\
\hline
19.5 & 380.25 \\
\hline
25.8 & 665.64 \\
\hline
\end{array}
\][/tex]

What is the approximate value of [tex]\( y \)[/tex] for [tex]\( x = 4 \)[/tex]?

A. 11
B. 16
C. 24
D. 43



Answer :

To estimate the value of [tex]\( y \)[/tex] for [tex]\( x = 4 \)[/tex] based on the given data, we can use polynomial regression. The data points provided are [tex]\( (x, y) \)[/tex]:

[tex]\[ (2.5, 6.25), (9.4, 88.36), (15.6, 243.63), (19.5, 380.25), (25.8, 665.64) \][/tex]

Here are the steps to determine the approximate value of [tex]\( y \)[/tex] at [tex]\( x = 4 \)[/tex]:

1. Fit a Quadratic Polynomial:
To fit a quadratic polynomial to the given data points, we determine the coefficients [tex]\((a, b, c)\)[/tex] of the quadratic equation [tex]\(y = ax^2 + bx + c\)[/tex].

2. Find the Polynomial Coefficients:
The coefficients obtained for the best-fit quadratic polynomial to these data points are:
[tex]\[ a = 0.99900978, \quad b = 0.02863292, \quad c = -0.08883057 \][/tex]

3. Construct the Polynomial Equation:
Using these coefficients, the quadratic polynomial equation becomes:
[tex]\[ y = 0.99900978x^2 + 0.02863292x - 0.08883057 \][/tex]

4. Calculate [tex]\( y \)[/tex] for [tex]\( x = 4 \)[/tex]:
Substitute [tex]\( x = 4 \)[/tex] into the polynomial equation:
[tex]\[ y = 0.99900978(4)^2 + 0.02863292(4) - 0.08883057 \][/tex]

5. Perform the Calculation:
[tex]\[ y = 0.99900978 \times 16 + 0.02863292 \times 4 - 0.08883057 \][/tex]
[tex]\[ y = 15.98415648 + 0.11453168 - 0.08883057 \][/tex]
[tex]\[ y \approx 16.0098576 \][/tex]

We find that the approximate value of [tex]\( y \)[/tex] for [tex]\( x = 4 \)[/tex] is:

[tex]\[ \boxed{16} \][/tex]