To find the approximate value of \( y \) for \( x = 4 \) using the given data points, we can use linear interpolation, an approach for estimating an intermediate value between two known values.
Given the data points:
[tex]\[
\begin{tabular}{|r|c|}
\hline
x & y \\
\hline
2.5 & 6.25 \\
\hline
9.4 & 88.36 \\
\hline
15.6 & 243.63 \\
\hline
19.5 & 380.25 \\
\hline
25.8 & 665.64 \\
\hline
\end{tabular}
\][/tex]
We want to find \( y \) when \( x = 4 \).
First, identify the interval where \( x = 4 \) lies, which is between \( x = 2.5 \) and \( x = 9.4 \).
The formula for linear interpolation between two points \((x_1, y_1)\) and \((x_2, y_2)\) is:
[tex]\[
y = y_1 + \frac{(x - x_1) \cdot (y_2 - y_1)}{(x_2 - x_1)}
\][/tex]
Here, \( (x_1, y_1) = (2.5, 6.25) \) and \( (x_2, y_2) = (9.4, 88.36) \).
Now, substitute the values:
[tex]\[
y = 6.25 + \frac{(4 - 2.5) \cdot (88.36 - 6.25)}{(9.4 - 2.5)}
\][/tex]
Calculating inside the fraction step-by-step:
[tex]\[
4 - 2.5 = 1.5
\][/tex]
[tex]\[
88.36 - 6.25 = 82.11
\][/tex]
[tex]\[
9.4 - 2.5 = 6.9
\][/tex]
Now, substitute and solve:
[tex]\[
y = 6.25 + \frac{1.5 \cdot 82.11}{6.9}
\][/tex]
Calculate the numerator:
[tex]\[
1.5 \cdot 82.11 = 123.165
\][/tex]
Now, divide by the denominator:
[tex]\[
\frac{123.165}{6.9} \approx 17.85
\][/tex]
Add to the initial \( y_1 \):
[tex]\[
6.25 + 17.85 = 24.10
\][/tex]
Therefore, the approximate value of [tex]\( y \)[/tex] for [tex]\( x = 4 \)[/tex] is [tex]\( \boxed{24} \)[/tex].
