To estimate the value of [tex]\(x\)[/tex] for [tex]\(y = 0.049\)[/tex], we first need to identify the interval in the given table where the [tex]\(y\)[/tex] value falls between the values provided.
Looking at the table:
[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
2.5 & 0.400 \\
\hline
9.4 & 0.106 \\
\hline
15.6 & 0.064 \\
\hline
19.5 & 0.051 \\
\hline
25.8 & 0.038 \\
\hline
\end{array}
\][/tex]
We see that [tex]\(0.049\)[/tex] lies between [tex]\(0.051\)[/tex] (at [tex]\(x = 19.5\)[/tex]) and [tex]\(0.038\)[/tex] (at [tex]\(x = 25.8\)[/tex]).
Next, we use linear interpolation to estimate the value of [tex]\(x\)[/tex] for [tex]\(y = 0.049\)[/tex].
The linear interpolation formula is:
[tex]\[
x = x_1 + \frac{(x_2 - x_1) \cdot (y - y_1)}{(y_2 - y_1)}
\][/tex]
For our specific case:
- [tex]\(x_1 = 19.5\)[/tex]
- [tex]\(x_2 = 25.8\)[/tex]
- [tex]\(y_1 = 0.051\)[/tex]
- [tex]\(y_2 = 0.038\)[/tex]
- [tex]\(y = 0.049\)[/tex]
Plugging in the values, we get:
[tex]\[
x = 19.5 + \frac{(25.8 - 19.5) \cdot (0.049 - 0.051)}{(0.038 - 0.051)}
\][/tex]
Simplifying the expression inside the fraction:
[tex]\[
x = 19.5 + \frac{6.3 \cdot (-0.002)}{-0.013}
\][/tex]
[tex]\[
x = 19.5 + \frac{-0.0126}{-0.013}
\][/tex]
[tex]\[
x = 19.5 + 0.9692307692307692
\][/tex]
[tex]\[
x \approx 20.469230769230766
\][/tex]
Rounding this to the nearest tenth, we get:
[tex]\[
x \approx 20.4
\][/tex]
Therefore, the correct answer is:
A. 20.4
