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Select the correct answer.

\begin{tabular}{|r|c|}
\hline[tex]$x$[/tex] & [tex]$y$[/tex] \\
\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}

The table lists the values for two parameters, [tex]$x$[/tex] and [tex]$y$[/tex], of an experiment. What is the approximate value of [tex]$y$[/tex] for [tex]$x=4$[/tex]?

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


Sagot :

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].
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