\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
No & \begin{tabular}{c}
concentración \\
{[tex]$\left[N O_2\right]\left(x_i\right)$[/tex]}
\end{tabular} & \begin{tabular}{c}
Absorción \\
[tex]$\left(y_i\right)$[/tex]
\end{tabular} & [tex]$\left(x_i-\bar{x}\right)$[/tex] & [tex]$\left(y_i-\bar{y}\right)$[/tex] & [tex]$\left(x_i-\bar{x}\right) \cdot\left(y_i-\bar{y}\right)$[/tex] & [tex]$\left(x_i-\bar{x}\right)^2$[/tex] \\
\hline
1 & 2 & 0.065 & & & & \\
\hline
2 & 6 & 0.205 & & & & \\
\hline
3 & 10 & 0.338 & & & & \\
\hline
4 & 14 & 0.474 & & & & \\
\hline
5 & 18 & 0.598 & & & & \\
\hline
Sumas & & & & & & \\
\hline
\end{tabular}



Answer :

To fill in the table and perform the necessary calculations, follow these steps:

1. Calculate the means of [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
[tex]\[ \bar{x} = 10.0 \quad \text{and} \quad \bar{y} = 0.336 \][/tex]

2. Compute [tex]\((x_i - \bar{x})\)[/tex] for each [tex]\(x_i\)[/tex]:
- For [tex]\(x_1 = 2\)[/tex]: [tex]\(x_1 - \bar{x} = 2 - 10 = -8.0\)[/tex]
- For [tex]\(x_2 = 6\)[/tex]: [tex]\(x_2 - \bar{x} = 6 - 10 = -4.0\)[/tex]
- For [tex]\(x_3 = 10\)[/tex]: [tex]\(x_3 - \bar{x} = 10 - 10 = 0.0\)[/tex]
- For [tex]\(x_4 = 14\)[/tex]: [tex]\(x_4 - \bar{x} = 14 - 10 = 4.0\)[/tex]
- For [tex]\(x_5 = 18\)[/tex]: [tex]\(x_5 - \bar{x} = 18 - 10 = 8.0\)[/tex]

3. Compute [tex]\((y_i - \bar{y})\)[/tex] for each [tex]\(y_i\)[/tex]:
- For [tex]\(y_1 = 0.065\)[/tex]: [tex]\(y_1 - \bar{y} = 0.065 - 0.336 = -0.271\)[/tex]
- For [tex]\(y_2 = 0.205\)[/tex]: [tex]\(y_2 - \bar{y} = 0.205 - 0.336 = -0.131\)[/tex]
- For [tex]\(y_3 = 0.338\)[/tex]: [tex]\(y_3 - \bar{y} = 0.338 - 0.336 = 0.002\)[/tex]
- For [tex]\(y_4 = 0.474\)[/tex]: [tex]\(y_4 - \bar{y} = 0.474 - 0.336 = 0.138\)[/tex]
- For [tex]\(y_5 = 0.598\)[/tex]: [tex]\(y_5 - \bar{y} = 0.598 - 0.336 = 0.262\)[/tex]

4. Compute [tex]\((x_i - \bar{x}) \cdot (y_i - \bar{y})\)[/tex] for each pair [tex]\((x_i, y_i)\)[/tex]:
- For [tex]\(i = 1\)[/tex]: [tex]\((-8.0) \cdot (-0.271) = 2.168\)[/tex]
- For [tex]\(i = 2\)[/tex]: [tex]\((-4.0) \cdot (-0.131) = 0.524\)[/tex]
- For [tex]\(i = 3\)[/tex]: [tex]\(0.0 \cdot 0.002 = 0.0\)[/tex]
- For [tex]\(i = 4\)[/tex]: [tex]\(4.0 \cdot 0.138 = 0.552\)[/tex]
- For [tex]\(i = 5\)[/tex]: [tex]\(8.0 \cdot 0.262 = 2.096\)[/tex]

5. Compute [tex]\((x_i - \bar{x})^2\)[/tex] for each [tex]\(x_i\)[/tex]:
- For [tex]\(i = 1\)[/tex]: [tex]\((-8.0)^2 = 64.0\)[/tex]
- For [tex]\(i = 2\)[/tex]: [tex]\((-4.0)^2 = 16.0\)[/tex]
- For [tex]\(i = 3\)[/tex]: [tex]\(0.0^2 = 0.0\)[/tex]
- For [tex]\(i = 4\)[/tex]: [tex]\(4.0^2 = 16.0\)[/tex]
- For [tex]\(i = 5\)[/tex]: [tex]\(8.0^2 = 64.0\)[/tex]

6. Calculate the sums:
- Sum of [tex]\((x_i - \bar{x}) \cdot (y_i - \bar{y})\)[/tex]: [tex]\(2.168 + 0.524 + 0.0 + 0.552 + 2.096 = 5.34\)[/tex]
- Sum of [tex]\((x_i - \bar{x})^2\)[/tex]: [tex]\(64.0 + 16.0 + 0.0 + 16.0 + 64.0 = 160.0\)[/tex]

Now we input these results into the table:

[tex]\[ \begin{tabular}{|c|c|c|c|c|c|c|} \hline \text{No} & \begin{tabular}{c} \text{concentración} \\ $\left[N O_2\right]\left(x_i\right)$ \end{tabular} & \begin{tabular}{c} \text{Absorción} \\ $\left(y_i\right)$ \end{tabular} & $\left(x_i-\bar{x}\right)$ & $\left(y_i-\bar{y}\right)$ & $\left(x_i-\bar{x}\right) \cdot\left(y_i-\bar{y}\right)$ & $\left(x_i-\bar{x}\right)^2$ \\ \hline 1 & 2 & 0.065 & -8.0 & -0.271 & 2.168 & 64.0 \\ \hline 2 & 6 & 0.205 & -4.0 & -0.131 & 0.524 & 16.0 \\ \hline 3 & 10 & 0.338 & 0.0 & 0.002 & 0.0 & 0.0 \\ \hline 4 & 14 & 0.474 & 4.0 & 0.138 & 0.552 & 16.0 \\ \hline 5 & 18 & 0.598 & 8.0 & 0.262 & 2.096 & 64.0 \\ \hline \text{Sumas} & & & & & 5.34 & 160.0 \\ \hline \end{tabular} \][/tex]

This completes the table with all the required values and sums.

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