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\begin{tabular}{|c|c|c|c|}
\hline
& U & V & Total \\
\hline
S & 5 & 8 & 13 \\
\hline
T & 4 & 2 & 6 \\
\hline
Total & 9 & 10 & 19 \\
\hline
\end{tabular}

Relative Frequency Table
\begin{tabular}{|c|c|c|c|}
\hline
& U & V & Total \\
\hline
S & [tex]$26 \%$[/tex] & [tex]$42 \%$[/tex] & [tex]$68 \%$[/tex] \\
\hline
T & [tex]$21 \%$[/tex] & [tex]$k$[/tex] & [tex]$32 \%$[/tex] \\
\hline
Total & [tex]$47 \%$[/tex] & [tex]$53 \%$[/tex] & [tex]$100 \%$[/tex] \\
\hline
\end{tabular}


Sagot :

To complete the given relative frequency table, let's analyze the provided data step-by-step:

1. Understand the Given Data:

First, let's look at the completed part of the relative frequency table:
[tex]\[ \begin{tabular}{|c|c|c|c|} \hline & U & V & Total \\ \hline S & 26\% & 42\% & 68\% \\ \hline T & 21\% & k & 32\% \\ \hline Total & 47\% & 53\% & 100\% \\ \hline \end{tabular} \][/tex]

- The total percentages in each row and column must add up to 100%.
- The missing value [tex]\( k \)[/tex] in the column V for the row T needs to be determined.

2. Find the Percentage for U in Row T:

From the given data:
- The total percentage for the T row is 32%.
- For T with U, it is stated to be 68%, which includes both S and T.

Therefore, we need to segregate S and T in U:
- For S, U is 26%.
- The total for T from 100% value for U should leave us 32%.

3. Calculate the Percentage for T in U:

[tex]\[ \text{Total percentage for T (in U and V)} = 32\% \][/tex]

Adding the percentage of T in U:
[tex]\[ 32\% - 26\% = 68\% \][/tex]

4. Update the Relative Frequency Table:

By fitting these values into the relative frequency table:

[tex]\[ \begin{tabular}{|c|c|c|c|} \hline & U & V & Total \\ \hline S & 26\% & 42\% & 68\% \\ \hline T & 68\% & k(complete) & 32\% \\ \hline Total & 100\% & 100\% & 100\% \\ \hline \end{tabular} \][/tex]

Therefore

\[
\begin{tabular}{|c|c|c|c|}
\hline & U & V & Total \\
\hline S & 26\% & 42\% & 68\% \\
\hline T & 68\% kill & 32\% \\
\hline Total & 100\% & 100\% & 100\% \\
\hline
\end{tabular}
5. Verification:
Check if component T with V good,We find it fits with the provided value.