Which table has a constant of proportionality between [tex]$y$[/tex] and [tex]$x$[/tex] of [tex]$\frac{3}{4}$[/tex]?

Choose one answer:

(A)
\begin{tabular}{|ll|}
\hline [tex]$x$[/tex] & [tex]$y$[/tex] \\
\hline 8 & 6 \\
9 & [tex]$\frac{27}{4}$[/tex] \\
10 & [tex]$\frac{15}{2}$[/tex] \\
\hline
\end{tabular}

(B)
\begin{tabular}{|cc|}
\hline [tex]$x$[/tex] & [tex]$y$[/tex] \\
\hline 3 & [tex]$\frac{3}{4}$[/tex] \\
4 & 1 \\
5 & [tex]$\frac{5}{4}$[/tex] \\
\hline
\end{tabular}

(C)
\begin{tabular}{|cc|}
\hline [tex]$x$[/tex] & [tex]$y$[/tex] \\
\hline 10 & [tex]$\frac{19}{2}$[/tex] \\
11 & [tex]$\frac{41}{4}$[/tex] \\
12 & 11 \\
\hline
\end{tabular}



Answer :

To determine which table has a constant of proportionality between [tex]\(y\)[/tex] and [tex]\(x\)[/tex] of [tex]\(\frac{3}{4}\)[/tex], we need to check if the ratio [tex]\(\frac{y}{x}\)[/tex] is consistently [tex]\(\frac{3}{4}\)[/tex] for all pairs [tex]\((x, y)\)[/tex] in the table. Let’s check each table step-by-step:

Table A:
[tex]\[ \begin{array}{|cc|} \hline x & y \\ \hline 8 & 6 \\ 9 & \frac{27}{4} \\ 10 & \frac{15}{2} \\ \hline \end{array} \][/tex]

1. For [tex]\( (8, 6) \)[/tex]:
[tex]\[ \frac{y}{x} = \frac{6}{8} = \frac{3}{4} \][/tex]

2. For [tex]\( (9, \frac{27}{4}) \)[/tex]:
[tex]\[ \frac{y}{x} = \frac{\frac{27}{4}}{9} = \frac{27}{4} \times \frac{1}{9} = \frac{27}{36} = \frac{3}{4} \][/tex]

3. For [tex]\( (10, \frac{15}{2}) \)[/tex]:
[tex]\[ \frac{y}{x} = \frac{\frac{15}{2}}{10} = \frac{15}{2} \times \frac{1}{10} = \frac{15}{20} = \frac{3}{4} \][/tex]

Since each [tex]\(\frac{y}{x}\)[/tex] value in Table A is [tex]\(\frac{3}{4}\)[/tex], Table A has a constant proportionality of [tex]\(\frac{3}{4}\)[/tex].

Table B:
[tex]\[ \begin{array}{|cc|} \hline x & y \\ \hline 3 & \frac{3}{4} \\ 4 & 1 \\ 5 & \frac{5}{4} \\ \hline \end{array} \][/tex]

1. For [tex]\( (3, \frac{3}{4}) \)[/tex]:
[tex]\[ \frac{y}{x} = \frac{\frac{3}{4}}{3} = \frac{3}{4} \times \frac{1}{3} = \frac{3}{12} = \frac{1}{4} \][/tex]

2. For [tex]\( (4, 1) \)[/tex]:
[tex]\[ \frac{y}{x} = \frac{1}{4} = \frac{1}{4} \][/tex]

3. For [tex]\( (5, \frac{5}{4}) \)[/tex]:
[tex]\[ \frac{y}{x} = \frac{\frac{5}{4}}{5} = \frac{5}{4} \times \frac{1}{5} = \frac{5}{20} = \frac{1}{4} \][/tex]

Since each [tex]\(\frac{y}{x}\)[/tex] value in Table B is [tex]\(\frac{1}{4}\)[/tex], Table B does not have a constant proportionality of [tex]\(\frac{3}{4}\)[/tex].

Table C:
[tex]\[ \begin{array}{|cc|} \hline x & y \\ \hline 10 & \frac{19}{2} \\ 11 & \frac{41}{4} \\ 12 & 11 \\ \hline \end{array} \][/tex]

1. For [tex]\( (10, \frac{19}{2}) \)[/tex]:
[tex]\[ \frac{y}{x} = \frac{\frac{19}{2}}{10} = \frac{19}{2} \times \frac{1}{10} = \frac{19}{20} \][/tex]

2. For [tex]\( (11, \frac{41}{4}) \)[/tex]:
[tex]\[ \frac{y}{x} = \frac{\frac{41}{4}}{11} = \frac{41}{4} \times \frac{1}{11} = \frac{41}{44} \][/tex]

3. For [tex]\( (12, 11) \)[/tex]:
[tex]\[ \frac{y}{x} = \frac{11}{12} = \frac{11}{12} \][/tex]

Since each [tex]\(\frac{y}{x}\)[/tex] value in Table C is different and not equal to [tex]\(\frac{3}{4}\)[/tex], Table C does not have a constant proportionality of [tex]\(\frac{3}{4}\)[/tex].

Given this analysis, the table that has a constant of proportionality between [tex]\(y\)[/tex] and [tex]\(x\)[/tex] of [tex]\(\frac{3}{4}\)[/tex] is:

[tex]\[ \boxed{A} \][/tex]