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Determine if the table shows a proportional relationship.

\begin{tabular}{|l|l|l|l|}
\hline
[tex]$x$[/tex] & 25.6 & 72.8 & 77.2 \\
\hline
[tex]$y$[/tex] & 6.4 & 18.2 & 19.3 \\
\hline
\end{tabular}

A. Yes, it is proportional because all [tex]$\frac{y}{x}$[/tex] ratios are equivalent to [tex]$\frac{1}{4}$[/tex].

B. Yes, it is proportional because all [tex]$\frac{y}{x}$[/tex] ratios are equivalent to [tex]$\frac{1}{3}$[/tex].

C. No, it is not proportional because [tex]$\frac{25.6}{6.4} \neq \frac{18.2}{72.8}$[/tex].

D. No, it is not proportional because [tex]$\frac{18.2}{72.8} \neq \frac{72.2}{19.3}$[/tex].



Answer :

GinnyAnswer
Let's determine if the table shows a proportional relationship by examining the ratios [tex]\(\frac{y}{x}\)[/tex] for each pair of values.

The table lists the following values:

| [tex]\(x\)[/tex] | 25.6 | 72.8 | 77.2 |
|------|------|------|------|
| [tex]\(y\)[/tex] | 6.4 | 18.2 | 19.3 |

To check if [tex]\(y\)[/tex] is proportional to [tex]\(x\)[/tex], we need to compute the ratios [tex]\(\frac{y}{x}\)[/tex] for each pair and see if they are all equivalent.

1. For the first pair ([tex]\(x = 25.6\)[/tex] and [tex]\(y = 6.4\)[/tex]):
[tex]\[ \frac{y}{x} = \frac{6.4}{25.6} = 0.25 \][/tex]

2. For the second pair ([tex]\(x = 72.8\)[/tex] and [tex]\(y = 18.2\)[/tex]):
[tex]\[ \frac{y}{x} = \frac{18.2}{72.8} = 0.25 \][/tex]

3. For the third pair ([tex]\(x = 77.2\)[/tex] and [tex]\(y = 19.3\)[/tex]):
[tex]\[ \frac{y}{x} = \frac{19.3}{77.2} = 0.25 \][/tex]

All the computed ratios [tex]\(\frac{y}{x}\)[/tex] are equal to [tex]\(0.25\)[/tex]. This means that the [tex]\(y\)[/tex]-values are consistently [tex]\(0.25\)[/tex] times their corresponding [tex]\(x\)[/tex]-values, indicating a consistent proportional relationship between [tex]\(x\)[/tex] and [tex]\(y\)[/tex].

Therefore, we can conclude that:
Yes, it is proportional because all [tex]\(\frac{y}{x}\)[/tex] ratios are equivalent to [tex]\(0.25\)[/tex].

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