Answer :
To determine which equation correctly models the relationship between \( y \) and \( x \) based on the given table of values, we need to evaluate each provided equation using the table's \( x \) values and check if the computed \( y \) values match the table's \( y \) values.
Here is the table of values:
[tex]\[ \begin{array}{cc} x & y \\ \hline 0 & 125 \\ 1 & 75 \\ 2 & 45 \\ 3 & 27 \\ \end{array} \][/tex]
Let's evaluate each equation:
Option (A): \( y = -50x + 125 \)
- For \( x = 0 \): \( y = -50(0) + 125 = 125 \), which matches the table's \( y \) value.
- For \( x = 1 \): \( y = -50(1) + 125 = 75 \), which matches the table's \( y \) value.
- For \( x = 2 \): \( y = -50(2) + 125 = 25 \), which does NOT match the table's \( y \) value of 45.
- For \( x = 3 \): \( y = -50(3) + 125 = -25 \), which does NOT match the table's \( y \) value of 27.
Since the equation does not hold true for all the \( x \) values, Option (A) is incorrect.
Option (B): \( y = -30x + 105 \)
- For \( x = 0 \): \( y = -30(0) + 105 = 105 \), which does NOT match the table's \( y \) value of 125.
- For \( x = 1 \): \( y = -30(1) + 105 = 75 \), which matches the table's \( y \) value.
- For \( x = 2 \): \( y = -30(2) + 105 = 45 \), which matches the table's \( y \) value.
- For \( x = 3 \): \( y = -30(3) + 105 = 15 \), which does NOT match the table's \( y \) value of 27.
Since the equation does not hold true for all the \( x \) values, Option (B) is incorrect.
Option (C): \( y = 75 \cdot \left(\frac{3}{5}\right)^x \)
- For \( x = 0 \): \( y = 75 \cdot \left(\frac{3}{5}\right)^0 = 75 \cdot 1 = 75 \), which does NOT match the table's \( y \) value of 125.
- For \( x = 1 \): \( y = 75 \cdot \left(\frac{3}{5}\right)^1 = 75 \cdot \frac{3}{5} = 45 \), which does NOT match the table's \( y \) value of 75.
- For \( x = 2 \): \( y = 75 \cdot \left(\frac{3}{5}\right)^2 = 75 \cdot \left(\frac{3}{5}\right)^2 = 27 \), which does NOT match the table's \( y \) value of 45.
- For \( x = 3 \): \( y = 75 \cdot \left(\frac{3}{5}\right)^3 = 75 \cdot \left(\frac{3}{5}\right)^3 = 16.2 \), which does NOT match the table's \( y \) value of 27.
Since the equation does not hold true for all the \( x \) values, Option (C) is incorrect.
Option (D): \( y = 125 \cdot \left(\frac{3}{5}\right)^x \)
- For \( x = 0 \): \( y = 125 \cdot \left(\frac{3}{5}\right)^0 = 125 \cdot 1 = 125 \), which matches the table's \( y \) value.
- For \( x = 1 \): \( y = 125 \cdot \left(\frac{3}{5}\right)^1 = 125 \cdot \frac{3}{5} = 75 \), which matches the table's \( y \) value.
- For \( x = 2 \): \( y = 125 \cdot \left(\frac{3}{5}\right)^2 = 125 \cdot \left(\frac{3}{5}\right)^2 = 45 \), which matches the table's \( y \) value.
- For \( x = 3 \): \( y = 125 \cdot \left(\frac{3}{5}\right)^3 = 125 \cdot \left(\frac{3}{5}\right)^3 = 27 \), which matches the table's \( y \) value.
Since this equation holds true for all the \( x \) values, Option (D) is the correct answer.
So, the correct equation that relates \( y \) to \( x \) for the values in the table is:
[tex]\[ \boxed{y = 125 \cdot \left(\frac{3}{5}\right)^x} \][/tex]
Here is the table of values:
[tex]\[ \begin{array}{cc} x & y \\ \hline 0 & 125 \\ 1 & 75 \\ 2 & 45 \\ 3 & 27 \\ \end{array} \][/tex]
Let's evaluate each equation:
Option (A): \( y = -50x + 125 \)
- For \( x = 0 \): \( y = -50(0) + 125 = 125 \), which matches the table's \( y \) value.
- For \( x = 1 \): \( y = -50(1) + 125 = 75 \), which matches the table's \( y \) value.
- For \( x = 2 \): \( y = -50(2) + 125 = 25 \), which does NOT match the table's \( y \) value of 45.
- For \( x = 3 \): \( y = -50(3) + 125 = -25 \), which does NOT match the table's \( y \) value of 27.
Since the equation does not hold true for all the \( x \) values, Option (A) is incorrect.
Option (B): \( y = -30x + 105 \)
- For \( x = 0 \): \( y = -30(0) + 105 = 105 \), which does NOT match the table's \( y \) value of 125.
- For \( x = 1 \): \( y = -30(1) + 105 = 75 \), which matches the table's \( y \) value.
- For \( x = 2 \): \( y = -30(2) + 105 = 45 \), which matches the table's \( y \) value.
- For \( x = 3 \): \( y = -30(3) + 105 = 15 \), which does NOT match the table's \( y \) value of 27.
Since the equation does not hold true for all the \( x \) values, Option (B) is incorrect.
Option (C): \( y = 75 \cdot \left(\frac{3}{5}\right)^x \)
- For \( x = 0 \): \( y = 75 \cdot \left(\frac{3}{5}\right)^0 = 75 \cdot 1 = 75 \), which does NOT match the table's \( y \) value of 125.
- For \( x = 1 \): \( y = 75 \cdot \left(\frac{3}{5}\right)^1 = 75 \cdot \frac{3}{5} = 45 \), which does NOT match the table's \( y \) value of 75.
- For \( x = 2 \): \( y = 75 \cdot \left(\frac{3}{5}\right)^2 = 75 \cdot \left(\frac{3}{5}\right)^2 = 27 \), which does NOT match the table's \( y \) value of 45.
- For \( x = 3 \): \( y = 75 \cdot \left(\frac{3}{5}\right)^3 = 75 \cdot \left(\frac{3}{5}\right)^3 = 16.2 \), which does NOT match the table's \( y \) value of 27.
Since the equation does not hold true for all the \( x \) values, Option (C) is incorrect.
Option (D): \( y = 125 \cdot \left(\frac{3}{5}\right)^x \)
- For \( x = 0 \): \( y = 125 \cdot \left(\frac{3}{5}\right)^0 = 125 \cdot 1 = 125 \), which matches the table's \( y \) value.
- For \( x = 1 \): \( y = 125 \cdot \left(\frac{3}{5}\right)^1 = 125 \cdot \frac{3}{5} = 75 \), which matches the table's \( y \) value.
- For \( x = 2 \): \( y = 125 \cdot \left(\frac{3}{5}\right)^2 = 125 \cdot \left(\frac{3}{5}\right)^2 = 45 \), which matches the table's \( y \) value.
- For \( x = 3 \): \( y = 125 \cdot \left(\frac{3}{5}\right)^3 = 125 \cdot \left(\frac{3}{5}\right)^3 = 27 \), which matches the table's \( y \) value.
Since this equation holds true for all the \( x \) values, Option (D) is the correct answer.
So, the correct equation that relates \( y \) to \( x \) for the values in the table is:
[tex]\[ \boxed{y = 125 \cdot \left(\frac{3}{5}\right)^x} \][/tex]
