Question
The table below shows the amount that must be repaid, y, when x dollars are borrowed.
Amount Borrowed Amount Repaid
$100 $135
$200 $260
$500 $635
$800 $1,010
(1 point)
Responses
y=45x
y is equal to 4 fifths x
y=54x+10
y is equal to 5 fourths x plus 10
y=45x+55
y is equal to 4 fifths x plus 55
y=54x
y is equal to 5 fourths x
Amount Borrowed Amount Repaid
$100 $135
$200 $260
$500 $635
$800 $1,010
(1 point)
Responses
y=45x
y is equal to 4 fifths x
y=54x+10
y is equal to 5 fourths x plus 10
y=45x+55
y is equal to 4 fifths x plus 55
y=54x
y is equal to 5 fourths x
Answers
GPT-4o mini
To find the correct relationship between the amount borrowed \( x \) and the amount repaid \( y \), we can analyze the data provided in the table.
We can first calculate the total amount repaid for each borrowed amount and then find a linear equation that best fits the data.
The given data is:
\[
\begin{align*}
&\text{Amount Borrowed (x)} & \text{Amount Repaid (y)} \\
100 & 135 \\
200 & 260 \\
500 & 635 \\
800 & 1010 \\
\end{align*}
\]
1. Calculate the difference between \( y \) and \( x \):
- For \( x = 100, y = 135 \): \( y - x = 135 - 100 = 35 \)
- For \( x = 200, y = 260 \): \( y - x = 260 - 200 = 60 \)
- For \( x = 500, y = 635 \): \( y - x = 635 - 500 = 135 \)
- For \( x = 800, y = 1010 \): \( y - x = 1010 - 800 = 210 \)
We can see that the amount repaid is greater than the amount borrowed by a varying amount.
2. Next, let's look for a pattern:
- For every increment in amounts that are multiples of $100, we can also compute the ratio of \( y \) to \( x \):
- For \( x = 100, y = 135 \): \( \frac{y}{x} = \frac{135}{100} = 1.35 \)
- For \( x = 200, y = 260 \): \( \frac{y}{x} = \frac{260}{200} = 1.3 \)
- For \( x = 500, y = 635 \): \( \frac{y}{x} = \frac{635}{500} = 1.27 \)
- For \( x = 800, y = 1010 \): \( \frac{y}{x} = \frac{1010}{800} = 1.2625 \)
These ratios suggest that the relation is not a simple \( y = kx \) for a constant \( k \).
3. A more fitting equation appears to be linear in form. To express \( y \) in terms of \( x \), we may try a function of the form \( y = mx + b \) where \( m \) and \( b \) are constants.
Given that the increase in repaid amount seems to be more than the borrowed amount directly, we can test the formulas offered:
- \( y = 1.35x \) from the first data point suggests \( y = \frac{135}{100}x \).
- Testing \( y = \frac{5}{4}x \) which is \( y = 1.25x \) does not fit as it isn't high enough.
To find the correct constants \( m \) and \( b \), we can examine:
From points \( (100, 135) \) and \( (200, 260) \), we can find slope:
Calculating it:
- Slope \( m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{260 - 135}{200 - 100} = \frac{125}{100} = 1.25 \) implies a consistent relationship.
Thus \( y = 1.35x - 10 \) with \( 135 = 100(1.35) + b \) verifies because of the constant change.
### Conclusion:
Considering the provided options in the question, none of them perfectly fit this derived relationship. Given choices generally imply a multiplier and constant addition. If \( y = 1.35x \) or close to multipliers as \( y \approx 1.25x\), mystery surrounding shard data can lead to estimates.
Final Selection: While not a perfect signature, **"y=54x"** may be best akin \( y \approx \frac{5}{4}x \) reflecting higher correlation directly observing raw \( 4^{\text{th}} \) related checks based on proximity reports.
Thus, best fitting responses depend more on defining relationships, analysis should stick with closest numeric ranges.
We can first calculate the total amount repaid for each borrowed amount and then find a linear equation that best fits the data.
The given data is:
\[
\begin{align*}
&\text{Amount Borrowed (x)} & \text{Amount Repaid (y)} \\
100 & 135 \\
200 & 260 \\
500 & 635 \\
800 & 1010 \\
\end{align*}
\]
1. Calculate the difference between \( y \) and \( x \):
- For \( x = 100, y = 135 \): \( y - x = 135 - 100 = 35 \)
- For \( x = 200, y = 260 \): \( y - x = 260 - 200 = 60 \)
- For \( x = 500, y = 635 \): \( y - x = 635 - 500 = 135 \)
- For \( x = 800, y = 1010 \): \( y - x = 1010 - 800 = 210 \)
We can see that the amount repaid is greater than the amount borrowed by a varying amount.
2. Next, let's look for a pattern:
- For every increment in amounts that are multiples of $100, we can also compute the ratio of \( y \) to \( x \):
- For \( x = 100, y = 135 \): \( \frac{y}{x} = \frac{135}{100} = 1.35 \)
- For \( x = 200, y = 260 \): \( \frac{y}{x} = \frac{260}{200} = 1.3 \)
- For \( x = 500, y = 635 \): \( \frac{y}{x} = \frac{635}{500} = 1.27 \)
- For \( x = 800, y = 1010 \): \( \frac{y}{x} = \frac{1010}{800} = 1.2625 \)
These ratios suggest that the relation is not a simple \( y = kx \) for a constant \( k \).
3. A more fitting equation appears to be linear in form. To express \( y \) in terms of \( x \), we may try a function of the form \( y = mx + b \) where \( m \) and \( b \) are constants.
Given that the increase in repaid amount seems to be more than the borrowed amount directly, we can test the formulas offered:
- \( y = 1.35x \) from the first data point suggests \( y = \frac{135}{100}x \).
- Testing \( y = \frac{5}{4}x \) which is \( y = 1.25x \) does not fit as it isn't high enough.
To find the correct constants \( m \) and \( b \), we can examine:
From points \( (100, 135) \) and \( (200, 260) \), we can find slope:
Calculating it:
- Slope \( m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{260 - 135}{200 - 100} = \frac{125}{100} = 1.25 \) implies a consistent relationship.
Thus \( y = 1.35x - 10 \) with \( 135 = 100(1.35) + b \) verifies because of the constant change.
### Conclusion:
Considering the provided options in the question, none of them perfectly fit this derived relationship. Given choices generally imply a multiplier and constant addition. If \( y = 1.35x \) or close to multipliers as \( y \approx 1.25x\), mystery surrounding shard data can lead to estimates.
Final Selection: While not a perfect signature, **"y=54x"** may be best akin \( y \approx \frac{5}{4}x \) reflecting higher correlation directly observing raw \( 4^{\text{th}} \) related checks based on proximity reports.
Thus, best fitting responses depend more on defining relationships, analysis should stick with closest numeric ranges.