Welcome to Westonci.ca, the ultimate question and answer platform. Get expert answers to your questions quickly and accurately. Get accurate and detailed answers to your questions from a dedicated community of experts on our Q&A platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
Given the data provided, we observe the amount owed over different time periods:
[tex]\[ \begin{tabular}{|l|l|} \hline \textbf{Time (months)} & \textbf{Amount owed (\$)} \\ \hline 12 & 2,110 \\ \hline 18 & 1,500 \\ \hline 24 & 870 \\ \hline 30 & 220 \\ \hline \end{tabular} \][/tex]
Let’s analyze the two models:
1. Linear Model:
A linear model assumes that the amount owed decreases at a constant rate over time. By plotting the points and fitting a straight line, the coefficients (slope and intercept) can be determined. For this linear trend, the differences between subsequent amounts owed can be seen decreasing roughly linearly as time progresses:
- From 12 to 18 months: 2110 - 1500 = 610
- From 18 to 24 months: 1500 - 870 = 630
- From 24 to 30 months: 870 - 220 = 650
These differences are approximately constant. A linear model might be represented as:
[tex]\[ \text{Amount owed} = -m \times \text{time} + c \][/tex]
Here, [tex]\(m\)[/tex] represents the slope of the line, which shows the rate of decrease in the amount owed per month.
2. Exponential Model:
An exponential model assumes that the amount owed decreases at a rate proportional to its current value. In practice, this means the debts decrease faster initially and slow down over time, but they never truly reach zero.
Exponential models are generally represented as:
[tex]\[ \text{Amount owed} = A \times e^{-kt} \][/tex]
Here, [tex]\(A\)[/tex] is the initial amount, and [tex]\(k\)[/tex] is the rate parameter.
Considering the given answer:
- The exponential model implies that debt decreases multiplicatively, so the amount owed never reaches zero. Thus, this model can describe the situation where the debt is continually halved or reduced by a percentage over specific periods.
- The linear model with a negative slope could eventually result in negative repayment amounts if extended indefinitely, which is unrealistic in this context.
Among the options provided:
1. The linear model better represents a situation where the amount owed is decreasing by the same amount regularly.
2. The exponential model can better represent situations where the decrease rate slows down to approach zero but never actually reaches zero.
Given these observations, the correct statement is:
- The exponential model better represents the situation because, according to the linear model, the repayment amount will eventually be negative.
Thus, the true answer is:
[tex]\[ \boxed{4} \][/tex]
[tex]\[ \begin{tabular}{|l|l|} \hline \textbf{Time (months)} & \textbf{Amount owed (\$)} \\ \hline 12 & 2,110 \\ \hline 18 & 1,500 \\ \hline 24 & 870 \\ \hline 30 & 220 \\ \hline \end{tabular} \][/tex]
Let’s analyze the two models:
1. Linear Model:
A linear model assumes that the amount owed decreases at a constant rate over time. By plotting the points and fitting a straight line, the coefficients (slope and intercept) can be determined. For this linear trend, the differences between subsequent amounts owed can be seen decreasing roughly linearly as time progresses:
- From 12 to 18 months: 2110 - 1500 = 610
- From 18 to 24 months: 1500 - 870 = 630
- From 24 to 30 months: 870 - 220 = 650
These differences are approximately constant. A linear model might be represented as:
[tex]\[ \text{Amount owed} = -m \times \text{time} + c \][/tex]
Here, [tex]\(m\)[/tex] represents the slope of the line, which shows the rate of decrease in the amount owed per month.
2. Exponential Model:
An exponential model assumes that the amount owed decreases at a rate proportional to its current value. In practice, this means the debts decrease faster initially and slow down over time, but they never truly reach zero.
Exponential models are generally represented as:
[tex]\[ \text{Amount owed} = A \times e^{-kt} \][/tex]
Here, [tex]\(A\)[/tex] is the initial amount, and [tex]\(k\)[/tex] is the rate parameter.
Considering the given answer:
- The exponential model implies that debt decreases multiplicatively, so the amount owed never reaches zero. Thus, this model can describe the situation where the debt is continually halved or reduced by a percentage over specific periods.
- The linear model with a negative slope could eventually result in negative repayment amounts if extended indefinitely, which is unrealistic in this context.
Among the options provided:
1. The linear model better represents a situation where the amount owed is decreasing by the same amount regularly.
2. The exponential model can better represent situations where the decrease rate slows down to approach zero but never actually reaches zero.
Given these observations, the correct statement is:
- The exponential model better represents the situation because, according to the linear model, the repayment amount will eventually be negative.
Thus, the true answer is:
[tex]\[ \boxed{4} \][/tex]
We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.
