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To determine the correct function that models the change between the number of days, [tex]\( t \)[/tex], and the number of orders remaining, let's analyze the situation and each given function.
### Problem Analysis
The store starts with 2,000 orders. Each day, the store owner completes one-third of the remaining orders from the previous day. That means two-thirds of the orders are completed each day, leaving one-third of the orders still remaining.
### Function Modeling
1. Initial Orders: On day 0, the number of orders is 2,000.
- [tex]\( f(0) = 2000 \)[/tex]
2. Day 1 Calculation:
- At the end of day 1, only one-third of the 2,000 orders should remain.
- Therefore, [tex]\( f(1) = 2000 \cdot \left(\frac{1}{3}\right) = 666.67 \)[/tex]
3. Day 2 Calculation:
- At the end of day 2, only one-third of the orders remaining at the end of day 1 will still be unfulfilled.
- Therefore, [tex]\( f(2) = 2000 \cdot \left(\frac{1}{3}\right)^2 \approx 222.22 \)[/tex]
This pattern continues, indicating that the number of orders remaining can be expressed by a geometric progression:
[tex]\[ f(t) = 2000 \cdot \left(\frac{1}{3}\right)^t \][/tex]
### Evaluating Function Choices
Given the following options:
- [tex]\( f(t) = 2000 \cdot \left(\frac{1}{3}\right)^t \)[/tex]
- [tex]\( f(t) = -2000 \cdot \left(\frac{1}{3}\right)^t \)[/tex]
- [tex]\( r(t) = 2000 \cdot 3^t \)[/tex]
- [tex]\( f(t) = -2000 \cdot 3t \)[/tex]
We can derive the correct option by methodically rejecting the incorrect ones:
1. [tex]\( f(t) = 2000 \cdot \left(\frac{1}{3}\right)^t \)[/tex]
- This correctly follows the geometric progression where each day one-third of the previous day's orders remain.
2. [tex]\( f(t) = -2000 \cdot \left(\frac{1}{3}\right)^t \)[/tex]
- A negative number of orders does not make sense in this context.
3. [tex]\( r(t) = 2000 \cdot 3^t \)[/tex]
- This would imply the number of orders increases exponentially, which contradicts the given problem.
4. [tex]\( f(t) = -2000 \cdot 3t \)[/tex]
- This is a linear function that gives negative values and also does not fit the progression model.
### Conclusion
The correct function that models the change between the number of days, [tex]\( t \)[/tex], and the number of orders remaining is:
[tex]\[ f(t) = 2000 \cdot \left(\frac{1}{3}\right)^t \][/tex]
Therefore, the correct answer is:
[tex]\[ 1 \][/tex]
### Problem Analysis
The store starts with 2,000 orders. Each day, the store owner completes one-third of the remaining orders from the previous day. That means two-thirds of the orders are completed each day, leaving one-third of the orders still remaining.
### Function Modeling
1. Initial Orders: On day 0, the number of orders is 2,000.
- [tex]\( f(0) = 2000 \)[/tex]
2. Day 1 Calculation:
- At the end of day 1, only one-third of the 2,000 orders should remain.
- Therefore, [tex]\( f(1) = 2000 \cdot \left(\frac{1}{3}\right) = 666.67 \)[/tex]
3. Day 2 Calculation:
- At the end of day 2, only one-third of the orders remaining at the end of day 1 will still be unfulfilled.
- Therefore, [tex]\( f(2) = 2000 \cdot \left(\frac{1}{3}\right)^2 \approx 222.22 \)[/tex]
This pattern continues, indicating that the number of orders remaining can be expressed by a geometric progression:
[tex]\[ f(t) = 2000 \cdot \left(\frac{1}{3}\right)^t \][/tex]
### Evaluating Function Choices
Given the following options:
- [tex]\( f(t) = 2000 \cdot \left(\frac{1}{3}\right)^t \)[/tex]
- [tex]\( f(t) = -2000 \cdot \left(\frac{1}{3}\right)^t \)[/tex]
- [tex]\( r(t) = 2000 \cdot 3^t \)[/tex]
- [tex]\( f(t) = -2000 \cdot 3t \)[/tex]
We can derive the correct option by methodically rejecting the incorrect ones:
1. [tex]\( f(t) = 2000 \cdot \left(\frac{1}{3}\right)^t \)[/tex]
- This correctly follows the geometric progression where each day one-third of the previous day's orders remain.
2. [tex]\( f(t) = -2000 \cdot \left(\frac{1}{3}\right)^t \)[/tex]
- A negative number of orders does not make sense in this context.
3. [tex]\( r(t) = 2000 \cdot 3^t \)[/tex]
- This would imply the number of orders increases exponentially, which contradicts the given problem.
4. [tex]\( f(t) = -2000 \cdot 3t \)[/tex]
- This is a linear function that gives negative values and also does not fit the progression model.
### Conclusion
The correct function that models the change between the number of days, [tex]\( t \)[/tex], and the number of orders remaining is:
[tex]\[ f(t) = 2000 \cdot \left(\frac{1}{3}\right)^t \][/tex]
Therefore, the correct answer is:
[tex]\[ 1 \][/tex]
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