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### Maximum Number of Running Shoe Inserts
From the given table:
- At combination A, the number of running shoe inserts is 15.
Thus,
The maximum number of running shoe inserts that can be produced is 15.
### At Possibility B
We know:
- At combination C, 9 running shoe inserts correspond to 4 hiking boot inserts.
- At combination E, 3 running shoe inserts correspond to 8 hiking boot inserts.
To find the number of running shoe inserts when 2 hiking boot inserts are produced, we can infer a linear relationship between the number of running shoe inserts and hiking boot inserts. By analyzing the changes:
- From C to E, the difference is 6 running shoe inserts (9 - 3) for 4 hiking boot inserts (8 - 4).
The rate of change (slope) is:
- [tex]\( \Delta \text{running shoe inserts} / \Delta \text{hiking boot inserts} = \frac{9 - 3}{8 - 4} = \frac{6}{4} = 1.5 \)[/tex].
When the number of hiking boot inserts decreases from 4 to 2:
- [tex]\( \Delta \text{hiking boot inserts} = 4 - 2 = 2 \)[/tex].
The change in running shoe inserts:
- [tex]\( \Delta \text{running shoe inserts} = 2 \times 1.5 = 3 \)[/tex].
Hence, the number of running shoe inserts at B:
- [tex]\( 9 (\text{from combination C}) - 3 (\text{change}) = 6 \)[/tex].
Thus,
At possibility B, 6 pairs of running shoe inserts can be produced.
### At Possibility D
We know:
- At combination C, 9 running shoe inserts correspond to 4 hiking boot inserts.
- At combination E, 3 running shoe inserts correspond to 8 hiking boot inserts.
When the number of running shoe inserts is 6, we will find the corresponding number of hiking boot inserts.
We use the same rate of change (slope):
- [tex]\( 1.5 \text{ running shoe inserts per hiking boot insert} \)[/tex].
From combination C:
- The change from 9 running shoe inserts to 6 is a decrease of 3.
This decrease corresponds to:
- [tex]\( \Delta \text{hiking boot inserts} = \frac{3}{1.5} = 2 \)[/tex].
Thus, the number of hiking boot inserts:
- [tex]\( 4 (\text{from combination C}) + 2 (\text{increase}) = 6 \)[/tex].
Therefore,
At possibility D, 6 pairs of hiking boot inserts can be produced.
Summarizing:
1. The maximum number of running shoe inserts that can be produced is 15.
2. At possibility B, 6 pairs of running shoe inserts can be produced.
3. At possibility D, 6 pairs of hiking boot inserts can be produced.
### Maximum Number of Running Shoe Inserts
From the given table:
- At combination A, the number of running shoe inserts is 15.
Thus,
The maximum number of running shoe inserts that can be produced is 15.
### At Possibility B
We know:
- At combination C, 9 running shoe inserts correspond to 4 hiking boot inserts.
- At combination E, 3 running shoe inserts correspond to 8 hiking boot inserts.
To find the number of running shoe inserts when 2 hiking boot inserts are produced, we can infer a linear relationship between the number of running shoe inserts and hiking boot inserts. By analyzing the changes:
- From C to E, the difference is 6 running shoe inserts (9 - 3) for 4 hiking boot inserts (8 - 4).
The rate of change (slope) is:
- [tex]\( \Delta \text{running shoe inserts} / \Delta \text{hiking boot inserts} = \frac{9 - 3}{8 - 4} = \frac{6}{4} = 1.5 \)[/tex].
When the number of hiking boot inserts decreases from 4 to 2:
- [tex]\( \Delta \text{hiking boot inserts} = 4 - 2 = 2 \)[/tex].
The change in running shoe inserts:
- [tex]\( \Delta \text{running shoe inserts} = 2 \times 1.5 = 3 \)[/tex].
Hence, the number of running shoe inserts at B:
- [tex]\( 9 (\text{from combination C}) - 3 (\text{change}) = 6 \)[/tex].
Thus,
At possibility B, 6 pairs of running shoe inserts can be produced.
### At Possibility D
We know:
- At combination C, 9 running shoe inserts correspond to 4 hiking boot inserts.
- At combination E, 3 running shoe inserts correspond to 8 hiking boot inserts.
When the number of running shoe inserts is 6, we will find the corresponding number of hiking boot inserts.
We use the same rate of change (slope):
- [tex]\( 1.5 \text{ running shoe inserts per hiking boot insert} \)[/tex].
From combination C:
- The change from 9 running shoe inserts to 6 is a decrease of 3.
This decrease corresponds to:
- [tex]\( \Delta \text{hiking boot inserts} = \frac{3}{1.5} = 2 \)[/tex].
Thus, the number of hiking boot inserts:
- [tex]\( 4 (\text{from combination C}) + 2 (\text{increase}) = 6 \)[/tex].
Therefore,
At possibility D, 6 pairs of hiking boot inserts can be produced.
Summarizing:
1. The maximum number of running shoe inserts that can be produced is 15.
2. At possibility B, 6 pairs of running shoe inserts can be produced.
3. At possibility D, 6 pairs of hiking boot inserts can be produced.
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