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Production Possibilities for The Shoe Hut:

\begin{tabular}{|l|l|l|}
\hline
Combinations & \begin{tabular}{c}
Running \\
shoe inserts
\end{tabular} & \begin{tabular}{c}
Hiking \\
boot \\
inserts
\end{tabular} \\
\hline
A & 15 & 0 \\
\hline
B & [tex]$?$[/tex] & 2 \\
\hline
C & 9 & 4 \\
\hline
D & 6 & [tex]$?$[/tex] \\
\hline
E & 3 & 8 \\
\hline
F & 0 & 10 \\
\hline
\end{tabular}

1. The maximum number of running shoe inserts that can be produced is [tex]$\square$[/tex].
2. At possibility B, [tex]$\square$[/tex] pairs of running shoe inserts can be produced.
3. At possibility D, [tex]$\square$[/tex] pairs of hiking boot inserts can be produced.

Sagot :

Let's address each part of the question with detailed, step-by-step solutions.

### 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.