To fill in the total weight for 9 apples and 1 plum, we first need to determine the weight of the apples and the plum from the provided data.
From the first line of the table:
- 10 apples weigh 50 oz. This means that the weight per apple is: \[ \text{Weight per apple} = \frac{50 \text{ oz}}{10 \text{ apples}} = 5 \text{ oz/apple} \]
Now, we can find the weight for 9 apples:
- 9 apples weigh: \[ 9 \text{ apples} \times 5 \text{ oz/apple} = 45 \text{ oz} \]
Next, let's assume the weight of the plum remains constant across the table. We can find the total weight of the other entries to better deduce the plum's weight.
Using the entries we currently have:
- For 10 apples (50 oz): 0 plums
- For 9 apples (unknown): 1 plum (this is what we wish to find)
- For 7 apples (unknown): 3 plums
- For 4 apples (unknown): 6 plums
- For 2 apples (unknown): 8 plums
- For 0 apples (unknown): 10 plums
To find the weight associated with the other combinations, we'll have to establish a pattern concerning the weights of plums and solve iteratively.
Looking at the weight reductions as plums increase:
- From 10 to 9 apples, 0 to 1 plum, there’s a weight decrease from 50 oz to Total Weight(9 apples + 1 plum).
- From 10 to 7 apples (3 plums), it's likely to subtract more weight; hence, we look at the pattern hierarchy.
Assuming that every time we trade off 2 apples (which contribute 10 oz) for plums the pattern is linear.
Let us calculate the most likely plum weight from the structure:
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When moving from 10 to 7 apples, we lose \(3 \times 5 \text{ oz} - 3 \times \text{Weight of a plum} = \text{Total loss in weight}\). It'll give us the weight differences iteratively.
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If we derive this from First:
- 10 apples + 0 plums = 50 oz
- 7 apples + 3 plums
\[ 50 oz - \text{Weight of 3 plums} = 7 \times 5 \text{ oz} \Rightarrow 50 oz - \text{Weight of 3 plums} = 35 oz \] Thus, \[ \text{Weight of 3 plums} = 50 oz - 35 oz = 15 oz \Rightarrow \text{Weight of plum} = 5 \text{ oz} \]
Continuing to plug that into the 9 and 1 section: 9 apples = 45 oz and if plum also consistently agreed to 5 oz, we have.
\[ \text{Total weight for 9 apples + 1 plum} = 45 oz (from 9 apples) + 5 oz = 50 oz \] So 9 apples and 1 plum equals 50 oz.
Now, rest of the table is: \[ \begin{array}{|c|c|c|} \hline \text{Number of apples} & \text{Number of plums} & \text{Total weight (oz)} \ \hline 10 & 0 & 50 \ 9 & 1 & 50 \ 7 & 3 & 45 \ 4 & 6 & 30 \ 2 & 8 & 20 \ 0 & 10 & 0 \ \hline \end{array} \]
Thus, the added portion would look like: \[ \text{For 9 apples and 1 Plum: Total weight = 50 oz} \]
Conclusively, the total weight in the table you've requested shall be filled now: 9 apples and 1 plum = 50 oz.
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