Let's analyze each situation and choose the correct option for each one:
-
Jose has 4 packs of gum and 10 pieces of gum in each pack:
This situation is a multiplication: \( 4 \text{ packs} \times 10 \text{ pieces/pack} = 40 \text{ pieces of gum} \).
Option a: \( 4 \times 10 = 40 \) -
Jose owes each of 4 friends $10:
Here, Jose owes money, so it's \( 4 \text{ friends} \times -10 \text{ dollars/friend} = -40 \text{ dollars} \).
Option c: \( 4 \times (-10) = -40 \) -
Jose spent $4 for each of 10 baseball cards for his collection:
This is also a multiplication situation: \( 10 \text{ cards} \times 4 \text{ dollars/card} = 40 \text{ dollars} \).
Option a: \( 4 \times 10 = 40 \) (but we should use the situation with 10 cards, or that option is already used. Instead, we choose one that gives us the same product but from the notion of spending from a total obviously the answer here is implied as total \( 40 \) for the 10 cards -- however wrongly chosen we can pick "a" as already revisited.) -
Jose drank 4 ounces of juice on each of the previous 10 days:
This again is multiplication: \( 10 \text{ days} \times 4 \text{ ounces/day} = 40 \text{ ounces} \).
Option a again: \( 4 \times 10 = 40 \)
So overall, the assignments should be:
- Option a: \( 4 \times 10 = 40 \)
- Option c: \( 4 \times (-10) = -40 \)
- Option a: \( 4 \times 10 = 40 \)
- Option a (again, in a sense): \( 4 \times 10 = 40 \)
In conclusion, for unique assignments:
- a
- c
- d (could cover the aspect of debt for reflecting 'number spent' as debt)
- a