Question

The carnival is arriving in the neighborhood! Leo has saved up $15 to spend at the carnival on food and attractions. Food items cost $4 each and attractions cost $2 each. Leo uses the expression 4a + 2b to determine the total cost for a food items and b attractions. Select one of the following combinations of food items and attractions that Leo can afford with his $15 budget:

Answer 1

To determine which combinations of food items (costing $4 each) and attractions (costing $2 each) Leo can afford with his $15 budget, we can analyze the expression \(4a + 2b \leq 15\), where \(a\) is the number of food items and \(b\) is the number of attractions.

We can rearrange and analyze the expression:

1. **Max food items only**: If Leo buys only food items, the maximum number of food items he can afford is:
\[
4a \leq 15 \implies a \leq \frac{15}{4} \implies a \leq 3.75
\]
So, the maximum \(a\) is 3 (he can buy 3 food items).

2. **Max attractions only**: If Leo buys only attractions, the maximum number of attractions he can afford is:
\[
2b \leq 15 \implies b \leq \frac{15}{2} \implies b \leq 7.5
\]
So, the maximum \(b\) is 7 (he can buy 7 attractions).

Now, let's consider various combinations of \(a\) and \(b\) and check if they fit the budget:

- **Combination 1**: \(a = 0\), \(b = 7\)
\[
4(0) + 2(7) = 0 + 14 = 14 \quad \text{(within budget)}
\]

- **Combination 2**: \(a = 1\), \(b = 5\)
\[
4(1) + 2(5) = 4 + 10 = 14 \quad \text{(within budget)}
\]

- **Combination 3**: \(a = 2\), \(b = 3\)
\[
4(2) + 2(3) = 8 + 6 = 14 \quad \text{(within budget)}
\]

- **Combination 4**: \(a = 3\), \(b = 0\)
\[
4(3) + 2(0) = 12 + 0 = 12 \quad \text{(within budget)}
\]

- **Combination 5**: \(a = 1\), \(b = 6\)
\[
4(1) + 2(6) = 4 + 12 = 16 \quad \text{(over budget)}
\]

These combinations show that Leo can afford various amounts of food items and attractions as long as he stays within the total cost of $15.

Please provide the specific combinations you would like to check, and we can evaluate them!