Question

The party will cost $150, plus $10.50 per child and $5.50 per adult. There needs to be at least one adult and one child at the party, and the cost cannot exceed $300. (14.286, 0) is a viable solution for the above situation. Consider the ordered pair as (c,a) where c represents number of children and a represents number of adults at the party. (1 point) True False

Answer 1

To evaluate whether the ordered pair (14.286, 0) is a viable solution for the given cost situation, we need to analyze the constraints provided:

1. The total cost equation is given by: $$\text{Total Cost}=150+10.50c+5.50a$$ where $c$ is the number of children and $a$ is the number of adults.

2. The conditions specify:

   • There must be at least one adult ($a\ge 1$).
   • There must be at least one child ($c\ge 1$).
   • The total cost cannot exceed $300.

Substituting the ordered pair (14.286, 0):

• $c=14.286$ (approximately 14 children)
• $a=0$ (0 adults)

Now verifying the conditions:

• Total cost: $$\text{Total Cost}=150+10.50(14.286)+5.50(0)=150+150.003=300.003$$ This exceeds $300, which is not allowed.

• Number of adults: The number of adults $a=0$, which violates the condition of having at least one adult.

• Number of children: The number of children can be roughly 14.286, which satisfies the condition of having at least one child.

Since $a$ must be at least 1 and the cost exceeds $300, (14.286, 0) is not a viable solution.

Thus, the answer is False.