a couple are celebrating their 30th anniversary by having a reception at a local reception hall. they have…

a couple are celebrating their 30th anniversary by having a reception at a local reception hall. they have budgeted $4,000 for their reception. if the reception hall charges a $50 cleanup fee plus $37 per person, find the greatest number of people that they may invite and still stay within their budget. the couple can invite at most □ people to the reception. (round down to the nearest whole person.)

a couple are celebrating their 30th anniversary by having a reception at a local reception hall. they have budgeted $4,000 for their reception. if the reception hall charges a $50 cleanup fee plus $37 per person, find the greatest number of people that they may invite and still stay within their budget. the couple can invite at most □ people to the reception. (round down to the nearest whole person.)

Answer

Explanation:

Step1: Set up the cost - equation

Let $x$ be the number of people. The total cost $C$ of the reception is given by the equation $C = 37x+50$. The budget is $4000$. So we have the inequality $37x + 50\leq4000$.

Step2: Solve the inequality for $x$

First, subtract 50 from both sides of the inequality: $37x+50 - 50\leq4000 - 50$, which simplifies to $37x\leq3950$. Then, divide both sides by 37: $x\leq\frac{3950}{37}\approx106.76$.

Step3: Round down to the nearest whole number

Since $x$ represents the number of people and we must stay within the budget, we round down the result of $\frac{3950}{37}$. So $x = 106$.

Answer:

106