Create the system of linear inequalities that models the following situation. The field hockey team at school is hosting a fundraiser. They hope to raise at least $600. They are selling chocolate for $4 and flowers for $5. The team captain estimates they will sell no more than 150 items, but they know that 50 flowers have already been preordered.(1 point) Responses 4x+5y≥600, x+y≥150, y≤50 4 x plus 5 y greater-than-or-equal-to 600 , x plus y greater-than-or-equal-to 150 , y less-than-or-equal-to 50 4x+5y≥600, x+y≤150, y≥50 4 x plus 5 y greater-than-or-equal-to 600 , x plus y less-than-or-equal-to 150 , y greater-than-or-equal-to 50 4x+5y>600, x+y≥150, y≥50 4 x plus 5 y greater than 600 , x plus y greater-than-or-equal-to 150 , y greater-than-or-equal-to 50 4x+5y≥600, x+y≥150, y≥50

Answer 1

To model the situation with a system of linear inequalities, we can define variables for the number of items sold:

• Let \( x \) be the number of chocolate items sold.
• Let \( y \) be the number of flower items sold.

Now let's formulate the inequalities based on the problem:

1. Total revenue from selling chocolates and flowers:
   The team wants to raise at least $600. The revenue from selling chocolates is \( 4x \) (since each chocolate costs $4) and from flowers is \( 5y \) (since each flower costs $5). Therefore, the inequality for the amount they want to raise can be expressed as: \[ 4x + 5y \geq 600 \]

2. Total items sold:
   The team captain estimates they will sell no more than 150 items in total. This can be represented as: \[ x + y \leq 150 \]

3. Preordered flowers:
   They know that 50 flowers have already been preordered, so they must sell at least those 50 flowers. This is given by: \[ y \geq 50 \]

Putting it all together, the complete system of inequalities is:

1. \( 4x + 5y \geq 600 \)
2. \( x + y \leq 150 \)
3. \( y \geq 50 \)

Therefore, among the provided options, the correct inequalities that model the situation are:

\[ 4x + 5y \geq 600, \quad x + y \leq 150, \quad y \geq 50 \]