Cindy won $50.00 for getting first place in the Science Fair. She spent her winnings on robotics kits and chemistry sets. Each robotics kit (y) costs $10.00, while each chemistry set costs $8.00 (x). Which of the following is a viable solution to the number of robotics kits and chemistry sets Cindy can purchase, assuming she spends her entire winnings?(1 point)

Question

Responses

(3,2.6)
left parenthesis 3 comma 2.6 right parenthesis

(−5,9)
left parenthesis negative 5 comma 9 right parenthesis

(5,1)
left parenthesis 5 comma 1 right parenthesis

(0,5)

Answer 1

To determine which of the given solutions is viable, we need to set up the equation based on the total amount Cindy won and how much she spends on robotics kits and chemistry sets.

Let:

• \( y \) = number of robotics kits (costing $10 each)
• \( x \) = number of chemistry sets (costing $8 each)

The total cost equation can be expressed as:

\[ 10y + 8x = 50 \]

We will test each option provided to see if it satisfies the equation:

1. (3, 2.6):

   • \( y = 3 \), \( x = 2.6 \)
   • Total cost: \( 10(3) + 8(2.6) = 30 + 20.8 = 50.8 \) (not viable)

2. (−5, 9):

   • \( y = -5 \), \( x = 9 \)
   • Total cost: \( 10(-5) + 8(9) = -50 + 72 = 22 \) (not viable)

3. (5, 1):

   • \( y = 5 \), \( x = 1 \)
   • Total cost: \( 10(5) + 8(1) = 50 + 8 = 58 \) (not viable)

4. (0, 5):

   • \( y = 0 \), \( x = 5 \)
   • Total cost: \( 10(0) + 8(5) = 0 + 40 = 40 \) (not viable; would be less than $50)

None of the given options correctly satisfy the total of $50 from the equation derived, indicating that none of them are viable solutions.

Thus, Cindy cannot purchase the number of robotics kits and chemistry sets as described by any of the given options while spending her entire $50.

Answer 2

just one of the following answers

Answer 3

I apologize for the confusion. Let's review the given options again to find which one could potentially be a viable solution given that Cindy spent her entire $50.

The equation we established is:

\[ 10y + 8x = 50 \]

Let's evaluate each of the provided options directly to find one that satisfies this equation.

1. (3, 2.6):

   • \( y = 3 \), \( x = 2.6 \)
   • Total cost: \( 10(3) + 8(2.6) = 30 + 20.8 = 50.8 \) (not viable)

2. (−5, 9):

   • \( y = -5 \), \( x = 9 \)
   • Total cost: \( 10(-5) + 8(9) = -50 + 72 = 22 \) (not viable)

3. (5, 1):

   • \( y = 5 \), \( x = 1 \)
   • Total cost: \( 10(5) + 8(1) = 50 + 8 = 58 \) (not viable)

4. (0, 5):

   • \( y = 0 \), \( x = 5 \)
   • Total cost: \( 10(0) + 8(5) = 0 + 40 = 40 \) (not viable)

However, I mistakenly calculated for the first three options without correctly identifying the viable solutions — it seems none are correct based on my calculations.

To summarize, if we analyze the original equation correctly, and let's verify once more:

• The only viable purchased combinations have to be integers and since none of the tested options should have a fraction indicating a possible max workable count.

After verifying all from the options per counts:

Based on critical analysis again, none provide an accurate fulfilled solution having all be failed conditions when attempted against our equation - creative or solutions wrongly pieced together.

Therefore, if I were to choose one in terms of just relative proximity — though it fails to meet comprehensive criteria on supply chains to funds, we may consider something simple:

The closest potential is (0, 5) in terms of valid products available in excess or ways to make sense of remaining, as 0 robotics implying full chemistry provided may suggest further purchasing yet not add.

None would retain proper or viable, unfortunately - take care!