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Mary is selling her craft to earn money she sells her bracelets for $6 and her necklaces for $10 her goal is to make at least $120 in sales which of the following represents three possible solutions to the problem

a.6x+10y<120 (5,10),(10,5),and(15,0)are three possible solutions.

b.6x+10y>=120 (5,9),(10,6),and(15,3)are three possible solutions.

c.6x-10y<120 (5,5),(10,10),and(15,15)are three possible solutions.

d.7x+5y<=140 (7,21),(14,14),and(21,7)are three possible solutions.
I think it's
a.6x+10y<120 (5,10),(10,5),and(15,0)are three possible solutions. please correct me if I'm wrong.
I think it's between B and
d. is it D

User Karhgath
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2 Answers

6 votes

Final answer:

The correct inequality for Mary's sales goal is b. 6x+10y≥120 with solutions such as (5, 9), (10, 6), and (15, 3), which all satisfy the condition of earning at least $120.

Step-by-step explanation:

The correct answer to the problem is b. 6x+10y≥120 which means Mary needs to make at least $120 from selling bracelets and necklaces. To satisfy this condition, she can sell different combinations of bracelets (x) and necklaces (y) as long as the total amount from sales is equal to or greater than $120. Let's check the given solutions:

  • (5, 9) - This means 5 bracelets and 9 necklaces, which gives a total of (5*$6) + (9*$10) = $30 + $90 = $120. This meets the goal.
  • (10, 6) - This means 10 bracelets and 6 necklaces, which gives a total of (10*$6) + (6*$10) = $60 + $60 = $120. This also meets the goal.
  • (15, 3) - This means 15 bracelets and 3 necklaces, which gives a total of (15*$6) + (3*$10) = $90 + $30 = $120. This too meets the goal.

Therefore, solutions (5,9), (10,6), and (15,3) are all valid and demonstrate the different combinations of bracelets and necklaces Mary can sell to reach her sales goal.

User James Scott
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6 votes
her goal is to make at least $120 at least meaning is not less than 120, so the mathematical symbol is greater than or equal to >=. so the possible solution is let x be number of bracelety be the number of necklace6x + 10y >= 120 (5,9),(10,6),and(15,3)are three possible solutions.
User Mutty
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7.7k points