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erik [133]
3 years ago
8

a jewelry store buys 24 necklaces for $300. the store wants to make a profit of $540. assuming it sells all 24 necklaces, how mu

ch should the store charge for each necklace?
Mathematics
1 answer:
Serga [27]3 years ago
3 0
If its meaning making a profit after the $300 then:
540+300=840
840/24= 33.60
They would charge $33.60 per necklace.

If you were meaning a profit of $540 total then:
540/24=22.5
They would charge $22.50 per necklace.
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Step-by-step explanation:

Let n(A) the number of elements in A.

Remember, the number of elements in A_1 \cup A_2 \cup A_3 satisfies

n(A_1 \cup A_2 \cup A_3)=n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)

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a) If A_1\subseteq A_2, n(A_1 \cap A_2)=n(A_1)=200, and if A_2\subseteq A_3, n(A_2\cap A_3)=n(A_2)=1000

Since A_1\subseteq A_2\; and \; A_2\subseteq A_3, \; then \; A_1\cap A_2 \cap A_3= A_1

So

n(A_1 \cup A_2 \cup A_3)=\\=n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)=\\=200+1000+5000-200-200-1000-200=4600

b) Since the sets are pairwise disjoint

n(A_1 \cup A_2 \cup A_3)=\\n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)=\\200+1000+5000-0-0-0-0=6200

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n(A_1 \cup A_2 \cup A_3)=\\=n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)=\\=200+1000+5000-2-2-2-1=6193

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