You are reaching the price of MP3 players.Yiu found an average price of $58.80. One MP3 player cost $56 and another cost $62. Find the price of the third MP3 player
Answers
The price of third MP3 Player is $58.4
Further explanation:
Given
Price of first MP3 player= p1= $56
Price of second MP3 Player = p2= $62
Price of third MP3 Player=?
Average = X = $58.80
The formula for average is:
We have 3 items.
Putting the values
The price of third MP3 Player is $58.4
Keywords: Average, Sum
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Related Questions
HELP PLEASEEEE!!The math team decided to sell customized pocket protectors as afundraiser. These were truly a hot item in our school, and $240 wasraised. Cecily had originally suggested a price that was $2 lowerfor each item, but if the team had agreed to that price they would havehad to sell 20 more pocket protectors to get the $240.How many pocket protectors did they actually sell? What was the price ofeach?
Withdrawals by the owner for personal use do not affect net income or net loss of the business. a. True b. False
Which steps should be followed to write an equivalent ratio to find 2% of 6700? Write 2% as the ratio StartFraction 200 Over 100 EndFraction. Write the equivalent ratio StartFraction question mark Over 6700 EndFraction. (100)(67) = 6700, so (20)(67) = 1340. Write 2% as the ratio StartFraction 20 Over 100 EndFraction. Write the equivalent ratio StartFraction 6700 Over question mark EndFraction. (20)(335) = 6700, so (100)(335) = 33,500. Write 2% as the ratio StartFraction 2 Over 100 EndFraction. Write the equivalent ratio StartFraction question mark Over 6700 EndFraction. (100)(67) = 6700, so (2)(67) = 134. Write 2% as the ratio StartFraction 2 Over 100 EndFraction. Write the equivalent ratio StartFraction question mark Over 6700 EndFraction. (100)(6.7) = 6700, so (20)(6.7) = 13.4
A school is having a canned food drive. Each class is challenged to collect 200 cans. If there are c classes in the school, which equation could be used to find the total number of cans, T, the school will collect if each class meets the challenge? A. T= c/200 B. T=200/cC. 200cD. 200+c
2.AC=FD
3.AF=CD
4.FA=CD
Answers
The answer to your question is AB = DE & BC = EF, so the only sides missing are AC = DF,
so Side AC is congruent to side DF is the answer.
I hope my answer helped you. Have a nice day!
Answer: AC=DF
Step-by-step explanation:
Took the test!
A.
(15 ÷ n) + 2
B.
(15 ÷ n) – 2
C.
15 ÷ (n – 2)
D.
15 ÷ (n + 2)
Answers
Because you have to divide the 15 and the number first then "decrease" it by 2
Answers
It is to be noted that the product of 2/3 and 36 is 24.
How to find the product
2/3 x 36
= 72/3
= 24
Therefore, the product of 2/3 and 36 is 24.
In mathematics, a product refers to the result obtained when two or more numbers or quantities are multiplied together.
It is the outcome of multiplication. The product is obtained by multiplying the multiplicands (the numbers being multiplied) and is represented by the product symbol (×) or by writing the numbers next to each other.
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Answers
The only possible option is that
Answers
Answers
Answer: Here's my answer
Step-by-step explanation:
The given information can be written as:
1. X^2 + k + l = 5x - 6
2. kl = 5x - 6
3. jk = 3x
Let's break it down step by step and analyze each equation:
1. X^2 + k + l = 5x - 6: This equation represents a quadratic equation, as it contains the term X^2. It relates the variables X, k, and l to the expression 5x - 6. To solve this equation, we would need additional information or constraints.
2. kl = 5x - 6: This equation shows that the product of k and l is equal to 5x - 6. It relates the variables k, l, and x. However, without specific values or constraints, we cannot determine the exact values of k, l, or x.
3. jk = 3x: This equation relates the variables j, k, and x. It states that the product of j and k is equal to 3x. Similarly to the previous equation, we need additional information or constraints to find specific values for j, k, or x.
In summary, the given equations represent relationships between different variables (X, k, l, j, and x), but without further information or constraints, we cannot solve them for specific values. It is important to have additional context or equations to determine the values of the variables or to find a specific solution.