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Strike441 [17]
1 year ago
9

Below are the average test scores of two different math class periods. Test 1 Test 2 Test 3 Test 4 Test 5 Test 6 83 87 82 89 4th

period 90 82 90 85 6th period 86 81 88 How do the median test scores of each class period compare? OA. The median test score of 4th period is equal to the median test score of 6th period. B. The median test score of 4th period is greater than the median test score of 6th period. C. The median test score of 6th period is greater than the median test score of 4th period. D. The median test score of 6th period is greater than the median test score of 4th ​
Mathematics
1 answer:
irinina [24]1 year ago
8 0

You need to understand that you're solving for the average, which you already know: 90.

Since you know the values of the first three exams, and you know what your final value needs to be, just set up the problem like you would any time you're averaging something.

Solving for the average is simple

<h3>What is the average?</h3>

Add up all of the exam scores and divide that number by the number of exams you took.

(87 + 88 + 92) / 3 = your average if you didn't count that fourth exam.

Since you know you have that fourth exam, just substitute it into the total value as an unknown, X

(87 + 88 + 92 + X) / 4 = 90

Now you need to solve for X, the unknown

87+88+92+X4 /(4) = 90/ (4)

Multiplying for four on each side cancels out the fraction.

So now you have

87 + 88 + 92 + X = 360

This can be simplified as

267 + X = 360

Negating the 267 on each side will isolate the X value, and give you your final answer

X = 93

Now that you have an answer, ask yourself does it make sense.

I say that it does because there were two tests that were below average, and one that was just slightly above average.

To learn more about the average visit:

brainly.com/question/10428039

#SPJ1

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Ash company reported sales of $580,000 for year 1, $630,000 for year 2, and $680,000 for year 3. Using year 1 as the base year,
Elenna [48]

Answer:

Year 2: 109.4%

Year 3: 118.9%

Step-by-step explanation:

<h3>Given;</h3>
  • Ash company reported sales of $580,000 for year 1
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  • $680,000 for year 3.

Now,

For the revenue trend percent for year 2

Year 2: $580,000 / $530,000 × 100 = 109.4%

For the revenue trend percent for year 3

Year 3: $630,000 / $530,000 × 100 = 118.9%

Thus, The answer is 109.4% and 118.9%.

<u>-TheUnknownScientist 72</u>

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Check whether the relation R on the set S = {1, 2, 3} is an equivalent
kozerog [31]

Answer:

R isn't an equivalence relation. It is reflexive but neither symmetric nor transitive.

Step-by-step explanation:

Let S denote a set of elements. S \times S would denote the set of all ordered pairs of elements of S\!.

For example, with S = \lbrace 1,\, 2,\, 3 \rbrace, (3,\, 2) and (2,\, 3) are both members of S \times S. However, (3,\, 2) \ne (2,\, 3) because the pairs are ordered.

A relation R on S\! is a subset of S \times S. For any two elementsa,\, b \in S, a \sim b if and only if the ordered pair (a,\, b) is in R\!.

 

A relation R on set S is an equivalence relation if it satisfies the following:

  • Reflexivity: for any a \in S, the relation R needs to ensure that a \sim a (that is: (a,\, a) \in R.)
  • Symmetry: for any a,\, b \in S, a \sim b if and only if b \sim a. In other words, either both (a,\, b) and (b,\, a) are in R, or neither is in R\!.
  • Transitivity: for any a,\, b,\, c \in S, if a \sim b and b \sim c, then a \sim c. In other words, if (a,\, b) and (b,\, c) are both in R, then (a,\, c) also needs to be in R\!.

The relation R (on S = \lbrace 1,\, 2,\, 3 \rbrace) in this question is indeed reflexive. (1,\, 1), (2,\, 2), and (3,\, 3) (one pair for each element of S) are all elements of R\!.

R isn't symmetric. (2,\, 3) \in R but (3,\, 2) \not \in R (the pairs in \! R are all ordered.) In other words, 3 isn't equivalent to 2 under R\! even though 2 \sim 3.

Neither is R transitive. (3,\, 1) \in R and (1,\, 2) \in R. However, (3,\, 2) \not \in R. In other words, under relation R\!, 3 \sim 1 and 1 \sim 2 does not imply 3 \sim 2.

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