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kiruha [24]
2 years ago
10

What is the interquartile range (IQR) of the data set below? 34, 36, 37, 29, 43, 48, 49

Mathematics
2 answers:
Leya [2.2K]2 years ago
7 0
The answer to your question is 14
yarga [219]2 years ago
4 0
The interquartile range is 14 :)
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In triangle EFG, m of E = 30 degrees, m of F = 60 degrees, and m of G = 90 degrees. Which of the following statements about tria
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Answer:

A. EG = √3 × FG

D. EG = √3/2 × EF

E. EF = 2 × FG

Step-by-step explanation:

∵ tan 60 = √3

∵ tan60 = EG/GF

∴ EG/GF = √3

∴ EG = √3 × GF ⇒ A

∵ m∠F = 60°

∵ sin60 = √3/2

∵ sin 60 = EG/EF

∴ √3/2 = EG/EF

∴ EG = √3/2 × EF ⇒ D

∵ cos60 = 1/2

∵ cos60 = GF/EF

∴ GF/EF = 1/2

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You are going to use an inclined plane to lift a heavy object to the top of a shelving unit with a height of 6 feet. The base of
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24000 is the answer
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A professor has recorded exam grades for 10 students in his​ class, but one of the grades is no longer readable. If the mean sco
Nostrana [21]

Answer:

unreadable score = 35

Step-by-step explanation:

We are trying to find the score of one exam that is no longer readable, let's give that score the name "x". we can also give the addition of the rest of 9 readable s scores the letter "R".

There are two things we know, and for which we are going to create equations containing the unknowns "x", and "R":

1) The mean score of ALL exams (including the unreadable one) is 80

so the equation to represent this statement is:

mean of ALL exams = 80

By writing the mean of ALL scores (as the total of all scores added including "x") we can re-write the equation as:

\frac{R+x}{10} =80

since the mean is the addition of all values divided the total number of exams.

in a similar way we can write what the mean of just the readable exams is:

\frac{R}{9} = 85\\ (notice that this time we don't include the grade x in the addition, and we divide by 9 instead of 10 because only 9 exams are being considered for this mean.

Based on the equation above, we can find what "R" is by multiplying both sides by 9:

\frac{R}{9} = 85\\R=85*9= 765

Therefore we can now use this value of R in the very first equation we created, and solve for "x":

\frac{R+x}{10} =80\\\frac{765+x}{10} =80\\765+x=80*10=800\\765+x=800\\x=800-765=35

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3 years ago
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