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jasenka [17]
3 years ago
14

What is the middle of 1/2 and 3/4

Mathematics
2 answers:
nirvana33 [79]3 years ago
8 0

Answer: Try 5/8

Step-by-step explanation: Because 1/2 and 3/4 can also be written as 4/8 and 6/8. And that's how I got 5/8.

cluponka [151]3 years ago
4 0

Hi! I believe that I have the answer.

My answer: 5/8.

1/2 can also be put as 4/8. 3/4 can also be put as 6/8. Right in the middle of 4/8 and 6/8 is 5/8.

I hope that helps!

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The IQR is 42.5

Step-by-step explanation:

Interquartile range is the difference of third and first quartile.

First of all we have to find the median for that purpose the data has to be arranged in ascending order. The data is already in ascending order.

As the number of values are odd

n=21

The median will be: (\frac{n+1}{2}) th\ term

Putting n=21

(\frac{21+1}{2})th\ term\\=(\frac{22}{2})th\ term\\= 11th\ term

The 11th term is 133

So median = 133

Now the data is divided into two halves

One is: 98, 100, 101, 102, 108, 109, 111,118, 129, 132

2nd is: 135, 135, 145, 146, 146, 156, 170 176, 180, 180

Q1 will be the median of first half and Q3 will be the median of 2nd half.

As now the halves contain even number of values, the medians will be the average of middle two values

<u>For First Half:</u>

98, 100, 101, 102, <u>108, 109</u>, 111,118, 129, 132

Q_1 = \frac{108+109}{2}\\Q_1 = \frac{217}{2}\\Q_1 = 108.5

<u>For Second Half:</u>

135, 135, 145, 146, 146, 156, 170 176, 180, 180

Q_2 = \frac{146+156}{2}\\Q_2 = \frac{302}{2}\\Q_2 = 151

Now

<u>Interquartile Range:</u>

IQR = Q_3-Q_1\\= 151-108.5\\=42.5

Hence,

The IQR is 42.5

Keywords: Median, IQR

Learn more about median at:

  • brainly.com/question/10940255
  • brainly.com/question/10941043

#LearnwithBrianly

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Anyone know this one?
asambeis [7]

Answer:

120

Step-by-step explanation:

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