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boyakko [2]
3 years ago
9

Aisha thinks 1,275 rounded to the nearest hundread is 1,200 because 2 is less than 5 what might aisha misunderstand about roundi

ng
Mathematics
2 answers:
lora16 [44]3 years ago
8 0
For rounding, you look at the number to the right of it.

In this case, it would be 7

because 7 is greater than 5, it is rounded up (remember, 5 and up is round up, while 4 and lower is round down)
So her answer should be 1300

hope this helps
djyliett [7]3 years ago
7 0
I believe Aisha is wrong because its rounding to the nearest hundred, not to the nearest thousandth 
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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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3 years ago
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Step-by-step explanation:

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