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mash [69]
3 years ago
8

Find the iqr of 4 5 6 8 9 10 11 12

Mathematics
2 answers:
lbvjy [14]3 years ago
8 0

Answer:

5

Step-by-step explanation:

4, 5, 6, 8, 9, 10, 11, 12

Median: 8.5

(IQR) interquartile range: 10.5 - 5.5 = 5

77julia77 [94]3 years ago
3 0

Answer:

The interquartile range is 5.

Step-by-step explanation:

Ah, a throwback to interquartile range... let me help :)

4,5,6,8,9,10,11,12

First, you need to know how to use the IQR. The interquartile range is basically known as the process of subtracting the upper quartile and the lower quartile of a set of data. The lower quartile should be written as Q1, and the upper quartile would be labeled as Q3. This would make the midpoint (median) data set Q2, and the highest possible point would be labeled Q4. Next, you have to always understand what you are looking at. For example, let's split the set 5,6,7,8,9,10,11,12 into groups. 5 and 6 would be Q1, 7 and 8 would be Q2, 9 and 10 would be Q3, and last but not least, 11 and 12 would be labeled as Q4. Now take Q1 and subtract it from Q3 and that is how you get your IQR.



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Answer:

A. 14x14x28

B. The maximum volume is 5488 cuibic inches

Step-by-step explanation:

The problem states that the box has square ends, so you can express volume with:

v=x^{2} y

Using the restriction stated in the problem to get another equation you can substitute in the one above:

4x+y=84\\\\

Substituting <em>y</em> whit this equation gives:

v=x^{2} (84-4x)\\\\v=84x^{2} -4x^{3}

Now find the limit of <em>x</em>:

\frac{84x^{2}-4x^{3}}{dx}=168x-12x^{2}\\\\x=\frac{168}{12}=14

Find the length:

y=84-4(14)=28

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v=(14)^{2}(28)= 5488

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hope this helped!

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