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2 years ago

What are the 1st and 3rd quartiles? what is the interquartile range?

Mathematics

2 answers:

PolarNik [594]2 years ago

8 0

Answer:

The first quartile is 5

The third is 8

The interquartile is 3

Step-by-step explanation:

boblepop8 is correct

FrozenT [24]2 years ago

7 0

Answer:

The first quartile is 5

The third is 8

The interquartile is the difference between the first and third quartile.

The interquartile is 3

Step-by-step explanation:

First, make a list of all the numbers. To find the first quartile, you have to first find the median. The median is the first 7 in the list.

Everything before 7 is the lower percentile of the range.

The firsr quartile is the median of the lower percentile.

The lower list is:

2,3,4,4,5,6,6,6

The median of this list is 5.

The process for finding the third quartile is the same except the list is the higher percentile which is the list of numbers above the median.

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(ab^2 x -b^4a^3)^2<br><br> What is the answers

Umnica [9.8K]

Answer:

$\left(ab^2x-b^4a^3\right)^2:\quad a^2b^4x^2-2a^4b^6x+a^6b^8$

Step-by-step explanation:

Given the expression

$\left(ab^2\:x\:-b^4a^3\right)^2$

solving the expression

$\left(ab^2\:x\:-b^4a^3\right)^2$

$\mathrm{Apply\:Perfect\:Square\:Formula}:\quad \left(a-b\right)^2=a^2-2ab+b^2$

$a=ab^2x,\:\:b=b^4a^3$

so the expression becomes

$=\left(ab^2x\right)^2-2ab^2xb^4a^3+\left(b^4a^3\right)^2$

$=a^2b^4x^2-2a^4b^6x+a^6b^8$

Thus,

$\left(ab^2x-b^4a^3\right)^2:\quad a^2b^4x^2-2a^4b^6x+a^6b^8$

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2 years ago

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Sati [7]

Answer:

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