What is the value of the interquartile range of the data below? 6 12 14 24

Mathematics

1 answer:

Strike441 [17]3 years ago

7 0

Answer:

Step-by-step explanation:

The interquartile range is the value of quartile 3 minus quartile 1.

The "box" part of this diagram has 3 lines, the left most, the middle, and the rightmost.

The leftmost line is Quartile 1. The right most is Quartile 3.

Hence

interquartile range = quartile 3 - quartile 1

Looking at the box plot, we can see that each small line in the number line is 2 units.

The leftmost line (quartile 1 ) is at 1 unit left of 30, so that is 30 -2 = 28

The rightmost line (quartile 3) is at 1 unit right of 40, so that is 40 + 2 = 42

Hence,

Interquartile range = 42 - 28 = 14

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Evaluate 30 - 12 / 3 * 2

Ivenika [448]

Answer:22

step by step:
so we have PE(MD)(AS)
(for multiplication and division it depends on left to right, same for addition and subtraction)

this could be read as
30 - [(12/3)*2]

first you would do 12/3 which is 4
then you do 4*2 which is 8
last you do 30-8 which is 22

hope that helps :)

3 0

3 years ago

How do you simplify: <img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B%28125%20%5E%7B2%7D%20%7D%20%29%20%20%20%5E%7B%20-%20

Nady [450]

Answer:

$\huge\boxed{\sqrt{(125^2)^{-\frac{1}{3}}}=\dfrac{1}{5}}$

Step-by-step explanation:

$\sqrt{(125^2)^{-\frac{1}{3}}}\qquad\text{use}\ (a^n)^m=(a^m)^n\\\\=\sqrt{\left(125^{-\frac{1}{3}\right)^2}\qquad\text{use}\ \sqrt{a^2}=a\ \text{for}\ a\geq0\\\\=125^{-\frac{1}{3}}\qquad\text{use}\ a^{-n}=\dfrac{1}{a^n}\\\\=\dfrac{1}{125^\frac{1}{3}}\qquad\text{use}\ a^\frac{1}{n}=\sqrt[n]{a}\\\\=\dfrac{1}{\sqrt[3]{125}}=\dfrac{1}{5}\qquad\text{because}\ 5^3=125$