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3 years ago

Study the following data set.

Mathematics

1 answer:

Vera_Pavlovna [14]3 years ago

6 0

Answer:

Step-by-step explanation:

The question is on interquartile range which is the median of the upper half of the data minus the median of the lower half of data

First arrange the data in an increasing order;

8,15,9,18,9,17,22,10,11,9,13

8,9,9,9,10,11,13,15,17,18,22

Find the median, which is the center value in the data set

8,9,9,9,10,11,13,15,17,18,22⇒the median is 11

Place brackets around the numbers above and below the median

(8,9,9,9,10)11 (13,15,17,18,22)

Find the median in the lower half of the data,Q1

(8,9,9,9,10) ⇒median is 9=Q1

Find the median in the upper half of the data,Q3

(13,15,17,18,22)⇒median is 17=Q3

Subtract Q1 from Q3 to get the interquartile range

Q3=17, and Q1=9

Q3-Q1=17-9=8

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Answer:

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Therefore $\left[\frac{(10^4)(5^2)}{(10^3)(5^3)}\right]^3=8$

Step-by-step explanation:

Given expression is (StartFraction (10 Superscript 4 Baseline) (5 squared) Over (10 cubed) (5 cubed)) cubed

Given expression can be written as below

$\left[\frac{(10^4)(5^2)}{(10^3)(5^3)}\right]^3$

To find the value of the given expression:

$\left[\frac{(10^4)(5^2)}{(10^3)(5^3)}\right]^3=\frac{((10^4)(5^2))^3}{((10^3)(5^3))^3}$

( By using the property ( $(\frac{a}{b})^m=\frac{a^m}{b^m}$ )

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( By using the property $(ab)^m=a^mb^m$ )

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( By using the property $(a^m)^n=a^{mn}$ )

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( By using the property $\frac{1}{a^m}=a^{-m}$ )

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$=(10^3)(5^{-3})$

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Therefore $\left[\frac{(10^4)(5^2)}{(10^3)(5^3)}\right]^3=8$

Therefore the value to the given expression is 8

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