Find the interquartile range (IQR) of the data in the box plot below.

Asap ASAP Please PLZ

Find The Interquartile Range (IQR) Of The Data In The Box Plot Below. Asap ASAP Please PLZ

Answers

Answer 1

Answer:

3

Step-by-step explanation:

IQR is the difference between the upper and lower medians so the IQR = 7 - 4 = 3

Answer 2

The interquartile range (IQR) of the data in the given box plot is: 3.

What is the Interquartile Range (IQR) of a Data?

In a box plot, the interquartile range (IQR) is the difference between the upper quartile and the lower quartile which are represented as each edge of the rectangular box.

Upper quartile = 7Lower quartile = 4

Interquartile range (IQR) = 7 - 4

Interquartile range (IQR) = 3

Learn more about Interquartile range (IQR) on:

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Related Questions

there are 12 inches in 1 foot
write an equation to find the number of inches in any number of feet

Answers

Answer:

divide the number of inches by 12

Step-by-step explanation:

1/12=x/12 then you would cross multiply it

A Bowling ball is rolled down the alley with the constant velocity of 1.6 m/sec at an angle of 85° to the starting line, the position of the person throwing a bowling ball can be represented by the point (0,0). where is the ball after six seconds?

Answers

We have been given that a Bowling ball is rolled down the alley with the constant velocity of 1.6 m/sec at an angle of 85° to the starting line, the position of the person throwing a bowling ball can be represented by the point (0,0).

We are asked to find the position of ball after 6 seconds.

We will use horizontal position and vertical position formula to solve our given problem.

Horizontal position is given by formula: [tex]x=v\cdot \text{cos}(\theta)\cdot t[/tex] and vertical position is given by formula: [tex]y=v\cdot \text{sin}(\theta)\cdot t[/tex],where,

v = Velocity,

t = Time,

[tex]\theta[/tex] = Angle to the starting line.

For our given problem [tex]v=1.6[/tex], [tex]t=6[/tex] and [tex]\theta=85^{\circ}[/tex].

[tex]x=1.8\cdot \text{cos}(85^{\circ})\cdot 6[/tex]

[tex]x=10.8 \cdot 0.087155742748[/tex]

[tex]x=0.9412820216784\approx 0.94[/tex]

[tex]y=1.8\cdot \text{sin}(85^{\circ})\cdot 6[/tex]

[tex]y=10.8\cdot 0.996194698092[/tex]

[tex]y=10.7589027\approx 10.76[/tex]

Therefore, the position of ball after 6 seconds would be [tex](0.94,10.76)[/tex].

Answer:

(0.8,9.6)

Step-by-step explanation:

x=1.6*cos85* 6=0.8

y=1.6*sin85 *6=9.6