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Paha777 [63]
3 years ago
10

Can someone explain INTERQUARTILE RANGE for me, and please give an example. Thank you!

Mathematics
2 answers:
Novosadov [1.4K]3 years ago
7 0
Https://www.khanacademy.org/math/cc-sixth-grade-math/cc-6th-data-statistics/cc-6th/v/calculating-interquartile-range-iqr This is a link that I gave to my little sister that helped her understand what interquartile range is. It has an example as well! I hope this helps!
AURORKA [14]3 years ago
3 0
Https://www.khanacademy.org/math/cc-sixth-grade-math/cc-6th-data-statistics/cc-6th/v/calculating-interquartile-range-iqr Link video explains everything with examples it helped me hope it helpes you
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(5x - 3) - (-4x + 1)
AveGali [126]

Answer:

9x-4

Step-by-step explanation:

add 5x and 4x

9x-3-1

subtract 1 from -3

9x-4

4 0
3 years ago
Find the area of the composite figure​
ehidna [41]
<h3>62.5 cm^2</h3>

The figure is a trapezoid, albeit not possessing the same lengths for the left and right sides. If I'm not mistaken, it has another name or term. But I forgot.

Anyway, to solve the problem, let us chop-chop or break the figure into two parts or two polygons that we are familiar with - a rectangle and a triangle. See the attached image for a clearer presentation.

After breaking the original polygon into two parts as portrayed by the red broken line, we are also going to divide the equivalent value of each side.

Remember, a rectangle has two pairs of common sides. Since the upper side of the polygon has a length of 10 cm, then, the length of the lower side (the side at the left side of the broken line) is 10, too. The same goes for the length of the left side which is 5 cm, and therefore the length of the broken line is 5 cm. Please refer to the attached image, particularly the figures in green.

From the standpoint of the triangle, based on thee figure, it has a height of 5 cm and a base of 5 cm (15 cm- 10 cm). Did you get it? If yes, let's continue.

Solving Time! :D

1. Solve for the area of the rectangle.

\begin{gathered}A_{rectangle}=L\times W\\=5\ cm \times 10\ cm\\=\bold{50\ cm^2}\end{gathered}

Arectangle=L×W=5 cm×10 cm=50 cm²

2. Solve for the area of the triangle.

\begin{gathered}A_{triangle}=\dfrac{b\times h}{2}\\=\dfrac{5\ cm\times5\ cm}{2}\\=\dfrac{25\ cm^2}{2}\\=\bold{12.5\ cm^2}\end{gathered}

Atriangle=2b×h=25 cm×5 cm=225 cm²=12.5 cm²

3. Combine (add) the two areas to find the area of the original figure.

\begin{gathered}A_{composite\ figure}=A_{rectangle}+A_{triangle}\\=50\ cm^2 +12.5\ cm^2\\=\bold{62.5\ cm^2}\end{gathered}

Acomposite figure=Arectangle+Atriangle=50 cm²+12.5 cm2=62.5 cm

If there's a Latex Error, Please click the second image attached below :)

4 0
2 years ago
Im stuck on this one​
azamat
What????????????????????????????
5 0
3 years ago
Give the radius or diameter. find the the circumference and area of each circle to the nearest tenth. Use 3.14 for pi
Vesnalui [34]

Answer:

A=π×r×r

r=D/2=24/2=12ft

A=3.14×12×12=452.16ft2

C=2πr

C=2×3.14×12

C=75.36ft

6 0
3 years ago
Varies directly with the number of
yulyashka [42]

Step-by-step explanation:

Let p and h represent pay and hours worked respectively,

p = kh

k = p/h = 148.5/18 = 8.25

p= 8.25h

90.75 = 8.25h

h = 11 hours.

Hope it helps! :)

3 0
2 years ago
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