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Orlov [11]
3 years ago
10

What is the interquartile range of this data?

Mathematics
1 answer:
jok3333 [9.3K]3 years ago
6 0
Hello!

To find the interquartile range you subtract the lower quartile by the higher quartile

33 - 25 = 8

The answer is C)8

Hope this helps!
You might be interested in
A construction crew is lengthening a road that originally measured 47 miles. The crew is adding one mile to the road each day. L
Dmitry_Shevchenko [17]

Answer:

78 miles

Step-by-step explanation:

Given that:

Original length, L = 47 miles

Additional length (miles) added per day, = 1 mile

Representing as an equation :

L(D) = original length + additional length per day * number of days

Let, D = number of days

L(D) = 47 + D

Length after 31 days :

L(31) = 47 + 31

= 78 miles

3 0
3 years ago
What is the gross pay per pay period?
Vinil7 [7]

Divide 71,290 with 2

71,290 / 2 = 35,645

Then, divide 35,645 with 12 ( because theres 12 months in 1 year )

35,645 / 12 = 2970.42

a) 2970.42 is your answer.

3 0
3 years ago
Which of the Following is the solution to 4|x+3| greater than equal to 8? Options are as followed: A.x less then equal to -5 or
Fynjy0 [20]
D x greater than equal to -5 and x greater y’all
6 0
3 years ago
Explain one way to add 3 digit numbers
Ray Of Light [21]
You can add them like this:

so if 156 + 213 + 109 was the equation then you would

first put them in order from biggest to smallest

213
156
+109
-----------

then add:

¹ ← ←
213 ↑
156 ↑
+109 ↑
---------↑ ( carry the 10 )
478 (add the ¹ )

then you get the answer 478!

hope that helps :)
4 0
3 years ago
Write an equation for an ellipse centered at the origin, which has foci at (\pm\sqrt{8},0)(± 8 ​ ,0)(, plus minus, square root o
Alchen [17]

Answer:

The equation of ellipse centered at the origin

\frac{x^2}{18} +\frac{y^2}{10} =1

Step-by-step explanation:

given the foci of ellipse (±√8,0) and c0-vertices are (0,±√10)

The foci are (-C,0) and (C ,0)

Given data (±√8,0)  

the focus has x-coordinates so the focus is  lie on x- axis.

The major axis also lie on x-axis

The minor axis lies on y-axis so c0-vertices are (0,±√10)

given focus C = ae = √8

Given co-vertices ( minor axis) (0,±b) = (0,±√10)

b= √10

The relation between the focus and semi major axes and semi minor axes are c^2=a^2-b^2

      a^{2} = c^{2} +b^{2}

a^{2} = (\sqrt{8} )^{2} +(\sqrt{10} )^{2}

a^{2} =18

a=\sqrt{18}

The equation of ellipse formula

\frac{x^2}{a^2} +\frac{y^2}{b^2} =1

we know that a=\sqrt{18} and b=\sqrt{10}

<u>Final answer:</u>-

<u>The equation of ellipse centered at the origin</u>

<u />\frac{x^2}{18} +\frac{y^2}{10} =1<u />

                                   

8 0
3 years ago
Read 2 more answers
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