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AlladinOne [14]
4 years ago
6

How do u find a association in a two way frequency table :) pls help I need the answer right now lol

Mathematics
1 answer:
Sedaia [141]4 years ago
7 0
Sometimes the best way to tell whether two variables are associated is to ask yourself whether they are not associated. Think backward. In a two-way frequency table, if the relative frequencies for one variable are the same (or close) for all categories of another variable, there is no (or little) association.
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Gerald has a punch bowl in the shape of a hemisphere as shown below.
zmey [24]

The closest to the maximum number of cups the punch bowl can hold is 30

<h3>How to determine the number of cups?</h3>

The given parameters are:

1 cup = 15 cubic inches

Diameter of bowl, d = 12 inches

The radius is the half of the diameter.

So, we have:

r = 6

The volume of the bowl is then calculated using:

V = \frac{2}{3}\pi r^3

This gives

V = \frac{2}{3}\pi * 6^3

Evaluate

V = 452.16

The maximum number of cups is then calculated using:

Cups = 452.16/15

Evaluate

Cups = 30.1444

Approximate

Cups = 30

Hence, the closest to the maximum number of cups the punch bowl can hold is 30

Read more about volumes at:

brainly.com/question/1972490

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6 0
2 years ago
12b−15&gt;21?? How is this done? Need help ASAP
dangina [55]
12b-15>21
+15 +15

12b > 36
----- ----
12 12

b = 3
3 0
4 years ago
Read 2 more answers
Passe os numeros a seguir para notação cientifica.
loris [4]

Answer:

ingles Hablas ingles?

5 0
3 years ago
Can somebody help me? Thanks!
ICE Princess25 [194]

Answer: both our A

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8 0
3 years ago
Solve for x9(x + 1) = 25 + x
vaieri [72.5K]
When you expand the equation, you will have
9x + 9=25 + x
Next you have to collect like terms
+x will go over the equality sign to become -x. The same applies to +9
9x - x=23 - 9
9x minus x is 8x
8x=14
Divide both sides by 8
x=14/8
You will have 1 6/8
6 is the remainder while 1 is the number of times 8 divided 14.
In decimal, it will be 1.75
Hope that helped. Good luck
8 0
3 years ago
Read 2 more answers
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