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astraxan [27]
3 years ago
13

Which better describes the data set, the mean or the median?

Mathematics
2 answers:
Xelga [282]3 years ago
8 0
The median...(hope it helps)
Serga [27]3 years ago
3 0
You would need to give us the data set to answer this question. I'm assuming this is a stats question which means I'd need to see what set of data you are working with to determine if the mean or median is best to describe the center. The meadian is resistant to outlying values while the mean is not. Therefore the meadian is a better describer for skewed distributions. Whenever you have outliers the dataset is skewed. Take a look at the shape of the data. Does it look skewed? You can make a simple graph with the data points to determine if there are outliars if you aren't able to tell just from looking at the data. (Again I don't know what data you are working with, but I hope this helped)
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Kobotan [32]
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4x + 6 + 3 = 17
4x + 9 = 17

Then, subtract 9 on each side:

4x + 9 - 9 = 17 - 9
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Then divide by 4 on each side to get the x by itself:

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Emmanuel carves a cone-shaped hollow out of a solid clay cylinder. What is the approximate volume of the solid portion of the cy
bixtya [17]

Answer:

3.14r^2(h-\frac{1}{3}h_1)

Step-by-step explanation:

Let h be the cylinders height and r the radius.

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V=\pi r^2h

-Since the cone is within the cylinder, it has the same radius as the cylinder.

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V=\pi r^2 \frac{h}{3}\\\\=\frac{1}{3}\pi r^2h_1

The volume of the  solid section of the cylinder is calculated by subtracting the cone's volume from the cylinders:

V=V_{cy}-V_{co}\\\\=\pi r^2h-\frac{1}{3}\pi r^2 h_1, \pi=3.14\\\\=3.14r^2(h-\frac{1}{3}h_1)

Hence, the approximate area of the solid portion is 3.14r^2(h-\frac{1}{3}h_1)

5 0
3 years ago
A rock weighs 80 g on a triple beam scale. The same rock is then gently placed inside a graduated cylinder containing 30 mL of w
Alexxandr [17]

Answer:

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Step-by-step explanation:

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<u>Density = mass / volume:</u>

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3 0
2 years ago
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