2, 4, 6, 7, 6, 6, 6,3 What is the Mean? What is the Median? What is the Mode? What is the Range?

Mathematics

2 answers:

kotegsom [21]3 years ago

7 0

Answer:

Mean: 5

Median: 6

Mode: 6

Range: 5

hope that helps

Nana76 [90]3 years ago

7 0

Step-by-step explanation:

Mean = 5

Median = 6

Mode = 6

Range = largest no.- smallest no.

= 7-2 = 5

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A triangle has an area of 221.16 in.² the height of the triangle is 29.1 inches what is the length of the base of the triangle

Zigmanuir [339]

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2 years ago

Let φ(x, y) = arctan (y/x) .

Alexandra [31]

Answer:

a) $\large F(x,y)=(-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2})$

b) $\large \mathbb{R}^2-\{(0,0)\}$

c) the points of the form (x, -x) for x≠0

Step-by-step explanation:

If φ(x, y) = arctan (y/x), the vector field F = ∇φ would be

$\large F(x,y)=(\frac{\partial \phi}{\partial x},\frac{\partial \phi}{\partial y})$

On one hand we have,

$\large \frac{\partial \phi}{\partial x}=\frac{\partial arctan(y/x)}{\partial x}=\frac{-y/x^2}{1+(y/x)^2}=-\frac{y/x^2}{1+y^2/x^2}=\\\\=-\frac{y/x^2}{(x^2+y^2)/x^2}=-\frac{y}{x^2+y^2}$

On the other hand,

$\large \frac{\partial \phi}{\partial y}=\frac{\partial arctan(y/x)}{\partial y}=\frac{1/x}{1+(y/x)^2}=\frac{1/x}{1+y^2/x^2}=\\\\=\frac{1/x}{(x^2+y^2)/x^2}=\frac{x}{x^2+y^2}$