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2 years ago

Draw a box and whisker plot of the following data 40,37,29,36,41,39,33

Mathematics

1 answer:

Setler79 [48]2 years ago

4 0

Answer:

Step-by-step explanation:

Its not the best drawing but here

median 37

minimum 29

maximum 41

first quartile 33

third quartile 40

IQR 7

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Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S.

sweet-ann [11.9K]

Answer:

2.794

Step-by-step explanation:

Recall that if G(x,y) is a parametrization of the surface S and F and G are smooth enough then

$\bf \displaystyle\iint_{S}FdS=\displaystyle\iint_{R}F(G(x,y))\cdot(\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y})dxdy$

F can be written as

F(x,y,z) = (xy, yz, zx)

and S has a straightforward parametrization as

$\bf G(x,y) = (x, y, 3-x^2-y^2)$

with 0≤ x≤1 and 0≤ y≤1

$\bf \displaystyle\frac{\partial G}{\partial x}= (1,0,-2x)\\\\\displaystyle\frac{\partial G}{\partial y}= (0,1,-2y)\\\\\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y}=(2x,2y,1)$

we also have

$\bf F(G(x,y))=F(x, y, 3-x^2-y^2)=(xy,y(3-x^2-y^2),x(3-x^2-y^2))=\\\\=(xy,3y-x^2y-y^3,3x-x^3-xy^2)$

and so

$\bf F(G(x,y))\cdot(\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y})=(xy,3y-x^2y-y^3,3x-x^3-xy^2)\cdot(2x,2y,1)=\\\\=2x^2y+6y^2-2x^2y^2-2y^4+3x-x^3-xy^2$

we just have then to compute a double integral of a polynomial on the unit square 0≤ x≤1 and 0≤ y≤1

$\bf \displaystyle\int_{0}^{1}\displaystyle\int_{0}^{1}(2x^2y+6y^2-2x^2y^2-2y^4+3x-x^3-xy^2)dxdy=\\\\=2\displaystyle\int_{0}^{1}x^2dx\displaystyle\int_{0}^{1}ydy+6\displaystyle\int_{0}^{1}dx\displaystyle\int_{0}^{1}y^2dy-2\displaystyle\int_{0}^{1}x^2dx\displaystyle\int_{0}^{1}y^2dy-2\displaystyle\int_{0}^{1}dx\displaystyle\int_{0}^{1}y^4dy+\\\\+3\displaystyle\int_{0}^{1}xdx\displaystyle\int_{0}^{1}dy-\displaystyle\int_{0}^{1}x^3dx\displaystyle\int_{0}^{1}dy-\displaystyle\int_{0}^{1}xdx\displaystyle\int_{0}^{1}y^2dy$

=1/3+2-2/9-2/5+3/2-1/4-1/6 = 2.794

5 0

2 years ago

Plz do this question if you did I will very appreciated you. plz.

Nastasia [14]

The area formula of a circle is $r^{2}$ × π.

With that being said, we can solve both problems!

Starting with part A:

-The radius we know is 21cm. From there, we can substitute in our formula with the new value.

$21^{2}$ x π

From there, we can square our radius:

(21 x 21) x π

(441) x π

And finally, we can multiply by pi to get our answer.

(441) x π

1385.44236023

Which rounded to the nearest tenth is 1385.4. Therefore, the area of the porthole is 1385.4 centimetres.

The second part can be followed with the same process, hopefully you understand the concept enough to where you're comfortable solving problems like this on your own. Feel free to reach out if you have any further trouble!

<em>Hope this helped! :)</em>

5 0

2 years ago

Can I get some help please?

Lynna [10]

The answer is 15. Just count the spaces between A and D.

3 0

2 years ago

Will Mark The Brainiest that is extra points aka FULL POINTS

valentina_108 [34]

Hello!

We can solve this by using substitution

we can plug y = x - 1 into the other equation

So 2x - 3( x - 1) = -1

We can distribute the -3 to (x - 1)

2x - 3x + 3 = -1

Combine like terms

-x + 3 = -1

Subtract 3 from both sides

-x = -4

Since x is negative we can multiply both sides by -1

x = 4

We can plug x into the first equation

y = (4) - 1

y = 3

The answers are x = 4 and y = 3

Hope this helps!

6 0

3 years ago

Read 2 more answers

What exactly is infinity? How far is it? I know it goes on forever, but does it stop? Have people ever counted so high?

defon

Infinity would be consider as an Unsolved problem it doesn't have an answer it Like an decimal that doesn't stop.

Example 0.3333333333333333333

6 0

2 years ago