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3 years ago

What can you tell about the mean of each distribution ?

Mathematics

2 answers:

postnew [5]3 years ago

7 0

There isn’t enough info provided

Gwar [14]3 years ago

4 0

You can tell this is like sharing something out among people like distributing

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Instructions: Find the surface area of the figure below. Round your answers to the nearest tenth, if necessary.

Y_Kistochka [10]

The surface area of the figure given is of: 61.3 yd².

<h3>What is the combined surface area of a figure?</h3>

The combined surface area of a figure is the area of each figure.

This figure is composed by:

A circle of radius 3 yd.

A triangle with base 6 yd and one leg 11.4 yd.

The area of the circle is given as follows:

$A = \pi r^2 = \pi \times 3^2 = 9\pi$

For the triangle, we need to find it's height, which is the <u>leg of a right triangle, in which one of the sides is 3 and the hypotenuse is 11.4</u>, hence:

h² + 3² = 11.4²

$h = \sqrt{11.4^2 - 3^2}$

h = 11 yd.

Hence the area of the triangle is:

A = 0.5 x 6 x 11 = 33 yd²

The surface area of the figure is:

9pi + 33 yd² = 61.3 yd².

More can be learned about the surface area of at brainly.com/question/21208177

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6 0

2 years ago

What is the volume of the square pyramid with base edges 12 cm and height 12 cm?

aliya0001 [1]

Answer:

V=144

Step-by-step explanation:

To find the volume of a pyramid, use the formula V=B•h, B being the base and h the height. So 12•12=144.

4 0

3 years ago

Read 2 more answers

Sketch the domain D bounded by y = x^2, y = (1/2)x^2, and y=6x. Use a change of variables with the map x = uv, y = u^2 (for u ?

cluponka [151]

Under the given transformation, the Jacobian and its determinant are

$\begin{cases}x=uv\\y=u^2\end{cases}\implies J=\begin{bmatrix}v&u\\2u&0\end{bmatrix}\implies|\det J|=2u^2$

so that

$\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\iint_{D'}\frac{2u^2}{u^2}\,\mathrm du\,\mathrm dv=2\iint_{D'}\mathrm du\,\mathrm dv$

where $D'$ is the region $D$ transformed into the $u$ - $v$ plane. The remaining integral is the twice the area of $D'$ .

Now, the integral over $D$ is

$\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\left\{\int_0^6\int_{x^2/2}^{x^2}+\int_6^{12}\int_{x^2/2}^{6x}\right\}\frac{\mathrm dx\,\mathrm dy}y$

but through the given transformation, the boundary of $D'$ is the set of equations,

$\begin{array}{l}y=x^2\implies u^2=u^2v^2\implies v^2=1\implies v=\pm1\\y=\frac{x^2}2\implies u^2=\frac{u^2v^2}2\implies v^2=2\implies v=\pm\sqrt2\\y=6x\implies u^2=6uv\implies u=6v\end{array}$

We require that $u>0$ , and the last equation tells us that we would also need $v>0$ . This means $1\le v\le\sqrt2$ and $0$ , so that the integral over $D'$ is

$\displaystyle2\iint_{D'}\mathrm du\,\mathrm dv=2\int_1^{\sqrt2}\int_0^{6v}\mathrm du\,\mathrm dv=\boxed6$

4 0

3 years ago

Someone help me with the steps please

Vesna [10]

Answer:

Let $u=3x^2-3x$

Solution: $\frac{1}{6} \sqrt{3x^2-3x}+C$

Step-by-step explanation:

See attached for step-by-step

8 0

2 years ago

Jacob signed up for a streaming music service that costs $6 per month. The service allows Jacob to listen to unlimited music, bu

tresset_1 [31]

Answer: $16

with the base being $6, you can write the equation 6 + .5x = y

“x” is represented by the number of songs downloaded, so we can replace x with 20

y is the total amount of money

6 + .5(20) = y

now, all you have to do is solve

6 + 10 = y

16 = y

5 0

2 years ago