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3 years ago

Which measure of center (median or mean) best describes the following data about prices of

Mathematics

1 answer:

krok68 [10]3 years ago

4 0

Answer:

$16

Step-by-step explanation:

mean is the average cost of those earrings

add all of them together and divide by how many earrings

4+50+10+12+10+20+12+10 = 128

128/8= 16

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A line has a slope of zero and passes through the point (-2, 3). Which of the following points must also lie on the line?

brilliants [131]

Answer:

(5,3)

Step-by-step explanation

Because the slope is 0, it means its a horizontal line. because the horizontal is on 3, then it doesnt matter what x is as long as y is 3 if that makes any sense.

7 0

3 years ago

C=2(y-k) solve for y

marshall27 [118]

Answer:

$\blue{y = \dfrac{C + 2k}{2}}$

Step-by-step explanation:

$C = 2(y - k)$

$C = 2y - 2k$

$C + 2k = 2y$

$\dfrac{C + 2k}{2} = y$

$y = \dfrac{C + 2k}{2}$

5 0

3 years ago

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

4x5 – 16x2 + 13x8 in standard form

FinnZ [79.3K]

Answer: I think standard form is when you you write it in thousands

Step-by-step explanation:

8 0

4 years ago

Solve using Quadratic Formula <br>x^2 + 3x - 3 = 0

katrin [286]

Answer:

a= 1 , b= 3 , c= –3

$x = \frac{ - b \frac{ + }{} \sqrt{ {b}^{2} - 4ac} }{2a} \\ x = \frac{ - 3 \frac{ + }{} {} \sqrt{ {3}^{2} - 4(1)( - 3)} }{2(1)} \\ = \frac{ - 3 \frac{ + }{} \sqrt{9 + 12} }{2} = \frac{ - 3 \frac{ + }{} \sqrt{21} }{2} \\ \\ x = \frac{ - 3 \frac{ + }{} \sqrt{21} }{2} \\ \\ x = 0.79 \\ x = - 3.79$

I hope I helped you^_^

4 0

3 years ago