Simplify 6t+t ????????

Identify the median and the lower quartile for this box plot 16,16.5,15.5,

MArishka [77]

The median should be 16

8 0

3 years ago

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Please help me, i promise its worth it!!!

Rashid [163]

$\\ \sf\longmapsto 2(L+B)=55$

$\\ \sf\longmapsto 2(\dfrac{4}{3}x+x)=55$

$\\ \sf\longmapsto \dfrac{8}{3}x+2x=55$

$\\ \sf\longmapsto \dfrac{8x+6x}{3}=55$

$\\ \sf\longmapsto \dfrac{14x}{3}=55$

$\\ \sf\longmapsto 14x=165$

$\\ \sf\longmapsto x=11.78$

$\\ \sf\longmapsto x\approx 12$

Now

B=x=12
L=4/3(12)=4(4)=16

5 0

2 years ago

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Let f(x,y,z) = ztan-1(y2) i + z3ln(x2 + 1) j + z k. find the flux of f across the part of the paraboloid x2 + y2 + z = 3 that li

Sophie [7]

Consider the closed region $V$ bounded simultaneously by the paraboloid and plane, jointly denoted $S$ . By the divergence theorem,

$\displaystyle\iint_S\mathbf f(x,y,z)\cdot\mathrm dS=\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV$

And since we have

$\nabla\cdot\mathbf f(x,y,z)=1$

the volume integral will be much easier to compute. Converting to cylindrical coordinates, we have

$\displaystyle\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV=\iiint_V\mathrm dV$
$=\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}\int_{z=2}^{z=3-r^2}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle2\pi\int_{r=0}^{r=1}r(3-r^2-2)\,\mathrm dr$
$=\dfrac\pi2$

Then the integral over the paraboloid would be the difference of the integral over the total surface and the integral over the disk. Denoting the disk by $D$ , we have

$\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-\iint_D\mathbf f\cdot\mathrm dS$

Parameterize $D$ by

$\mathbf s(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf j+2\,\mathbf k$
$\implies\mathbf s_u\times\mathbf s_v=u\,\mathbf k$

which would give a unit normal vector of $\mathbf k$ . However, the divergence theorem requires that the closed surface $S$ be oriented with outward-pointing normal vectors, which means we should instead use $\mathbf s_v\times\mathbf s_u=-u\,\mathbf k$ .

Now,

$\displaystyle\iint_D\mathbf f\cdot\mathrm dS=\int_{u=0}^{u=1}\int_{v=0}^{v=2\pi}\mathbf f(x(u,v),y(u,v),z(u,v))\cdot(-u\,\mathbf k)\,\mathrm dv\,\mathrm du$
$=\displaystyle-4\pi\int_{u=0}^{u=1}u\,\mathrm du$
$=-2\pi$

So, the flux over the paraboloid alone is

$\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-(-2\pi)=\dfrac{5\pi}2$

6 0

3 years ago

Pam and Amanda and Mike work at different clothing stores. Amanda made twice what Pam earned and Julie

Dmitriy789 [7]

Answer:

Pam: $181

Amanda: $362

Julie: $452

Step-by-step explanation:

(What does Mike have to do with this problem?)

Let a = Amanda's pay

Let p = Pam's pay

Let j = Julie's pay

"Amanda made twice what Pam earned"

a = 2p

"Julie made $90 more than Amanda"

j = a + 90

j = 2p + 90

Pam earned p

Total salary

a + p + j = 2p + p + 2p + 90

Total salary

$995

2p + p + 2p + 90 = 995

5p = 905

p = 181

a = 2p = 2(181) = 362

j = 2p + 90 = 362 + 90 = 452

Answer:

Pam: $181

Amanda: $362

Julie: $452

8 0

3 years ago

Help please............................

Vesnalui [34]

Step-by-step explanation:

csc x / (cot x + tan x)

Write in terms of sine and cosine.

(1 / sin x) / [(cos x / sin x) + (sin x / cos x)]

Multiply top and bottom by sin x.

1 / [cos x + (sin²x / cos x)]

Multiply top and bottom by cos x.

cos x / (cos²x + sin²x)

Use Pythagorean identity.

cos x / 1

cos x

3 0

3 years ago