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3 years ago

6 for $36.00 unit rate

Mathematics

2 answers:

Pavel [41]3 years ago

8 0

Yes, so our unit rate is 1:6

RUDIKE [14]3 years ago

3 0

The unit rate is $6 for 1

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Write the ratio as a fraction in simplest form, with whole numbers in the numerator and denominator.

MrRissso [65]

Answer:

2/5

Step-by-step explanation:

8 to 20 is 8/20=4/10=2/5

4 0

3 years ago

( 2x + 1 ) + ( 3x+2 ) =

MrRa [10]

Answer: 5x+3

Let's go over this problem step by step:

First, you need to combine all the line terms.

2x + 1 + 3x + 2

Next, you get this equation:

( 2x + 3x ) + ( 1 + 2 )

2x and 3x is 5x. 1 and 2 is 3.

So your final answer is 5x+3.

7 0

3 years ago

I need help please!!!!!!!

KATRIN_1 [288]

Answer:

Step-by-step explanation:

I don't know for sure, but I am pretty sure this is it.

4 0

2 years ago

Read 2 more answers

Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of

tresset_1 [31]

Because I've gone ahead with trying to parameterize $S$ directly and learned the hard way that the resulting integral is large and annoying to work with, I'll propose a less direct approach.

Rather than compute the surface integral over $S$ straight away, let's close off the hemisphere with the disk $D$ of radius 9 centered at the origin and coincident with the plane $y=0$ . Then by the divergence theorem, since the region $S\cup D$ is closed, we have

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iiint_R(\nabla\cdot\vec F)\,\mathrm dV$

where $R$ is the interior of $S\cup D$ . $\vec F$ has divergence

$\nabla\cdot\vec F(x,y,z)=\dfrac{\partial(xz)}{\partial x}+\dfrac{\partial(x)}{\partial y}+\dfrac{\partial(y)}{\partial z}=z$

so the flux over the closed region is

$\displaystyle\iiint_Rz\,\mathrm dV=\int_0^\pi\int_0^\pi\int_0^9\rho^3\cos\varphi\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=0$

The total flux over the closed surface is equal to the flux over its component surfaces, so we have

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iint_S\vec F\cdot\mathrm d\vec S+\iint_D\vec F\cdot\mathrm d\vec S=0$

$\implies\boxed{\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=-\iint_D\vec F\cdot\mathrm d\vec S}$

Parameterize $D$ by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec k$

with $0\le u\le9$ and $0\le v\le2\pi$ . Take the normal vector to $D$ to be

$\vec s_u\times\vec s_v=-u\,\vec\jmath$

Then the flux of $\vec F$ across $S$ is

$\displaystyle\iint_D\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^9\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_u\times\vec s_v)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{2\pi}\int_0^9(u^2\cos v\sin v\,\vec\imath+u\cos v\,\vec\jmath)\cdot(-u\,\vec\jmath)\,\mathrm du\,\mathrm dv$

$=\displaystyle-\int_0^{2\pi}\int_0^9u^2\cos v\,\mathrm du\,\mathrm dv=0$

$\implies\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\boxed{0}$

8 0

3 years ago

A batch consists of 12 defective coils and 88 good ones. Find the probability of getting two good coils when two coils are rando

Allushta [10]

Answer:

The probability of getting two good coils when two coils are randomly selected if the first selection is replaced before the second is made is 0.7744

Solution:

Total number of coils = number of good coils + defective coils = 88 + 12 = 100

p(getting two good coils for two selection) = p( getting 2 good coils for first selection ) $\times$ p(getting 2 good coils for second selection)

p(first selection) = p(second selection) = $\frac{\text { number of good coils }}{\text { total number of coils }}$

Hence, p(getting 2 good coil for two selection) = $\frac{88}{100} \times \frac{88}{100} =\bold{0.7744}$

5 0

3 years ago