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4 years ago

Explain why P(AD) and P(DIA) from the table below are not equal.

Mathematics

1 answer:

Jet001 [13]4 years ago

6 0

Answer:

P(A) ≠ P(D)

Step-by-step explanation:

P(A|D) = P(AD)/P(D)

P(D|A) = P(AD)/P(A)

The two denominators are not equal, so the quotients cannot be equal.

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Write a new function that represents a transformation to the parent function if the new function shifts right 5 units and up 3 u

irga5000 [103]

Answer:

Let's define two transformations.

Vertical translation.

If we have a function f(x), a vertical translation of N untis is written as:

g(x) = f(x) + N

If N is positive, then the translation is upwards

If N is negative, then the translation is downwards.

Horizontal translation.

If we have a function f(x), a horizontal translation of N units is written as:

g(x) = f(x - N)

if N is positive, then the translation is to the right

If N is negative, then the translation is to the left.

Now we have a function g(x) that is a transformation of a parent function f(x) (we actually do not know which parent function, so i assume f(x) = x^2) such that we have a shift right 5 units and up 3 units.

Then:

g(x) = f(x - 5) + 3

and again, using f(x) = x^2

g(x) = (x - 5)^2 + 3

6 0

3 years ago

Use stoke's theorem to evaluate∬m(∇×f)⋅ds where m is the hemisphere x^2+y^2+z^2=9, x≥0, with the normal in the direction of the

ludmilkaskok [199]

By Stokes' theorem,

$\displaystyle\int_{\partial\mathcal M}\mathbf f\cdot\mathrm d\mathbf r=\iint_{\mathcal M}\nabla\times\mathbf f\cdot\mathrm d\mathbf S$

where $\mathcal C$ is the circular boundary of the hemisphere $\mathcal M$ in the $y$ - $z$ plane. We can parameterize the boundary via the "standard" choice of polar coordinates, setting

$\mathbf r(t)=\langle 0,3\cos t,3\sin t\rangle$

where $0\le t\le2\pi$ . Then the line integral is

$\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=\int_{t=0}^{t=2\pi}\mathbf f(x(t),y(t),z(t))\cdot\dfrac{\mathrm d}{\mathrm dt}\langle x(t),y(t),z(t)\rangle\,\mathrm dt$
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We can check this result by evaluating the equivalent surface integral. We have

$\nabla\times\mathbf f=\langle1,0,0\rangle$

and we can parameterize $\mathcal M$ by

$\mathbf s(u,v)=\langle3\cos v,3\cos u\sin v,3\sin u\sin v\rangle$

so that

$\mathrm d\mathbf S=(\mathbf s_v\times\mathbf s_u)\,\mathrm du\,\mathrm dv=\langle9\cos v\sin v,9\cos u\sin^2v,9\sin u\sin^2v\rangle\,\mathrm du\,\mathrm dv$

where $0\le v\le\dfrac\pi2$ and $0\le u\le2\pi$ . Then,

$\displaystyle\iint_{\mathcal M}\nabla\times\mathbf f\cdot\mathrm d\mathbf S=\int_{v=0}^{v=\pi/2}\int_{u=0}^{u=2\pi}9\cos v\sin v\,\mathrm du\,\mathrm dv=9\pi$

as expected.

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3 years ago

128.16÷1.2<br>128.16 / 1.2

Dafna1 [17]

Answer:

106.8

Step-by-step explanation:

126.16÷106.8

thank you

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crimeas [40]

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4 years ago

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alexandr1967 [171]

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4 years ago