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3 years ago

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Assume the random variable x is normally distributed with mean u = 82 and standard deviation o =5. Find the indicated probabilit

y

1 answer:

jasenka [17]3 years ago

7 0

P(
x86)
P(x < 86) =
(Round to four decimal places as needed

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PLEASE ANSWER ASAP!!

Rufina [12.5K]

Answer:

The instantaneous rate of increase of f(x) at $x_{0} = 3$ is 3.
One possible equation of this line is y = 3x - 16.
The line is tangent to the graph of y = f(x). The slope of the line is the same as the instantaneous rate of increase of f(x) at $x_{0} = 3$ .

Step-by-step explanation:

<h3>1.</h3>

$\begin{aligned}&\lim_{h\to 0}{\frac{f(3+h)-f(h)}{h}}\\ =&\lim_{h\to 0}\frac{1}{h} \cdot [(3^{3} + 3 \times 3^{2}h +3\times 3h^{2} + h^{3})- 4(3^{2} + 2\times 3h + h^{2}) + 2 \\[-1em] &\phantom{\lim_{h\to 0}\frac{1}{h}\cdot []} -(3^{3} -4\times 3^{2} + 2)]\\ = &\lim_{h \to 0}\frac{h^{3} + 5h^{2} +3h}{h}\\=& \lim_{h \to 0}{h^{2}} + \lim_{h \to 0}{5h} + \lim_{h\to 0}{3}\\ =& 3\end{aligned}$ .

<h3>2.</h3>

$f(3) = 3^{3} - 4\times 3^{2} + 2 =-7$ .

In other words, the graph of y = f(x) passes through the point (3, -7) where $x = 3$ .

The point-slope form of a line in a cartesian plane is:

$y - y_0 = m (x - x_0)$ .

For this line,

$(3, -7)$ is the point on the line, while

$3$ is the slope of the line.

The equation of this line will thus be

$y - (-7) = 3(x - 3)$ .

That's equivalent to

$y = 3x - 16$ .

<h3>3.</h3>

Refer to the diagram attached. The line touches the graph of y = f(x) at x = 3 without crossing it. The line here is thus a tangent to the graph of y = f(x) at x = 3. The slope of the line represents the instantaneous rate of increase of f(x) at $x_{0} = 3$ .

4 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

$\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=\int_{t=0}^{t=2\pi}\mathbf f(x(t),y(t),z(t))\cdot\dfrac{\mathrm d}{\mathrm dt}\langle x(t),y(t),z(t)\rangle\,\mathrm dt$
$=\displaystyle\int_0^{2\pi}\langle0,0,3\cos t\rangle\cdot\langle0,-3\sin t,3\cos t\rangle\,\mathrm dt=9\int_0^{2\pi}\cos^2t\,\mathrm dt=9\pi$

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

Note: Enter your answer and show all the steps that you use to solve this problem in the space provided.

agasfer [191]

Answer:

rectangle

Step-by-step explanation:

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3 years ago

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The current price of a pair if snowshoes is 129.99$. The price went down 13% and there is a tax of 5% what is the answer?

Doss [256]

129.99*0.13
129.99-16.90
113.09*0.05
5.65+113.09
118.74

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3 years ago

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31600 rounded to the nearest thousand is what

Zigmanuir [339]

32000 is your answer

hope this helps

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3 years ago