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3 years ago

If the relative frequencies are 0.48 and 0.52, which conclusion is most likely supported by the data?

2 answers:

8 0

I think the closest possible answer to this question is There is likely an association between the categorical variables because the relative frequencies are both close to 0.50.Thank you for your question. Please don't hesitate to ask in Brainly your

Anna007 [38]3 years ago

4 0

The right answer for the question that is being asked and shown above is that: "D.) There is likely an association between the categorical variables because the relative frequencies are both close to 0.50." Given that a relative frequencies of 0.48 and 0.52, there will be an association between the categorical variables.

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Use Stokes' Theorem to evaluate C F · dr where C is oriented counterclockwise as viewed from above. F(x, y, z) = 2yi + xzj + (x

Lady_Fox [76]

By Stokes' theorem, the line integral of $\vec F$ over $C$ is given by the surface integral of the curl of $\vec F$ over $S$ , where $S$ is the region of intersection of the plane $z=y+6$ and the cylinder $x^2+y^2=1$ with $S$ having positive/upward orientation.

Parameterize $S$ by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+(u\sin v+6)\,\vec k$

with $0\le u\le1$ and $0\le v\le2\pi$ .

Take the normal vector to $S$ to be

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

The curl of $\vec F$ is

$\nabla\times\vec F(x,y,z)=(1-x)\,\vec\imath-\vec\jmath+(z-2)\,\vec k$

Then the line integral is equivalent to

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

$=\displaystyle\int_0^{2\pi}\int_0^1\bigg((1-u\cos v)\,\vec\imath-\vec\jmath+(u\sin v+4)\,\vec k\bigg)\cdot\bigg(-u\,\vec\jmath+u\,\vec k\bigg)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{2\pi}\int_0^1(5u+u^2\sin v)\,\mathrm du\,\mathrm dv=\boxed{5\pi}$

4 0

3 years ago

Help me show work please

masha68 [24]

Step-by-step explanation:

The point-slope form of an equation of a line:

$y-y_1=m(x-x_1)$

m - slope

(x₁, y₁) - given point

$\bullet\ (2,\ 7),\ m=-4\\\\y-7=-4(x-2)\\\\\bullet(-6,\ -2),\ m=3\\\\y-(-2)=3(x-(-6))\to y+2=3(x+6)\\\\\bullet(12,\ 5),\ m=-3\\\\y-5=-3(x-12)\\\\\bullet(7,\ -6),\ m=\dfrac{1}{2}\\\\y-(-6)=\dfrac{1}{2}(x-7)\to y+6=\dfrac{1}{2}(x-7)\\\\\bullet(4,\ -5),\ m=6\\\\y-(-5)=6(x-4)\to y=y+5=6(x-4)\\\\\bullet(-8,\ 2),\ m=-\dfrac{3}{4}\\\\y-2=-\dfrac{3}{4}(x-(-8))\to y-2=-\dfrac{3}{4}(x+8)$

8 0

3 years ago

The average fine for entering a National Forest Wilderness without a permit increased from

galina1969 [7]

Fine increase = 78 - 65 = 13
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Discounted amount = 560 - 476 = 84
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5 0

3 years ago

the cat is a high speed catamaran auto ferry that operates between City A and City B. The cat can make the trip in 1 1/2 hours a

malfutka [58]

City A and City B are about 76.5 miles apart because 51*1.5 = 76.5

3 0

3 years ago

The polygons in each pair are similar. find the missing side length

balandron [24]

Answer:

Missing length = 20

Step-by-step explanation:

Ratio is 16/20=4/5

25×4/5=20

8 0

3 years ago