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Dmitry_Shevchenko [17]
3 years ago
6

I cannot figure out what I am doing wrong. Can someone please help me with this problem?

Mathematics
2 answers:
AveGali [126]3 years ago
7 0
You are getting the median wrong. No worry lol, trust me I used to do it alot myself. The median is 33 
WITCHER [35]3 years ago
5 0
Okay, so you have the mean, mode, and range correct. To find the medium you have to put the numbers in order from smallest to largest. Then the number that is in the middle would be the answer. So for this problem, the medium would be 33.
Plus the mode would be 33 and 34.
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4 years ago
Find the line integral of f? around the perimeter of the rectangle with corners (3,0, (3,2, (?2,2, (?2,0, traversed in that orde
satela [25.4K]
Without knowing exactly what f is, this is impossible to do. So let's assume f(x,y)=1. Then the line integral over the given rectangle will correspond to the "signed" perimeter of the region.

You don't specify that the loop is complete, so in fact the integral will only give the "signed" length of three sides.

Parameterize the region by first partitioning the contour into three sub-contours:

C_1:\mathbf r_1(t)=(3,0)(1-t)+(3,2)t=(3,2t)\implies\dfrac{\mathrm d\mathbf r_1}{\mathrm dt}=(0,2)
C_2:\mathbf r_2(t)=(3,2)(1-t)+(-2,2)t=(3-5t,2)\implies\dfrac{\mathrm d\mathbf r_2}{\mathrm dt}=(-5,0)
C_3:\mathbf r_3(t)=(-2,2)(1-t)+(-2,0)t=(-2,2-2t)\implies\dfrac{\mathrm d\mathbf r_3}{\mathrm dt}=(0,-2)

where 0\le t\le1 for each sub-contour. Then the line integral is given by

\displaystyle\int_Cf\,\mathrm dS=\int_{C_i}f(\mathbf r_i(t))\cdot\frac{\mathrm d\mathbf r_i}{\mathrm dt}

with i\in\{1,2,3\}. You have

\displaystyle\int_{C_1}f\,\mathrm dS=\int_0^1(1,1)\cdot(0,2)\,\mathrm dt=2
\displaystyle\int_{C_2}f\,\mathrm dS=\int_0^1(1,1)\cdot(5,0)\,\mathrm dt=5
\displaystyle\int_{C_1}f\,\mathrm dS=\int_0^1(1,1)\cdot(0,-2)\,\mathrm dt=-2

Then the integral over the entire contour would be 2+5-2=5. Note that if the loop is complete, then the last leg of the contour would evaluate to -5, and so the total would end up as 0. This result would also follow from the fact that f(x,y) is conservative, i.e. f(x,y)=\nabla g(x,y) for some scalar field g, and so the line integral is path independent. Its value would depend only on the endpoints of the contour, which in the case of a closed loop would simply be 0.
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3 years ago
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sammy [17]
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