anonymous
  • anonymous
I'm lost, Estimate the area of the shaded region in the graph by using the Trapezoidal Rule with n=4.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
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amistre64
  • amistre64
trap rule eh ...
anonymous
  • anonymous
Yeah, that one...

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amistre64
  • amistre64
the area of a trap is average of the height * width f(1)+f(2) ------- * 1 if the width is 1 that is 2
amistre64
  • amistre64
do it for each section
amistre64
  • amistre64
\[\frac{f(0)+f(1)}{2}*\frac{b-a}{4}+\frac{f(1)+f(2)}{2}*\frac{b-a}{4}+\frac{f(2)+f(3)}{4}*\frac{b-a}{4}+\frac{f(3)+f(4)}{4}*\frac{b-a}{4}\] \[\frac{b-a}{2*4}(f(0)+f(1)+f(1)+f(2)+f(2)+f(3)+f(3)+f(4))\] \[\frac{b-a}{2*4}(f(0)+2f(1)+2f(2)+2f(3)+f(4))\]
anonymous
  • anonymous
But how? They don't even label the graph,
amistre64
  • amistre64
0, 6, 9.5 .. just count and estimate the best you can with the intersections
amistre64
  • amistre64
0, 6, 9.5 ,4, 2 0 + 12 + 19 + 8 + 2 41/8 = 5*(b-a)
amistre64
  • amistre64
b-a looks to be 16
amistre64
  • amistre64
41/8 * 16 = 41*2 = 82 or so? does that fit an answer?
anonymous
  • anonymous
Possible choices are: 12 10.5 6.9 8.5 7.4 12.5
anonymous
  • anonymous
Kinda accurate for an estimate
amistre64
  • amistre64
i counted each little blip as 4 so im a little off
amistre64
  • amistre64
each blip might be 1/4
amistre64
  • amistre64
0+2*1.5+2*2.5+2*1+.5 3+5+2+.5 = 10.5 4 ----- 42.0/(2*4) 42/8 = 5.somehting as far as I can tell
anonymous
  • anonymous
Which still isn't close to any of their answers, hmm
amistre64
  • amistre64
whered you get this program at; biglots?
anonymous
  • anonymous
what?
amistre64
  • amistre64
biglots is a store that has a rep of subpar merchandise :)
anonymous
  • anonymous
Yeah, lol I know that, I was wondering what you meant by "program"
amistre64
  • amistre64
:) i assumed you were working this from a program
anonymous
  • anonymous
Ah, no I was trying to do it by good ole' paper :P
amistre64
  • amistre64
well, the concept I believe is sound: b-a/n = width and the average of yor heights is the height b*h = area for each section
anonymous
  • anonymous
I found one similar online where I believe they got 11.5, http://www.csuchico.edu/~rford/MATH7BWEB/Math7BQuizzes/Math%207B%20Exam%202%20Review
amistre64
  • amistre64
do we have numbers to go with yours?
anonymous
  • anonymous
nope...
amistre64
  • amistre64
then all we can do is assume a basic interval if 1 per "tic" that i know of
anonymous
  • anonymous
but then one box is over 16
amistre64
  • amistre64
4 per box would be 4 tics wide
amistre64
  • amistre64
0 to 6 has an avg of 3 6 to 10 has an avg of 8 10 to 4 has an avg of 7 4 to 2 has an avg of 3 4(3+3+8+7) 4(21) = 84 total
amistre64
  • amistre64
other than that, its just a guessing game to figure our what it is they are scaling hteir tic marks as
anonymous
  • anonymous
hmm, I do believe that I got points taken off in Physics for not labeling my graphs like that, lol
anonymous
  • anonymous
anonymous
  • anonymous
Hey, I called him first :P

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