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anonymous

  • 4 years ago

I'm lost, Estimate the area of the shaded region in the graph by using the Trapezoidal Rule with n=4.

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  1. anonymous
    • 4 years ago
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  2. amistre64
    • 4 years ago
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    trap rule eh ...

  3. anonymous
    • 4 years ago
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    Yeah, that one...

  4. amistre64
    • 4 years ago
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    the area of a trap is average of the height * width f(1)+f(2) ------- * 1 if the width is 1 that is 2

  5. amistre64
    • 4 years ago
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    do it for each section

  6. amistre64
    • 4 years ago
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    \[\frac{f(0)+f(1)}{2}*\frac{b-a}{4}+\frac{f(1)+f(2)}{2}*\frac{b-a}{4}+\frac{f(2)+f(3)}{4}*\frac{b-a}{4}+\frac{f(3)+f(4)}{4}*\frac{b-a}{4}\] \[\frac{b-a}{2*4}(f(0)+f(1)+f(1)+f(2)+f(2)+f(3)+f(3)+f(4))\] \[\frac{b-a}{2*4}(f(0)+2f(1)+2f(2)+2f(3)+f(4))\]

  7. anonymous
    • 4 years ago
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    But how? They don't even label the graph,

  8. amistre64
    • 4 years ago
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    0, 6, 9.5 .. just count and estimate the best you can with the intersections

  9. amistre64
    • 4 years ago
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    0, 6, 9.5 ,4, 2 0 + 12 + 19 + 8 + 2 41/8 = 5*(b-a)

  10. amistre64
    • 4 years ago
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    b-a looks to be 16

  11. amistre64
    • 4 years ago
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    41/8 * 16 = 41*2 = 82 or so? does that fit an answer?

  12. anonymous
    • 4 years ago
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    Possible choices are: 12 10.5 6.9 8.5 7.4 12.5

  13. anonymous
    • 4 years ago
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    Kinda accurate for an estimate

  14. amistre64
    • 4 years ago
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    i counted each little blip as 4 so im a little off

  15. amistre64
    • 4 years ago
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    each blip might be 1/4

  16. amistre64
    • 4 years ago
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    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

  17. anonymous
    • 4 years ago
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    Which still isn't close to any of their answers, hmm

  18. amistre64
    • 4 years ago
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    whered you get this program at; biglots?

  19. anonymous
    • 4 years ago
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    what?

  20. amistre64
    • 4 years ago
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    biglots is a store that has a rep of subpar merchandise :)

  21. anonymous
    • 4 years ago
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    Yeah, lol I know that, I was wondering what you meant by "program"

  22. amistre64
    • 4 years ago
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    :) i assumed you were working this from a program

  23. anonymous
    • 4 years ago
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    Ah, no I was trying to do it by good ole' paper :P

  24. amistre64
    • 4 years ago
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    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

  25. anonymous
    • 4 years ago
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    I found one similar online where I believe they got 11.5, http://www.csuchico.edu/~rford/MATH7BWEB/Math7BQuizzes/Math%207B%20Exam%202%20Review

  26. amistre64
    • 4 years ago
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    do we have numbers to go with yours?

  27. anonymous
    • 4 years ago
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    nope...

  28. amistre64
    • 4 years ago
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    then all we can do is assume a basic interval if 1 per "tic" that i know of

  29. anonymous
    • 4 years ago
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    but then one box is over 16

  30. amistre64
    • 4 years ago
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    4 per box would be 4 tics wide

  31. amistre64
    • 4 years ago
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    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

  32. amistre64
    • 4 years ago
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    other than that, its just a guessing game to figure our what it is they are scaling hteir tic marks as

  33. anonymous
    • 4 years ago
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    hmm, I do believe that I got points taken off in Physics for not labeling my graphs like that, lol

  34. anonymous
    • 4 years ago
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    i need help on http://openstudy.com/study#/updates/4f3084efe4b0fc09381eed4a asap pretty please!!!!

  35. anonymous
    • 4 years ago
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    Hey, I called him first :P

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spraguer (Moderator)
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is replying to Can someone tell me what button the professor is hitting...

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