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anonymous

  • one year ago

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  1. anonymous
    • one year ago
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    This is an easy one.

  2. anonymous
    • one year ago
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    Ok first you try finding the deer population in the first year. Do you know what you would do?

  3. anonymous
    • one year ago
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    um no not sure

  4. anonymous
    • one year ago
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    Well as you said, the rate of increase is 3%. All you would do for the first year, is to multiply the original amount (1253) by 3% or .03

  5. anonymous
    • one year ago
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    Does that make sense on why you would do that?

  6. anonymous
    • one year ago
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    oh okay yeah that makes sense.

  7. anonymous
    • one year ago
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    what did you get for your answer?

  8. anonymous
    • one year ago
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    Ok good, now since the deer population is INCREASING do you have an idea on what you would do with that number?

  9. anonymous
    • one year ago
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    umm something with the 6?

  10. anonymous
    • one year ago
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    I'll get to the 6 later. Since we did the first year, we will add 37.59 to the ORIGINAL 1253 deer.

  11. anonymous
    • one year ago
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    does that make sense?

  12. anonymous
    • one year ago
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    so 37.59 + 1253?

  13. anonymous
    • one year ago
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    correct. remember its INCREASING, so you will add.

  14. anonymous
    • one year ago
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    I got 1290.59

  15. anonymous
    • one year ago
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    yes! OK we just finished year 1, now since its increasing by 3% every year, you must do the same thing we just did for the next year. what do you think that is?

  16. anonymous
    • one year ago
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    37.59 + 1290.59 ?

  17. anonymous
    • one year ago
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    You have the right idea. You will take 1290.59 and take 3% of that number.

  18. anonymous
    • one year ago
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    oh okay

  19. anonymous
    • one year ago
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    you take 3% of the previous year, and add that to the new population.

  20. anonymous
    • one year ago
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    what level of schooling is it though? Is it possible they want you to use an exponential function? likeP= A^(r t) where P is population A is starting amount r is the rate t is the term?

  21. anonymous
    • one year ago
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    or is Rstones method okay? because they will both work.

  22. anonymous
    • one year ago
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    this is 8th year and I think they want me to write a function that represents the deer population.

  23. anonymous
    • one year ago
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    that models the deer population actually.

  24. anonymous
    • one year ago
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    do you know if they want you to use e like eulers number?

  25. anonymous
    • one year ago
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    yeah I think they want that exponential function in there because that's what the lesson is about

  26. anonymous
    • one year ago
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    well that would've helped

  27. anonymous
    • one year ago
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    ill let hugh explain it.

  28. anonymous
    • one year ago
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    its cool man, you can do it if you like, didn't mean to hijack your help

  29. anonymous
    • one year ago
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    sorry I mean it didn't specifically say to use that, but since that was what some of the lesson talked about... I didn't know if it needed to be used

  30. anonymous
    • one year ago
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    these are questions at the end of the lessons

  31. anonymous
    • one year ago
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    were they talking about continuous compounding ?

  32. anonymous
    • one year ago
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    I barely know what any of this means, I missed the class

  33. anonymous
    • one year ago
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    umm nope I do think it just needs to be written as exponential function to get the answer

  34. anonymous
    • one year ago
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    ok 1 sec... lol.. dusting off my cobwebs...

  35. anonymous
    • one year ago
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    my book has this function y=y(0)e^kt

  36. anonymous
    • one year ago
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    if that helps at all

  37. anonymous
    • one year ago
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    When critters and bacteria multiply they use a compounding exponential function.. and the euler number. the function would look like. yah awesome... thats it..

  38. anonymous
    • one year ago
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    so your y(0) is your starting population

  39. anonymous
    • one year ago
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    so y(0)= 1253 then

  40. anonymous
    • one year ago
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    good, the k is the rate at which they grow.. as a decimal of 1..

  41. anonymous
    • one year ago
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    so not 3 .. but ..?

  42. anonymous
    • one year ago
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    k = 3% or 0.03

  43. anonymous
    • one year ago
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    good one

  44. anonymous
    • one year ago
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    and then t is your term :)

  45. anonymous
    • one year ago
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    term? the 6? that's the only one left

  46. anonymous
    • one year ago
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    that's it.. a term of 6 years.

  47. anonymous
    • one year ago
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    okay

  48. anonymous
    • one year ago
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    well wait

  49. anonymous
    • one year ago
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    to be a function... then you would want a function of t... so...

  50. anonymous
    • one year ago
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    f[t] = y(0) E^(k t) and you would plug in your value for y(0)

  51. anonymous
    • one year ago
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    y(6) ?

  52. anonymous
    • one year ago
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    or 1253

  53. anonymous
    • one year ago
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    and your constant for the rate of growth (k)

  54. anonymous
    • one year ago
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    yes starting population = y(0)

  55. anonymous
    • one year ago
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    1253

  56. anonymous
    • one year ago
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    So you should have something that looks like f(t) = 1253 e^(0.03 t)

  57. anonymous
    • one year ago
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    so 1253 e^(0.03*6)

  58. anonymous
    • one year ago
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    you ahve the correct equation.. and if you need to make it a formula instead.. then you keep the t, and set a parameter in your function for it.

  59. anonymous
    • one year ago
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    i'm getting 1500

  60. anonymous
    • one year ago
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    I have a choice that is 1456. That's the closest to that

  61. anonymous
    • one year ago
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    looks good..

  62. anonymous
    • one year ago
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    oh its multipole choice?

  63. anonymous
    • one year ago
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    yeah there's a few to pick from 1044 9134 1777 1496

  64. anonymous
    • one year ago
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    f[6] = 1253 E^(0.03 (6)) = 1500.11

  65. anonymous
    • one year ago
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    yeah so I guess it's all just approximate

  66. anonymous
    • one year ago
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    wow that's sneaky hey ... yeah I'd go with the closest value..

  67. anonymous
    • one year ago
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    In the question is subtly uses "about how many dear" so I guess it doesn't have to be exact.

  68. anonymous
    • one year ago
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    thats like 1252 deer hey.. I guess one died.

  69. anonymous
    • one year ago
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    yeah lol

  70. anonymous
    • one year ago
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    so thanks for helping me it's really appreciated!

  71. anonymous
    • one year ago
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    no worries.. good luck

  72. anonymous
    • one year ago
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    thank you :) I understand this more thanks to you

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