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anonymous

  • one year ago

The Mendes family bought a new house 11 years ago for $100,000. The house is now worth $196,000. Assuming a steady rate of growth, what was the yearly rate of appreciation? Round your answer to the nearest tenth of a percent (1.2% etc)

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  1. anonymous
    • one year ago
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    @saseal

  2. anonymous
    • one year ago
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    growth : y = a(1+r)^2

  3. anonymous
    • one year ago
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    yea

  4. anonymous
    • one year ago
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    \[196000 = 100000 (1 + \frac{ r }{ 100 } ) ^{11}\]

  5. anonymous
    • one year ago
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    would i multiplu 100 to both sides ?

  6. anonymous
    • one year ago
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    divide by 100000 on both sides

  7. anonymous
    • one year ago
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    1.96

  8. phi
    • one year ago
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    you probably should be using the continuous interest formula http://cs.selu.edu/~rbyrd/math/continuous/ \[ P= P_0 e^{r \ t} \]

  9. anonymous
    • one year ago
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    \[1.96 = (1+\frac{ r }{ 100 })^{11}\]

  10. anonymous
    • one year ago
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    now can i divide by 100

  11. anonymous
    • one year ago
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    no you need to take log

  12. phi
    • one year ago
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    if you assume compounding at a yearly rate, your answer will be off by a few tenths of a percent

  13. anonymous
    • one year ago
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    yea it asked for yearly rate of appreciation

  14. anonymous
    • one year ago
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    log1.96=log (1+r/100)^2

  15. anonymous
    • one year ago
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    ^11****

  16. anonymous
    • one year ago
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    no

  17. anonymous
    • one year ago
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    how can there be power after you log

  18. anonymous
    • one year ago
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    im confused

  19. anonymous
    • one year ago
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    \[\log_{11} (1 + \frac{ r }{ 100 }) = 1.96 \]

  20. anonymous
    • one year ago
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    \[\log_{11} (1.96) = 0.28064\]

  21. phi
    • one year ago
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    ***Assuming a steady rate of growth*** means continuous they want the annual interest rate assuming continuous compounding.

  22. anonymous
    • one year ago
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    yea thats what we've been using

  23. anonymous
    • one year ago
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    okay i about to solve it tell me if im correct

  24. anonymous
    • one year ago
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    2.8%?

  25. anonymous
    • one year ago
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    I believe you are to assume annual compounding. By "steady rate of growth" I believe the author simply means that the annual rate of appreciation is constant over the duration of the problem.

  26. anonymous
    • one year ago
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    no not 2.8

  27. anonymous
    • one year ago
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    dayum if thats the case its simple interest formula

  28. anonymous
    • one year ago
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    you move the decimal two places forward or back

  29. anonymous
    • one year ago
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    And base 11 logs are not required. You have\[1.96=\left( 1+\frac{ r }{ 100 } \right)^{11}\]Using base 10 logs\[\log1.96 = 11 \log \left( 1+\frac{ r }{ 100 } \right)\]\[\frac{ \log 1.96 }{ 11 } = \log \left( 1+\frac{ r }{ 100 } \right)\]\[0.02657 = \log \left( 1+\frac{ r }{ 100 } \right)\]\[1+\frac{ r }{ 100 } = 10^{0.02657} = 1.0631\]Can you take it from here and solve for \(r\)?

  30. anonymous
    • one year ago
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    no i dont understand

  31. anonymous
    • one year ago
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    would it be .0631/100 ?

  32. anonymous
    • one year ago
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    yea

  33. anonymous
    • one year ago
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    0.000631 = r

  34. anonymous
    • one year ago
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    So close. \[1+\frac{ r }{ 100 } = 0.0631\]Subtract 1 from both sides\[\frac{ r }{ 100 } = 0.0631\]How do you solve for r?

  35. anonymous
    • one year ago
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    6.31

  36. anonymous
    • one year ago
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    Sorry. First line should read\[1+\frac{ r }{ 100 } = 1.0631\]

  37. anonymous
    • one year ago
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    6.31%

  38. anonymous
    • one year ago
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    You are correct. Now round to the number of decimal places required. Well done?

  39. anonymous
    • one year ago
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    Well done!

  40. anonymous
    • one year ago
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    now you need to round off somemore

  41. anonymous
    • one year ago
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    can yall help me with another ?

  42. anonymous
    • one year ago
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    Delete this question and post the other.

  43. anonymous
    • one year ago
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    it said that was wrong . it says to round to the nearest tenth of percent

  44. anonymous
    • one year ago
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    6.3% correct ??

  45. anonymous
    • one year ago
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    yea

  46. anonymous
    • one year ago
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    i told you to round off somemore lol

  47. anonymous
    • one year ago
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    it still says its wrong

  48. anonymous
    • one year ago
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    We both indicated that you were to round to the correct number of decimal places :)

  49. anonymous
    • one year ago
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    its still wrong . i put 6.3%

  50. anonymous
    • one year ago
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    Well. I don't know what the issue is. \[196000 = 100000\times 1.603^{11}\]so the math checks out.

  51. anonymous
    • one year ago
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    Sorry. 1.063^11

  52. anonymous
    • one year ago
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    wanna try simple interest method?

  53. anonymous
    • one year ago
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    I can't imagine that it's continuous compounding but, if so, you'll have to use natural logs.\[F=Pe^{rt}\]\[196000 = 100000e^{11r}\]\[1.96 = e^{11r}\]\[\ln 1.96 = 11r\]\[r=\frac{ \ln 1.96 }{ 11 }\]

  54. anonymous
    • one year ago
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    okay .

  55. anonymous
    • one year ago
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    .06117

  56. anonymous
    • one year ago
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    Convert to percent by multiplying by 100% and round to the correct number of decimal places.

  57. anonymous
    • one year ago
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    hate that kinda english-testing maths questions

  58. phi
    • one year ago
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    yes, you get 6.1%

  59. anonymous
    • one year ago
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    Me too. Just finished a graduate course in financial mathematics and "steady rate of growth" was never used to indicate continuous compounding. @phi , if this is the correct answer, I owe you an apology. Please forgive my doubt.

  60. anonymous
    • one year ago
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    it was correct

  61. anonymous
    • one year ago
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    Congratualtions

  62. anonymous
    • one year ago
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    sorta can't understand how e comes into play in continuous compounding

  63. anonymous
    • one year ago
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    gratz

  64. anonymous
    • one year ago
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    THANK YALL !!!! . can yall help me with the next one

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