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anonymous

  • 5 years ago

S(t) = 68 − 20 log (t + 1), t ≥ 0 What was the average score after 4 months? after 24 months? After what time t was the average score 50%?

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  1. Owlfred
    • 5 years ago
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    Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

  2. anonymous
    • 5 years ago
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    Round to the nearst two decimal places

  3. anonymous
    • 5 years ago
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    calculator exercise. replace t by 4, and 24. i will write it if you like

  4. anonymous
    • 5 years ago
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    is this log base ten?

  5. anonymous
    • 5 years ago
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    i dont get it

  6. anonymous
    • 5 years ago
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    t = time in months yes?

  7. anonymous
    • 5 years ago
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    so for the first one put \[S(4)=68-20log(4+1)=68-20log(5)\]

  8. anonymous
    • 5 years ago
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    in other words put 4 where you see a t and then compute

  9. anonymous
    • 5 years ago
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    you need a calculator to find log(5)

  10. anonymous
    • 5 years ago
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    i got 54.02

  11. anonymous
    • 5 years ago
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    let me try

  12. anonymous
    • 5 years ago
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    \[S(24)=68-20log(25)=...\]

  13. anonymous
    • 5 years ago
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    i let me know if you have an answer

  14. anonymous
    • 5 years ago
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    okay me to so for 24 months its 40.04

  15. anonymous
    • 5 years ago
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    sound right

  16. anonymous
    • 5 years ago
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    hold on i will compute

  17. anonymous
    • 5 years ago
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    yup that is what i got

  18. anonymous
    • 5 years ago
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    now we have to find when S = 50. do you know how to do that one?

  19. anonymous
    • 5 years ago
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    no

  20. anonymous
    • 5 years ago
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    first replace S(t) by 50 \[50=68-20log(t+1)\] then get the log by itself on one side \[20log(t+1)=68-50=18\]

  21. anonymous
    • 5 years ago
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    ok so far?

  22. anonymous
    • 5 years ago
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    no , why did you put 68-50

  23. anonymous
    • 5 years ago
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    ok let me go step by step. \[50=68-20log(t+1)\] add \[20log(t+1)\] to both sides to get \[20log(t+1)+50=68\] then subtract 50 from both sides to get \[20log(t+1) = 68-50=18\] then divide by 20 to get \[log(t+1)=\frac{18}{20}=\frac{9}{10}=.9\]

  24. anonymous
    • 5 years ago
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    i did not need to add log(t+1) to both sides, but i prefer dealing with positive things when i am trying to solve. i could have subtracted 68 from both sides and gotten the same answer working with negative numbers. matter of preference.

  25. anonymous
    • 5 years ago
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    so we have arrived via some algebra at \[log(t+1)=.9\] is that ok so far?

  26. anonymous
    • 5 years ago
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    yes i think im getting it

  27. anonymous
    • 5 years ago
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    these were just algebra steps to get log(t+1) by itself. same steps i would have used to solve for x if the equation was \[50=68-20x\]

  28. anonymous
    • 5 years ago
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    you would get \[x=.9\] and here \[x=log(t+1)\] so \[log(t+1)=.9\]

  29. anonymous
    • 5 years ago
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    we are not done yet, though because we have to solve for t.

  30. anonymous
    • 5 years ago
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    okay im following you

  31. anonymous
    • 5 years ago
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    \[log(x)=y \] \[10^y=x\]

  32. anonymous
    • 5 years ago
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    what i meant to say was \[log(x)=y\] means the same thing as \[10^y=x \]

  33. anonymous
    • 5 years ago
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    here we have \[log(t+1)=.9\] so \[10^{.9}=t+1\]

  34. anonymous
    • 5 years ago
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    to find \[10^{.9}\] in need a calculator. then to get t by itself subtract 1.

  35. anonymous
    • 5 years ago
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    let me know what you get.

  36. anonymous
    • 5 years ago
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    6.94 sound right

  37. anonymous
    • 5 years ago
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    its wrong???

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