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Posted by on Wednesday, December 17, 2008 at 3:42pm.

I really need help with my math homework today, because I am totally confused! Please give me the answer, but also walk me through it, too, so I can fully understand it, because just the answer won't really help.

#1: A math teacher wants to curve a set of grades. *Just a note, I don't even understand what curve means* She wants to create a formula to turn an old grade, s, into a new grade, t, where t = As + B for some constants A and B. She wants this formula to give a 100 to a student who originally scored 100, and a score of 81 to a student who originally scored 62.

a. How can we view the pairs of original and new scores as points on a line?
b. Use your answer to part a. to determine A and B in the formula.
c. What grade should a student who originally scored 74 receive?

I will post more questions in a little bit. Thank you so much, any help is appreciated, even if you can only answer part of it.

Thanks!

  • Algebra--please help! - , Wednesday, December 17, 2008 at 3:47pm

    well, if an original 100 gives a 100

    100=A*100+B
    and if a 62 is turned into and 81 (magic wand).
    81=A*62+B
    You have two equations, two unknowns.

    First, subtract the second equation from the first.
    100-81=A(100-62) And you solve for A.
    Then, put that A into either equation and solve for B.

  • Algebra--please help! - , Wednesday, December 17, 2008 at 3:56pm

    Thanks for replying so fast!

    Once you got to "subtract the second equation," I got a little confused.

    So basically its, 100-81=A(100-63), like you said. Does the B stay put?

    And what do you mean by "solve for A" and "put A into either equation?"

    Thanks so much!

  • Algebra--please help! - , Wednesday, December 17, 2008 at 4:02pm

    When you subtract the second equation fromthe first, you have B-B which is zero, so B disappears.

    Solving for A?
    100-81=A(100-63),
    19=A*37
    A=19/37

    Now put that into any equation..
    100=100A+B
    100=100(19/37) + B
    B= 100(1-19/37)=100(18/37)
    and you have A and B
    check my arithmetic

  • Algebra--please help! - , Wednesday, December 17, 2008 at 4:06pm

    Oh, I see... when I got 19/37 when I solved, I thought I was incorrect.

    One last thing-- why did you write
    B= 100(1-19/37)=100(18/37)? Why did you take away 1?

    Thank you!

  • Algebra--please help! - , Wednesday, December 17, 2008 at 4:09pm

    100=100(19/37) + B
    subtract 100(19/37) from each side.

  • Algebra--please help! - , Wednesday, December 17, 2008 at 4:14pm

    I hate to ask so many questions, but why did you write one, though?

    Thanks!

  • Algebra--please help! - , Wednesday, December 17, 2008 at 4:16pm

    100-100(19/37)=
    100(1-19/37)

    It is easier to do this than multiply 100 Times by 19 then divide by 37, then subtract from 100.

  • Algebra--please help! - , Wednesday, December 17, 2008 at 4:20pm

    Thank you so much!

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