week 3 group math presentation

# week 3 group math presentation - You start out with the...

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Group A

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Aids Cases 1993 year the to s correspond 0 x where , 430 76 2 N by ed approximat be can ds in thousan cases aids of N number cumulative the 2003 to 1993 from 2 = + + - = x x
Use the equation to find N for each year in the table . 1 422 1993, putting x=0, N = -2(0)^2+76(0)+430 = 430 4 1995, putting x=2, N = -2(2)^2+76(2)+430 = 574 4 1997, putting x=4, N = -2(4)^2+76(4)+430 = 702 4 1999, putting x=6, N = -2(6)^2+76(6)+430 = 814 4 2001, putting x=8, N = -2(8)^2+76(8)+430 = 910 4 2003, putting x=10, N=-2(10)^2+76(10)+430 = 990 1993 1995 1997 1999 2001 2003 422 565 677 762 844 930

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Discuss how well this equation approximates the data . The equation approximates to a good degree to start of with in 1993, but for further dates are from 1993, the approximation gradually becomes a little more off the actual number of cases.
Rewrite the equation with the right side completely factored

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Unformatted text preview: You start out with the equation -2x^2+76x+430 Since 38 is half of 76, and 215 is half of 430 you would say. . -2(x^2-38x-215) Next, 5+38=43 so. .. -2(x^2-43x+5x-215) then, take your double x's and write like so. . -2(x(x-43)+5(x-43) (since 5(-43)=-215) Finally, -2 (x+5)(x-43) is the answer because. ... (-215(-2)=430) Use your equation from part (c) to find N for each year in the table . For 1993, putting x=0, N=-2(0-43)(0+5)= 430 4 For 1995, putting x=2, N=-2(2-43)(2+5) = 574 4 For 1997, putting x=4, N=-2(4-43)(4+5) = 702 4 For 1999, putting x=6, N=-2(6-43)(6+5) = 814 4 For 2001, putting x=8, N=-2(8-43)(8+5) = 910 4 For 2003, putting x=10, N=-2(10-43)(10+5) = 990 Are the Answers the Same Yes they are both the same Questions...
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## week 3 group math presentation - You start out with the...

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