SOLUTION: In 2002 a cable company charged $34.99 per month. In 2005 it cost $43.43. What is the company's equation? Use C for the cost per month and t for time in years since 2002. (2003=1)

Algebra ->  Linear-equations -> SOLUTION: In 2002 a cable company charged $34.99 per month. In 2005 it cost $43.43. What is the company's equation? Use C for the cost per month and t for time in years since 2002. (2003=1)      Log On


   

Question 1131105: In 2002 a cable company charged $34.99 per month. In 2005 it cost $43.43.
What is the company's equation? Use C for the cost per month and t for time in years since 2002. (2003=1)
Found 2 solutions by josmiceli, stanbon:
Answer by josmiceli(19441) About Me  (Show Source):
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
In 2002 a cable company charged $34.99 per month. In 2005 it cost $43.43.
What is the company's equation? Use C for the cost per month and t for time in years since 2002. (2003=1)
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slope = 12(43.43-34.99)/(4-1) = 12*8.44/3 = 33.76 (amt. of cost gain per year)
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intercept:: C(1 yr.) = 12*34.99 so C(0) = 12*34.99-33.76 = 386.12
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Equation::
C(t) = 33.76t + 386.12
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Cheers,
Stan H.
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