SOLUTION: My problem: The demand (d), for DVD players sold in 1 monthe is a linear function of price (p), for $150 <= p <= $400. If the price is $200, then 50 DVD players will be sold each

Algebra ->  Linear-equations -> SOLUTION: My problem: The demand (d), for DVD players sold in 1 monthe is a linear function of price (p), for $150 <= p <= $400. If the price is $200, then 50 DVD players will be sold each      Log On


   



Question 260793: My problem: The demand (d), for DVD players sold in 1 monthe is a linear function of price (p), for $150 <= p <= $400. If the price is $200, then 50 DVD players will be sold each month, if the price is $300, only 30 will be sold. a) using order pairs of the form, (p,d), write an equation for the demand(d) as a function of price,(p). I did this as 300-200/30-50, and ended up with -5. I wrote my equation as d(p)= -5p + 200, which is wrong. The book has the equation as d(p)= -0.20p + 90 and I have no idea how they got that equation. Thanks for your help..
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Since the demand (d) relies on the price (p), the price (p) is the independent variable and the demand (d) is the dependent variable. You might have learned earlier that of the ordered pair (x,y), 'x' is the independent variable and 'y' is the dependent variable.


From the slope formula m=%28y%5B2%5D-y%5B1%5D%29%2F%28x%5B2%5D-x%5B1%5D%29, we see that the y variables are placed in the numerator. This means that the dependent variables are in the numerator.

So this means that you should have the slope of m=%2830-50%29%2F%28300-200%29=-20%2F100=-1%2F5=-0.20 (since the dependent variable 'd' should be in the numerator). Let me know if this helps.