Hi Kelly,
a) Let the number of riders be a function of the fare charged. Graph the function, identify the graph of the function (line, parabola, hyperbola, or exponential), and find the slope of the graph.
Let y = the number of riders
Let x = ticket price, in dollars
y = -160x + 1160
This is a linear equation which will graph as a straight line.
The slope of -160 indicates they will lose 160 riders for every $1 increase in ticket price. (That's the same as a decrease of 40 riders for every increase of 25¢.) 1160 is the maximum number of riders if the ride was free.
Graph: Graph points at (0,1160), (2, 840), and (7,40). Draw a straight line through them.
Graph Type: linear (line)
What is the slope of the graph? -160
b) The bus company has determined that even if they set the price very low, there is a maximum number of riders permitted each day. If the price is $0 (free), how many riders are permitted each day? 1160, which is the y intercept of the equation (0,1160) means at $0 there are 1,160 riders.
Answer: 1160, which is the y intercept of the equation above. (0,1160) means at $0 there are 1,160 riders.
Show work in this space:
From the points (2.25,800) and (2.50,760), the slope can be found by:
Y2 - Y1
---------- = m
X2 - X1
Using (2.25,800) and (2.50,760), it becomes:
800 - 760
---------------- = 40/-.25 = -160 = m
2.25 - 2.5
Using the point-slope formula: Y- y1 = m(X - x1) gives the equation:
Y - 800 = -160(X - 2.25)
Y - 800 = -160X + 360
Y = -160X + 1160
c) If the bus company sets the price too high, no one will be willing to ride the bus. Beginning at what ticket price will no one be willing to ride the bus?
Answer: $7.25
Show work in this space:
Using the equation Y = -160X + 1160, let the number of riders, Y, = 0. Then solve for x.
Y = -160X + 1160
0 = -160X + 1160
160x = 1160
x = 7.25
At a price of $7.25, no one will ride the bus.
I hope that helps!1 :-)
2007-06-07 17:10:25
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answer #1
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answered by Pi R Squared 7
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If they can only raise the fare in $0.25 increments, then the bus company will maximize their revenue at either $3.50 per ride, which will have a corresponding ridership of 600 passengers, resulting in revenue of $2,100 per day, OR, they can raise the fare to $3.75 per ride, at which point ridership will drop to 560 passengers, but revenue will stay at $2,100 per day.
At $3.25 and at $4.00, ridership will be at 640 or 520 passengers respectively, and the revenue will drop to $2,080 per day.
2007-06-07 14:01:15
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answer #2
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answered by 2007_Shelby_GT500 7
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