Per Set
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Set of 6 Wilkinson 45° Button Knob Chrome Machine Heads 3 Per Side - Pure Tone | ![]() |
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US $27.96 | 7d 10h 52m |
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NEW Grover Gold 3X3 Tuners (406G) - 3 Per Side - Sets or Singles | ![]() |
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US $59.00 | 26d 4h 12m |
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NEW Grover Chrome 3X3 Tuners (305C) - 3 Per Side - Sets or Singles | ![]() |
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US $47.00 | 26d 4h 2m |
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NEW Grover Chrome 3X3 Tuners (406C) - 3 Per Side - Sets or Singles | ![]() |
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US $45.00 | 26d 4h 19m |
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Set of 6 Wilkinson 2 Prong Hex Knob Chrome Machine Heads 3 Per Side - Pure Tone | ![]() |
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US $27.96 | 7d 10h 29m |
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NEW Grover Chrome 3X3 Tuners (102-18C) - 3 Per Side - Sets or Singles | ![]() |
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US $47.00 | 26d 4h 40m |
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set of 6 GOLD GUITAR TUNERS TULIP shape keys 3x3 3 per side lp style w/ screws | ![]() |
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US $29.99 | 2d 1h 45m |
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NEW Grover Chrome 3X3 Tuners (205C) - 3 Per Side - Sets or Singles | ![]() |
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US $47.00 | 10d 6h 26m |
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Per Set

The US open will allow players with more challenges per set, is this good or bad for tennis?
this is to tennis man...unlimited challenges might be bad because some players will abuse it by disrupting the flow of others.
I thought they were sticking to three. it is not the slams that have changed but the other tournaments so now in tourneys like Indian Wells instead of two challenges it will be three like the slams. it will all be universal now.
To answer the question yes I think it is better.
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No items matching your keywords were found.
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![]() |
Set of 6 Wilkinson 45° Button Knob Chrome Machine Heads 3 Per Side - Pure Tone | ![]() |
![]() |
US $27.96 | 7d 10h 52m |
![]() |
NEW Grover Gold 3X3 Tuners (406G) - 3 Per Side - Sets or Singles | ![]() |
![]() |
US $59.00 | 26d 4h 12m |
![]() |
NEW Grover Chrome 3X3 Tuners (305C) - 3 Per Side - Sets or Singles | ![]() |
![]() |
US $47.00 | 26d 4h 2m |
![]() |
NEW Grover Chrome 3X3 Tuners (406C) - 3 Per Side - Sets or Singles | ![]() |
![]() |
US $45.00 | 26d 4h 19m |
![]() |
Set of 6 Wilkinson 2 Prong Hex Knob Chrome Machine Heads 3 Per Side - Pure Tone | ![]() |
![]() |
US $27.96 | 7d 10h 29m |
![]() |
NEW Grover Chrome 3X3 Tuners (102-18C) - 3 Per Side - Sets or Singles | ![]() |
![]() |
US $47.00 | 26d 4h 40m |
![]() |
set of 6 GOLD GUITAR TUNERS TULIP shape keys 3x3 3 per side lp style w/ screws | ![]() |
![]() |
US $29.99 | 2d 1h 45m |
![]() |
NEW Grover Chrome 3X3 Tuners (205C) - 3 Per Side - Sets or Singles | ![]() |
![]() |
US $47.00 | 10d 6h 26m |
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The King of Grass -- Roger Federer vs Roddick
A store sells an average of one television set per day, everyday of the year. It costs $70 to store a televisi?
A store sells an average of one television set per day, everyday of the year. It costs $70 to store a television set for one year. It costs $100 to place an order for more sets. How oftend do you place an order to minimize the costs? What is the best strategy to order?"
Let x be the number of TVs ordered at a time.
They will be sold in x days (1 per day). So on average they will be stored x/2 days. The cost to store 1 TV for 1 day is 70/365 = 14/71, so the average storage cost per TV is x/2 * 14/71 = x * 7/71.
Each batch of x TVs cost $100 to order, so the ordering cost per TV is 100/x.
So you want to find x such that cost per TV (y) is minimised, i.e. you want to minimise
y = storage_cost_per_TV + ordering_cost_per_TV
= (x * 7/71) + (100/x)
This will be at a minimum when dy/dx = 0
So:
* find dy/dx
* find x such that dy/dx = 0
* This x will not be a whole number. Check y for the two whole numbers nearest to that x, that will give you the optimum x, i.e. the best number of TVs to order at a time.
Aug172010
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