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 Topic: labour
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 How much money do I make per sq ft? 2017-03-28 From jackie:If I make \$12.00 per hour and I cover 750 sq ft in one hour, how much money do I make per sq ft?Answered by Penny Nom. How much labor should the firm employ? 2006-10-28 From Christy:A dressmaking firm has a production function of Q=L-L(squared)/800. Q is the number of dresses per week and L is the number of labor hours per week. Additional cost of hiring an extra hour of labor is \$20. The fixed selling price is P=\$40. How much labor should the firm employ? What is the resulting output and profit? I am having a difficult time with this, HELP!Answered by Stephen La Rocque. What is the total cost of a driveway ... 2006-08-29 From Ismael:What is the total cost of a driveway 15' wide, 40' long and 4" thick if the concrete costs \$60.00 per cubic yard and the labor costs \$1.25 per square foot?Answered by Stephen La Rocque. Labour efficiency 2005-08-23 From Rob: The problem, on the surface, seems very simple and yet has created some controversy among a group of accountants. The problem itself has to do with labour efficiency rates and only involves two variables; standard working hours, and actual working hours. The difficulty lies in deriving an efficiency % from these two numbers. Standard working hours or the targeted number of labour hours required to produce one widget, which I will represent as "s". Actual working hour or the actual number of labour hours require to produce one widget, which I will represent as "a". Labour efficiency I will represent with "E". The prevailing calculation with which I have a problem with is this: s/a=E or if s=3000, and a=4000 then 3000/4000=75% What bothers me about the calculation is that the standard hours get represented as a percentage of the actual hours and in my opinion changes the focus of the calculation from standard or target, where it should be, to the actual hours. I cannot define why, but this just seems inherently wrong to me. The calculation that I use: (1+((s-a)/s))=E or if s=3000, and a=4000 then (1+((3000-4000)/3000))=66.67% My calculation is like a %change from standard calculation. However, there is something that also concerns me about my calculation. If you substitute 100 for a and 50 for s, then you come to a quandary, because if you plug those numbers into the second equation the result is of course zero % efficient which doesn't sit right with me either. If you plug them into the first calculation you get 50% efficiency which doesn't really seem to work either, because you require 100% more hours to do the same work in this case. ??? Is the first calculation correct? Am I missing something altogether? Are both calculations off base?Answered by Harley Weston.

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