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Quandaries & Queries
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 Topic: linear functions
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 A linear relationship 2016-10-16 From Bianca:x represents the number of hours since 8am. y represents the number of children the school nurse has seen during the school day. The nurse has seen 3 children by 8am and 27 children by noon. How many children has the nurse seen at 2pm? If the nurse has seen 18 children, what time is it?Answered by Penny Nom. The temperature inside and outside a greenhouse 2016-07-22 From Mehzad:Every day, the temperature in a greenhouse is at its low temperature, 70 degrees Fahrenheit, at 2 a.m. and at its high temperature, 84 degrees Fahrenheit, at 2pm. Its temperature increases linearly between 2 am and 2pm , and decreases linearly from 2 pm and to 2am. The outside temperature follows the same linear patterns, but has a low temperature of 60 degrees Fahrenheit at 2 am and a high temperature of 78 degrees Fahrenheit at 2 pm. At which of the following times will the temperature inside and outside the greenhouse be the same? a) 8:00 am b) 12:00 noon c) 4:00 pm d) The two temperatures will never be the sameAnswered by Penny Nom. 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. Graphing a linear function 2000-05-17 From Chelsea:I need help with grahing linear functions.If you could e-mail me back the basics and how tos I would be much appriciative.Answered by Penny Nom.

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