a concession stand vendor counts the money in the register. after 2 h it has $82.75. after 6 h is has $ 360.75. how much money will be in the register at the end of an 8h shift?
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well we can assume it is linear I geuss? Use the points (2,82.75) and (6,360.75) and find the equation of the line through them. Then plug 8 into your equation to estimate the amount at the end of 8 hours, assuming that this relationship is linear
what would the solution be?
do you know how to find slope, and use point slope form for the equation of a line
?
or just find money per hour: (360.75-82.75)/4=69.5 per hour
did you get 489.25?
then in two more hours there will be 139 more. 360.25+139=499.25
i get 499.25
woops i made a mistake
its 360.75+139=499.75
thank you so much!
no rsvitale, you are incorrect. because initially the vendor had to start at 0,0. and your line doesn't pass through that
you would start with some money in the register for change
there was nothing saying you start with no money
there is nothing to suggest its linear either.
so there is not really enough information,
yup, thats my sense too.
there are a lot of functions that go through 2 points
exactly.
the initial conditions should have been specified, or at least the peak purchase times.
i assumed since it is algebra 2 it was linear, i don't remember doing anything more advanced than that when I took that class
oh I see. then it should be all right.
actually, rsvitale, if you plot the graph, at hour 0, the vendor has negative money. so your argument that he has some money for change is not valid.
well rsvitale was right because my quiz showed i got that question right
I am sure he was right. but the problem doesn't give enough information. I think for the class you are taking, you assume that it is linear, but you cannot make that assumption as you study more advanced topics.
true
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