Computer stuff and linear equations

redsoxnation

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Yes, this is a real algebra question.

I've been tasked with calculating the slope and intercept for the following equation:

total cost = (cost of a computer printer) + (ink cost per page) x (pages printed)

NO SPECIFIC INFORMATION HAS BEEN GIVEN.

I'm thinking that, if this is y = mx + b, then m = about $200 x = 1.2 cents/page (based on a cartridge costing $59.99 with a yield of 5,000 pages - smart, huh?). "b" would be the pages printed over the life of the printer, yes? But I can't find out that information anywhere. So, here are my questions -

a) Am I on track?

b) How can I figure out the number of pages printed over the life of a printer?

If anyone can help me, you guys can! Thanks so much.
 

Subhotosh Khan

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redsoxnation said:
Yes, this is a real algebra question.

I've been tasked with calculating the slope and intercept for the following equation:

total cost = (cost of a computer printer) + (ink cost per page) x (pages printed)

NO SPECIFIC INFORMATION HAS BEEN GIVEN.

I'm thinking that, if this is y = mx + b, then m = about $200 x = 1.2 cents/page (based on a cartridge costing $59.99 with a yield of 5,000 pages - smart - too expensive you must not shop at Costco or Sam ,huh?). "b" would be the pages printed over the life of the printer, yes? But I can't find out that information anywhere. So, here are my questions -

a) Am I on track? Not quite.

b) How can I figure out the number of pages printed over the life of a printer?

Why do you want to do that? That is not what the "problem" asks (you've been tasked to "calculating the slope and intercept for the following equation")

If anyone can help me, you guys can! Thanks so much.

b = cost of the printer = $200

m = $60/5000 = 0.012 $/page (= 1.2 cents/page)

Make sure you either exclusively work with dollars or exclusively work with pennies.
 
J

JeffM

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redsoxnation said:
Yes, this is a real algebra question.

I've been tasked with calculating the slope and intercept for the following equation:

total cost = (cost of a computer printer) + (ink cost per page) x (pages printed)

NO SPECIFIC INFORMATION HAS BEEN GIVEN.

I'm thinking that, if this is y = mx + b, then m = about $200 x = 1.2 cents/page (based on a cartridge costing $59.99 with a yield of 5,000 pages - smart, huh?). "b" would be the pages printed over the life of the printer, yes? But I can't find out that information anywhere. So, here are my questions -

a) Am I on track?

No. As S Khan pointed out, you are not being asked to SOLVE the equation for the total cost of the printer, but to FORMULATE a linear equation for the total cost. And they have given you the equation in words. All you need to do is to translate it into math symbols. I know it sounds picky, but it helps to write down what concept each symbol stands for. It keeps you from drifting in your thought.

So y = total cost.

You take it from there. Equate a symbol to every variable relevant to the problem. Then build your equation or formula from those symbols.


b) How can I figure out the number of pages printed over the life of a printer? Well, you would need that (and the price of ink) to SOLVE the equation. As it is, you do not need either to FORMULATE the equation. The purpose of the exercise is to get you used to translating a problem in words into a mathematical form. The resulting equation will work for any type of printer, no matter what its purchase price, its life in pages, or the price of ink catridges. This is the mental economy inherent in math. You create a general analysis that you can use over and over again for different problems of the same kind. You may have overthought this problem. If so, take a trip to Fenway.

If anyone can help me, you guys can! Thanks so much.
 

Subhotosh Khan

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By the way, the equation defines total cost without including cost of paper (and the value of time spent while praying that the paper does not jam)
 

Subhotosh Khan

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JeffM said:
You may have overthought this problem. If so, take a trip to Fenway.[/color]
And watch Bosox play !!! the way they are playing these days - might increase the stress level.....
 

redsoxnation

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Subhotosh Khan said:
Make sure you either exclusively work with dollars or exclusively work with pennies.
Yes, of course. Units, units, units!

Thank you, All, for your help and insight. Now, our illustrious Professor has asked us "What fraction of the cost is the initial purchase price?". So, I was probably overthinking the problem initially, but now do I have to know how many darn pages the stupid ("Don't call me stupid!") printer prints over its lifetime??
JeffM said:
You may have overthought this problem. If so, take a trip to Fenway.
We banished the Sox to California for a couple of games. That'll learn 'em!
 
J

JeffM

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redsoxnation said:
Subhotosh Khan said:
Make sure you either exclusively work with dollars or exclusively work with pennies.
Yes, of course. Units, units, units!

Thank you, All, for your help and insight. Now, our illustrious Professor has asked us "What fraction of the cost is the initial purchase price?". So, I was probably overthinking the problem initially, but now do I have to know how many darn pages the stupid ("Don't call me stupid!") printer prints over its lifetime??
JeffM said:
You may have overthought this problem. If so, take a trip to Fenway.
We banished the Sox to California for a couple of games. That'll learn 'em!
Red

You are thinking too concretely. What is the ONE of the words used to describe the x, y and z in algebra, etc. It is "unknown." But at this stage of your mathematical career, your unknowns are numbers. If you knew the concrete numbers, how would you answer the question? Well, doesn't that same process work when you have to use letters instead of numbers? You can reason with numbers even when you do not know what they are. Life is full of problems where you are missing enough information to give a specific answer. That doesn'y mean you can't get somewhere with the problem.
 

redsoxnation

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redsoxnation said:
You are thinking too concretely.
JeffM - Yes, you're right. That's my issue with this particular math class; it's requiring me to look a t math problems in a new way. I'm used to numbers being numbers (or at least letter equivalents). But I'm trying. I refuse to succumb to the "old dog/new trick" school of thought.

redsoxnation said:
Life is full of problems where you are missing enough information to give a specific answer. That doesn'y mean you can't get somewhere with the problem.
Very true!

Thanks for your input and words of wisdom!
 
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