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Mathematics 21 Online
OpenStudy (anonymous):

Help! Will fan and medal! A pharmacy claims that the average medication costs $32 but it could differ as much as $8. Write and solve an absolute value inequality to determine the range of medication costs at this pharmacy. |x − 32| ≥ 8; The medication costs range from $24 to $40 |x − 32| ≥ 8; The medications cost less than $24 or greater than $40. |x − 32| ≤ 8; The medication costs range from $24 to $40 |x − 32| ≤8; The medications cost less than $24 or greater than $40.

OpenStudy (anonymous):

got a guess?

OpenStudy (anonymous):

I dont know really.B looks like it might be good. But not quite sure.

OpenStudy (anonymous):

three of these answers do not make any sense at all the numbers are not even important, just the logic

OpenStudy (anonymous):

Well which one is most logical?

OpenStudy (anonymous):

let me ask you, what does "The medications cost less than $24 or greater than $40. " mean?

OpenStudy (anonymous):

how can something cost less than $24 or more than $40?

OpenStudy (anonymous):

like you might be $20 or else $45? does that make sense?

OpenStudy (anonymous):

Err.. Not sure. But I'm understanding now.

OpenStudy (anonymous):

yeah it doesn't make any sense at all if something cost between $8 or $32, the most it could cost is \(32+8=40\) and the least would be \(32-8=24\) in other words in simple english it cost between $24 and $40

OpenStudy (anonymous):

in math you would write this as \(24\leq x\leq 40\) if \(x\) was the cost

OpenStudy (anonymous):

in simple english "The medication costs range from $24 to $40"

OpenStudy (anonymous):

Well yeah. Thats what I was thinking. So is it maybe A?

OpenStudy (anonymous):

maybe , but no

OpenStudy (anonymous):

C?

OpenStudy (anonymous):

to say the difference between \(x\) and \(32\) is LESS THAN OR EQUAL TO $8, you would write \[|x-32|\leq 8\]

OpenStudy (anonymous):

So it has to be either C or D?

OpenStudy (anonymous):

I think its c.

OpenStudy (anonymous):

It was c. Thanks @satellite73 you rock :)

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