SOLUTION: A scale is accurate within .5 pound how do I write an absolute value inequality if a person weighs 95 pounds to express the possible error in the weight (w) given by the scale. I t

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Question 892655: A scale is accurate within .5 pound how do I write an absolute value inequality if a person weighs 95 pounds to express the possible error in the weight (w) given by the scale. I then have to solve the inequality
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
i believe your solution is going to be:

|x - 95| <= .5

x represents the weight of the person as measured by the scale.

here's how it works.

suppose the person weighs 95.5 pounds.

equation becomes |95.5 - 95| <= .5 which becomes |.5| <= .5 which becomes .5 <= .5 which is within the prescribed limits.

suppose the person weighs 94.5 pounds.

equation becomes |94.5 - 95| <= .5 which becomes |-.5| <= .5 which becomes .5 <= .5 which is also good.

since the maximum error of the scale is +/- .5, the scale will not measure a 95 pound person outside those limits.

you can expect that |x - 95| <= .5

suppose the scale measures 94.8 pounds.

the difference is -.2 which becomes .2 which is less than .5.

suppose the scale measures 95.2 pounds.

the difference is +.2 which becomes .2 which is less than .5.

the absolute value automtically takes care of the plus and minus aspects of the difference.