Ask your own question, for FREE!
Mathematics 7 Online
OpenStudy (anonymous):

help please! i will medal! A company did a quality check on all the packs of nuts it manufactured. Each pack of nuts is targeted to weigh 18.25 oz. A pack must weigh within 0.36 oz of the target weight to be accepted. What is the range of rejected masses, x, for the manufactured nuts? A) x < 17.89 or x > 18.61 because |x - 18.25| > 0.36 B) x < 17.89 or x > 18.61 because |x - 0.36| + 18.25 > 0 C) x < 18.25 or x > 18.61 because |x - 18.25| > 0.36 D) x < 18.25 or x > 18.61 because |x - 0.36| + 18.25 > 0

jimthompson5910 (jim_thompson5910):

If x is the mass of some unknown nut, then x-18.25 is the distance from the target mass of 18.25 oz Make this distance positive (since negative distance makes no sense) to get |x-18.25|

jimthompson5910 (jim_thompson5910):

Now if x = 18.25, then |x-18.25| becomes |18.25-18.25| = |0| = 0 so that distance is 0 (ie we're right on target) if x was say x = 20, then |x-18.25| = |20 - 18.25| = |1.75| = 1.75 that distance, or error, is 1.75 oz we're off by 1.75 oz if x = 20

jimthompson5910 (jim_thompson5910):

the same can be done with values below 18.25, say x = 15 |x-18.25| = |15 - 18.25| = |-3.25| = 3.25 now we're off by 3.25 oz

jimthompson5910 (jim_thompson5910):

the values of x that we want (or accept) are ones that make |x-18.25| equal to 0.36 or less than 0.36 in other words, the distance or error must be at most 0.36. This is the max tolerated error we're accepting

jimthompson5910 (jim_thompson5910):

so we accept values of x that make |x-18.25| <= 0.36 true flip this around to get |x-18.25| > 0.36 and that represents all of the x values we reject because they are too far from 18.25

jimthompson5910 (jim_thompson5910):

solving |x-18.25| > 0.36 for x will give you the two boundaries that tell you how far you can go (when you change from acceptance to rejection)

OpenStudy (anonymous):

so it's A or C

jimthompson5910 (jim_thompson5910):

correct

jimthompson5910 (jim_thompson5910):

do you know how to solve |x-18.25| > 0.36 for x?

OpenStudy (anonymous):

i get what you explain thank you for that but i dont get the beginning either the 17.89

OpenStudy (anonymous):

and no i dont know how to do that

jimthompson5910 (jim_thompson5910):

it turns out that if you have something of the form |x| > k for some positive number k, then it breaks down into the following x < -k or x > k

jimthompson5910 (jim_thompson5910):

so what that means in this case, |x-18.25| > 0.36 breaks down into the two following inequalities x-18.25 < -0.36 or x-18.25 > 0.36

jimthompson5910 (jim_thompson5910):

solve each of them and tell me what you get

OpenStudy (anonymous):

ok one second

jimthompson5910 (jim_thompson5910):

alright

OpenStudy (anonymous):

x<18.61 or x<17.89

OpenStudy (anonymous):

@jim_thompson5910

jimthompson5910 (jim_thompson5910):

very good, x-18.25 < -0.36 or x-18.25 > 0.36 x < -0.36+18.25 or x > 0.36+18.25 x < 17.89 or x > 18.61

jimthompson5910 (jim_thompson5910):

I got the same thing, just in a different order

jimthompson5910 (jim_thompson5910):

oh and you should have x > 18.61 not x < 18.61

OpenStudy (anonymous):

oh yeah i read it too fast

jimthompson5910 (jim_thompson5910):

that's alright

OpenStudy (anonymous):

so A

jimthompson5910 (jim_thompson5910):

correct

OpenStudy (anonymous):

thank you for the help you help me a lot

jimthompson5910 (jim_thompson5910):

you're welcome

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!