anonymous
  • anonymous
A company did a quality check on all the packs of roasted almonds it manufactured. Each pack of roasted almonds is targeted to weigh 21.25 oz. A pack must weigh within 0.24 oz of the target weight to be accepted. What is the range of rejected masses, x, for the manufactured roasted almonds? x < 21.01 or x > 21.49 because |x - 0.24| + 21.25 > 0 x < 21.25 or x > 21.49 because |x - 21.25| > 0.24 x < 21.01 or x > 21.49 because |x - 21.25| > 0.24 x < 21.25 or x > 21.49 because |x - 0.24| + 21.25 > 0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
@jim_thompson5910
jim_thompson5910
  • jim_thompson5910
Draw a number line and make sure 21.25 is labeled on it |dw:1434586836395:dw|
jim_thompson5910
  • jim_thompson5910
the accepted range is within 0.24 oz so add 0.24 to 21.25 to get 21.49 |dw:1434586879589:dw|

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jim_thompson5910
  • jim_thompson5910
subtract 0.24 from 21.25 to get 21.01 |dw:1434586898947:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1434586921667:dw|
anonymous
  • anonymous
A
jim_thompson5910
  • jim_thompson5910
|dw:1434586955603:dw|
jim_thompson5910
  • jim_thompson5910
Why A?
anonymous
  • anonymous
NVM i dont know
jim_thompson5910
  • jim_thompson5910
In general, let's call T the target we want to hit In this case, T = 21.25
jim_thompson5910
  • jim_thompson5910
E is the accepted error E = 0.24 in this case
jim_thompson5910
  • jim_thompson5910
The rejected range will be found by solving |x-T| > E where x is any random item's mass
jim_thompson5910
  • jim_thompson5910
The accepted range can be found by solving \(\Large |x-T| \le E\)
jim_thompson5910
  • jim_thompson5910
which basically boils down to \[\Large T - E \le x \le T+E\]
anonymous
  • anonymous
OK

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