Mathswatch Bounds / Measures of Accuracy

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idek23
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Can someone help me with this question? Im unsure how to do it.
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Y1_UniMaths
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This question is all about bounds. For 17 disc cases to definitely fit, you want the largest volume of your box and the smallest volume of your disc cases. So work out the upper bound of each measurement on the box and calculate the volume then divide this by the volume using the lower bound of each measurement on the disc case
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idek23
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(Original post by Y11_Maths)
This question is all about bounds. For 17 disc cases to definitely fit, you want the largest volume of your box and the smallest volume of your disc cases. So work out the upper bound of each measurement on the box and calculate the volume then divide this by the volume using the lower bound of each measurement on the disc case
I tried out what you said and it came out as incorrect. Is there anything that I would need to do or correct here?
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Prasiortle
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(Original post by idek23)
Can someone help me with this question? Im unsure how to do it.
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(Original post by Y11_Maths)
This question is all about bounds. For 17 disc cases to definitely fit, you want the largest volume of your box and the smallest volume of your disc cases. So work out the upper bound of each measurement on the box and calculate the volume then divide this by the volume using the lower bound of each measurement on the disc case
A volume calculation won't be enough: you will need to consider all three dimensions (length, width, height) separately.
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Y1_UniMaths
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(Original post by idek23)
I tried out what you said and it came out as incorrect. Is there anything that I would need to do or correct here?
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Show each length of the disc case is less than the length of the box
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idek23
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(Original post by Prasiortle)
A volume calculation won't be enough: you will need to consider all three dimensions (length, width, height) separately.
What would I need to consider when it comes to the three dimensions, if I wouldn't just need volume. I'm bamboozled
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idek23
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(Original post by Y11_Maths)
Show each length of the disc case is less than the length of the box
How would I write that in the answer box?
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Prasiortle
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(Original post by idek23)
What would I need to consider when it comes to the three dimensions, if I wouldn't just need volume. I'm bamboozled
Try to understand why considering volume isn't sufficient: if you have a 30 cm x 20 cm x 10 cm box, it has a volume of 600 cm^3. A box with dimensions 90 cm x 2 cm x 3 cm has volume 540 cm^3, which is less than 600 cm^3, and yet it's obvious that this box won't fit inside the other box because the 90 cm length is longer than all three dimensions (30 cm, 20 cm, and 10 cm) of the other box, so no matter which way you rotate it, it won't fit in. In other words, we see that the volume being less is a necessary condition (i.e. if the volume is larger, we know that it certainly won't fit), but not a sufficient condition (so if the volume is smaller, that still isn't enough to tell us that it certainly will fit).

Hence here, you need to actually look at the packing arrangement shown in the diagram. It shows that the 19.3 cm length is stacked against the 20 cm length, and the bounds are 19.25 to 19.35, so this is definitely less than 20; the 14.2 cm length is stacked against the 15 cm length, and the bounds are 14.15 to 14.25, so this is definitely less than 15; seventeen copies of the 1.5 cm length are stacked against the 28 cm length, and the bounds are 1.45 to 1.55, so 17 copies of this length will be between 17*1.45 = 24.65 and 17*1.55 = 26.35, so the value will definitely be less than 28. Thus by considering the specific packing arrangement in all three dimensions, we have proved that the cases will fit.
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thekidwhogames
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Look at bounds of individual lengths.
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Prasiortle
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(Original post by idek23)
How would I write that in the answer box?
Ignore Y11_Maths as he had the wrong idea to begin with. See post #8 (above) - my full explanation of how to solve your question - if you haven't already.
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Iahmed512
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Done this exact question, but Prasiortle
post to the solution is probably correct.
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SPatel2005
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how do I write the answer in the box plz reply so I can just copy and paste the answer
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Kalgore
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The way to answer this is not as is mentioned above you need to look specifically at the two areas you are interested in which is the width of the box versus the width of the cases.So for the width 28cm the error interval is 27.5cm to 28.5cmThe error interval for the cases is 1.45 to 1.551.55x17= 26.35This is less than 27.5 so therefore the cases will fit.If you post this in your answer you will find that it is correct and you will get 4 marks.
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