Mathswatch Bounds / Measures of Accuracy
Watch
Announcements
Page 1 of 1
Skip to page:
Report
#2
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
0
reply
(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
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
0
reply
Report
#4
(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
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
0
reply
Report
#5
(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?
I tried out what you said and it came out as incorrect. Is there anything that I would need to do or correct here?
0
reply
(Original post by Prasiortle)
A volume calculation won't be enough: you will need to consider all three dimensions (length, width, height) separately.
A volume calculation won't be enough: you will need to consider all three dimensions (length, width, height) separately.
0
reply
(Original post by Y11_Maths)
Show each length of the disc case is less than the length of the box
Show each length of the disc case is less than the length of the box
0
reply
Report
#8
(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
What would I need to consider when it comes to the three dimensions, if I wouldn't just need volume. I'm bamboozled
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.
0
reply
Report
#10
(Original post by idek23)
How would I write that in the answer box?
How would I write that in the answer box?
0
reply
Report
#11
Done this exact question, but Prasiortle
post to the solution is probably correct.
post to the solution is probably correct.
0
reply
Report
#12
how do I write the answer in the box plz reply so I can just copy and paste the answer
Last edited by SPatel2005; 10 months ago
0
reply
Report
#13
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.
0
reply
X
Page 1 of 1
Skip to page:
Quick Reply
Back
to top
to top