To an appropriate degree of accuracy- huh?

3 significant figures is usually appropriate or use the same as they've done in the question.

Reply 2

For the lower bound you would indeed use 13.5

As for an appropriate degree of accuracy, the area you're using is to 3.s.f, so I'd give the answer in 3.s.f. as well.

Reply 3

Find the lower bound, find the upper bound, and use your head to work out what an appropriate degree of accuracy is. For example, if the lower bound is 5.12984... and the upper bound is 5.13469..., it makes no sense to give more than 3sf, because after that they start disagreeing, and it makes no sense to give less than 3sf, because you know you have at least 3sf right!

Reply 4

CHEM1STRY

OP

11

Sorry, I think I need to explain it further.

It states we write all the figures on the display. It wants a value of the area we can use which is appropriate in the equation

\sqrt{\displaystyle\frac{A}{\pi}} = r

where A is the area and r is the radius. We need to use an appropriate value of A and then give all the figures on the display for r. Why is their appropriate value the upper bound? Surely it should be 14? i.e. they say we need to use A=14.5; why is this?

OP

Sorry

OP

I hope I have explained it well. The answer is 2.1.

Reply 7

Erm, we can't view that.

Reply 8

CHEM1STRY

OP

11

Sorry,

I just don't understand:
a) Why they round the answer to 2.1 (2sf)?
b) Why they use 14.5 as the value for the area?

Reply 9

a) I guess because for the area they used an area that's to 1.d.p, so figured the answer should be to that degree of accuracy as well.
b) Because they are wrong.

Reply 10

SCE1912

You know that a lower bound for r is 2.07
Use A = 14.5 is find an upper bound for r.

can use say with certainty what r must be to 1 dp?
can use say with certainty what r must be to 2 or more dp?

Reply 11

CHEM1STRY

OP

11

SCE1912

You know that a lower bound for r is 2.07
Use A = 14.5 is find an upper bound for r.

can use say with certainty what r must be to 1 dp?
can use say with certainty what r must be to 2 or more dp?

Ah so basically because they quote the area as 14, we should round to two s.f?

So...

\sqrt{\displaystyle\frac{14}{\pi}} = 2.1 (2sf)

That works. Thanks!

Reply 12

Lysdexia

a) I guess because for the area they used an area that's to 1.d.p, so figured the answer should be to that degree of accuracy as well.
b) Because they are wrong.

For ****'s sake.

Disregard the above, it's nonsense. You have lower and upper bounds for the area from the information you're given (i.e. 13.5 < area < 14.5), so you work out lower and upper bounds for the radius and work out how many significant figures they agree to.

CHEM1STRY

Ah so basically because they quote the area as 14, we should round to two s.f?

So...

\sqrt{\displaystyle\frac{14}{\pi}} = 2.1 (2sf)

That works. Thanks!

No, this isn't right. See above.

By taking the area to be 13.5 or higher we get that the radius is 2.07... or more, and by taking the area to be 14.5 or less we get that the radius is 2.14... or less, so the radius is clearly 2.1 to 2sf no matter what the area is (because the upper and lower bounds of the radius are both 2.1 to 2sf). We can't give it to 3sf because we don't know it to 3sf, and giving it to 1sf is stupid.

Reply 13

Alright, he said himself he wasn't being very clear, no need to jump down my ignorant little throat.

Reply 14

Perhaps it will help if I explain a bit more. If we knew, for instance, that the area was 14.00 to 4sf, we would know that the area was somewhere between 13.995 and 14.005. We can therefore find that the radius is somewhere between 2.110627... and 2.111381..., both of which are 2.111 to 4sf. If we knew the area was 14.00000 to 7sf, the area would be between 13.999995 and 14.000005, so the radius would be between 2.1110037458... and 2.1110044997..., both of which are 2.111004 to 7sf.

The fact that both figures are coming out to similar accuracy is no accident, but it's not just some silly hand-wavey "we should always give things to the same degree of accuracy" nonsense. If we knew the area was 10 to 1sf, then we'd know it was between 5 and 15, so the radius would be between 1.26... and 2.18..., so we can't even give that to 1sf.

Reply 15

Christ you're arrogant.

Reply 16

Lysdexia

Alright, he said himself he wasn't being very clear, no need to jump down my ignorant little throat.

I gave the correct solution earlier on in the thread and an explanation; I don't understand why you contradicted both me and the textbook, while saying things like "they gave the area to 1dp" and "the area we're using is to 3sf" (neither of which was true). :confused:

Reply 17

The whole question was very unclear and above everything else my original post was before yours.

Reply 18

Lysdexia

Christ you're arrogant.

Call me what you like; I'm not giving him the wrong answer or saying things that don't make any sense / aren't true.

Reply 19