To an appropriate degree of accuracy- huh?

Question on IGCSE 2007 3H.

The area of a circle is 14cm² to 2sf.

Find the lower bound of the radius.

\sqrt{\displaystyle\frac{13.5}{\pi}} = r = 2.072964897.

That bit is fine. Then it says:

Give the radius of the circle to an appropriate degree of accuracy. I wouldv'e thought you would use the 14cm² for the area, but you use 14.5cm² according to the markscheme. Why is this?

Thanks!

Reply 20

Lysdexia

I don't know how the hell you've got so much reputation. You seem to be one of the rudest people on this forum.

Is this what Cambridge does to you?

Reply 21

generalebriety

16

Lysdexia

No, the temper is purely my own.

As far as I'm concerned, when someone comes along and says "the area is 14 but I don't understand why they've used 14.5" and you say "the area is given to 3sf / 1dp", you're either being an idiot or you haven't bothered to read the question. The area was given to 2sf and 0dp, and that was clear right from the start. When I corrected you (and the OP) and gave an explanation as to why my answer was right, you (and the OP) ignored it, and you went on to say the same thing again, contradicting not only me but also the textbook (edit: or mark scheme or whatever it was). That's arrogant. And as someone who's spent years on this site selflessly helping people out, that rather understandably pisses me off. Now, stop derailing the thread.

Reply 22

CHEM1STRY

OP

11

generalebriety

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.

Ah yes, both the lower and upper bounds give us 2.1 (2sf), and because of the fact that we only have the area to 2sf, we can only quote our answer as 2.1.

Reply 23

Lysdexia

I'm derailing nothing. The thread ended when you insulted three of the people in this topic, had a hissy fit because we were ignoring you/didn't read your post (the latter is what I did), and then carried on your obnoxious little rampage, giving me orders and jumping up on your high horse. A simple "That's not right" is what most people with manners would have said, but simple, polite social skills clearly escape you.

Anyway, you don't have to worry about my "derailing" this thread anymore because I'm out of it. Have the final say if you want.

Reply 24

CHEM1STRY

OP

11

Sorry I didn't understand the post generalebriety, but I get it now and you have helped alot! Thanks :smile:

Reply 25

generalebriety

16

CHEM1STRY

because of the fact that we only have the area to 2sf

No, the actual statement of the accuracy to which you know the area is irrelevant. You use the bounds you have on the area to find bounds on the radius, and you are then allowed to quote the radius to as many significant figures as its lower and upper bounds agree on. If knowing the area to 2sf meant that the lower and upper bounds on the radius agreed to 4sf (ok, it wouldn't, but hypothetically...), you could give it to 4sf instead.

Reply 26

Siddd

BULLLLLLLLLLLLLLLLLLSSSSHHHHHHHHHHHIIIIIIIIIIIIIIIIITTTTTTTTTTTEEEEEEEEEE :wink:

yeh

Reply 27

CHEM1STRY

OP

11

generalebriety

No, the actual statement of the accuracy to which you know the area is irrelevant. You use the bounds you have on the area to find bounds on the radius, and you are then allowed to quote the radius to as many significant figures as its lower and upper bounds agree on. If knowing the area to 2sf meant that the lower and upper bounds on the radius agreed to 4sf (ok, it wouldn't, but hypothetically...), you could give it to 4sf instead.