# Maths bounds question, HELP!Watch

#1

I'm usually content with bounds and I find them very easy for a higher tier topic, but I have never encountered a bounds question like this before. The wording has put me off and I'm really not sure how to present my answer.
0
2 years ago
#2
(Original post by Samii123)

I'm usually content with bounds and I find them very easy for a higher tier topic, but I have never encountered a bounds question like this before. The wording has put me off and I'm really not sure how to present my answer.
Let's say the lower bound of a value x is 3.2431 and the upper bound is 3.2453. We want to write x to the best degree of accuracy possible.

So we could say x = 3 (to the nearest whole number) since anything between the LB and UB will round to 3.

But we can do better than that.

We can say x = 3.2 (1 d.p) since both the LB and UB round to this.

Can we do better than that?

The LB rounds to 3.24 and the UB rounds to 3.25 to 2 d.p. so we can't write x to 2.d.p.

So to the best degree of accuracy, we can say x = 3.2 (1 d.p.).
1
2 years ago
#3
(Original post by Samii123)

I'm usually content with bounds and I find them very easy for a higher tier topic, but I have never encountered a bounds question like this before. The wording has put me off and I'm really not sure how to present my answer.
If it is correct to 2 decimal places, that means the third decimal place can vary.

6.43 can be 4.425 to 4.435 since any number in the interval (that is between) 4.425 and 4.435 will end up rounding to 6.43. Example: 4.427 rounds to 4.43.

So 6.43 is correct to 2 d.p means a lower bound of 4.425 and an upper bound of 4.435.

Do the same for the other question.

Now, to find the maximum (upper bound) of v, you want the square root of (max a)/(min b) and the minimum (lower bound) is square root (min a)/(max b).

Makes sense?

Edit: damn... notnek got there first!
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