Rounded value

[darkyadoo]

Good Afternnon,
I am not very familiar with the GCSE and maths in English. Below my reasonnng.

It seems that $x$ and $y$ have been truncated so we have $$5.43\le x<5.44\\4.514\le y<4.515$$Therefore $$\sqrt{\dfrac{5.43}{4.515}} < \sqrt{\dfrac{x}{y} }=w<\sqrt{\dfrac{5.44}{4.514}}$$
Once computing, I give the values truncated with 4 decimal places
$$1.09665 < w <1.09778$$So the lower bound : $$B_L=1.09665$$ and the upper bound : $$B_U=1.09778$$The last question, do they ask the most accurate rounded value of $w$?
See What I did :
$$B_U-B_L=0.0013$$ so it can not be rounded to 3 decimal places, so let's try to 2 decimal places, therefore $$w\approx1.10$$
since $$[B_L;B_U)\subset [1.095; 1.105)$$, this rounded value to 2 decimal places is the most accurate.

 

[lookagain]

I seems that $x$ and $y$ have been truncated so we have $$5.43\le x<5.44\\4.514\le y<4.515$$

Or, would x and y be rounded to their respective number of places to use in the calculation?

x between 5.4250 and 5.4349... ?
y between 4.51350 and 4.51449... ?

 

[darkyadoo]

if x is between 5.4250 and 5.4349... that means that 5.426...could be the true value of x, therefore it is said that x=5.43 correct to 2 decimal places, which means for me that the tenth and the hundredth of the correct value is 4 and 3....maybe I missed something...

 

[darkyadoo]

so "correct to 2 decimal places" means "rounded to 2 decimal places" and not truncated, right?

 

[blamocur]

so "correct to 2 decimal places" means "rounded to 2 decimal places" and not truncated, right?

I believe you are right. At least that's what Google returned on the search for "what it means that a number is correct to two places"

 

[darkyadoo]

Below is my corrected answer :

$$5.425\le x<5.435\\4.5135\le y <4.5145 \iff \dfrac{1}{4.5145}<\dfrac{1}{y} \le \dfrac{1}{4.5135}$$
then $$\sqrt{\dfrac{5.425}{4.5145}}<\sqrt{\dfrac{x}{y}}=w < \sqrt{\dfrac{5.435}{4.5135}} \\ L_B=1.0962132< w <1.0973447=U_B$$
Since $U_B-L_B>0.00113>0.001$, It can not be rounded to 3 decimal places, therefore $w$ wil be rounded to 2 decimal places :$$w\approx 1.10$$ since $(L_B;U_B)\subset [1.095;1.105)$

Now I would like to know if I have answered the following question correctly : "Work out the value of $w$ with a suitable degree of accuracy"

 

[blamocur]

I agree.