Rounded value

[darkyadoo]

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

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

 

[lookagain]

I seems that [imath]x[/imath] and [imath]y[/imath] have been truncated so we have [math]5.43\le x<5.44\\4.514\le y<4.515[/math]

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?