Determine the 3 digit chopping and rounding approximately of the number 0.9

We want to "chop and round" the periodic number 0.9 at 3 digits.

First, remember that we can write our periodic number as:

0.9 = 0.9999...

Such that the nine repeats infinitely.

Now we want to chop it at the third digit, and then round.

To chop at the third digit we just "cut" the number at the third digit after the decimal point, we will get:

0.9 ≈ 0.999

Now we round.

Remember that to round a number, we need to look at the last digit after the decimal point.

If the last digit is 5 or larger, we round up, adding 1 to the previous decimal.

If the last digit is 4 or smaller, we round down, don't adding anything to the previous decimal.

in our number:

0.999

The last digit is a 9, so we round up.

Then we add "1" to the previous decimal

but the previous decimal is a 9, so when we add 1, it will transform into a 10.

So it also adds one to the previous decimal, which also is a 9, so the process repeats, this time adding 1 to the decimal to its left, in this case, that decimal will be the first decimal at the left of the decimal point.

So after rounding, we will get:

0.999 ≈ 1

Then we can conclude that:

he 3 digit chopping and rounding approximately of the number 0.9 is 1"

d i think

step-by-step explanation:

i think asking google will you get the answer to that question