Answer: |x-430|≤20
Explanation:
Let's call x₀=430grams and let's call δ= 20grams. So x₀
is the recommended weight but it is allowed to vary + or - 20 grams. We can see it as an interval of real numbers whose center is 430 and whose ratio is 20.
If we solve this inequation we will obtain -20≤ x-430≤20.
If we add both sides 430, we will obtain -20+430≤x≤430+20.
This will be mean that the weight vary from 410grams to 450 grams.