Answer:
All real numbers greater than or equal to 4,999.85 millimeters and less than or equal to 5,000.15 millimeters
Step-by-step explanation:
Let
x -----> values for the center thickness of the lens
we know that
The absolute value that represent this problem is
[tex]\left|x-5,000\right|\le 0.150[/tex]
Solve the absolute value
First case (positive)
[tex]+(x-5,000)\le 0.150[/tex]
Adds 5,000 both sides
[tex]x\le 0.150+5,000[/tex]
[tex]x\le 5,000.15\ mm[/tex]
Second case (negative)
[tex]-(x-5,000)\le 0.150[/tex]
Multiply by -1 both sides
[tex](x-5,000)\ge -0.150[/tex]
Adds 5,000 both sides
[tex]x\ge -0.150+5,000[/tex]
[tex]x\ge 4,999.85\ mm[/tex]
therefore
The extreme acceptable values for the center thickness of the lens is the interval [4,999.85,5,000.15]
All real numbers greater than or equal to 4,999.85 millimeters and less than or equal to 5,000.15 millimeters