#### Answer

$|d-0.11885|\le0.00015$

#### Work Step by Step

Half-way between the given values, $
0.1187\text{ and } 0.1190
,$ is
\begin{array}{l}\require{cancel}
\dfrac{0.1187+0.1190}{2}
\\\\=
\dfrac{0.2377}{2}
\\\\=
0.11885
.\end{array}
The distance from $
0.11885
$ to $0.1187$ and $0.1190$ is $
0.00015
.$
Since $|x|\lt c$ implies $-c\lt x\lt c$ (or $|x|\le c$ implies $-c\le x\le c$), then the given inequality, $
0.1187\le d \le0.1190
,$ is equivalent to
\begin{array}{l}\require{cancel}
|d-0.11885|\le0.00015
.\end{array}