To solve this problem, we need to find the equivalence of bippities (B), bops (P), and boos (O) in the same unit and then sum them up.
### Step-by-Step Solution:
1. Convert Bippities to Bops:
- We are given that \(\(6\)\) bippities equal \(\(3\)\) bops.
- Therefore, \(\(1\)\) bippity equals \(\(\frac{3}{6} = \frac{1}{2}\)\) bops.
2. Convert Bops to Boos:
- We are also given that \(\(6\)\) bops equal \(\(2\)\) boos.
- Therefore, \(\(1\)\) bop equals \(\(\frac{2}{6} = \frac{1}{3}\)\) boos.
3. Convert Bippities to Boos:
- From the conversion above, we have found that \(\(1\)\) bippity = \(\(\frac{1}{2}\)\) bops.
- Since \(\(1\)\) bop equals \(\(\frac{1}{3}\)\) boos, we can substitute this into the bippity-to-bops conversion:
- Hence, \(\(1\)\) bippity = \(\(\frac{1}{2} \times \frac{1}{3} = \frac{1}{6}\)\) boos.
4. Summing up One Bippity, One Bop, and One Boo:
- We now have the following equivalences:
- \(\(1\)\) bippity = \(\(\frac{1}{6}\)\) boos.
- \(\(1\)\) bop = \(\(\frac{1}{3}\)\) boos.
- \(\(1\)\) boo = \(\(1\)\) boo (this is already in boos).
- Adding these together:
\(\[
\frac{1}{6} \text{ boos (from 1 bippity)} + \frac{1}{3} \text{ boos (from 1 bop)} + 1 \text{ boo}
\]\)
- To add these fractions, we find a common denominator, which is \(\(6\)\):
\(\[
\frac{1}{6} + \frac{2}{6} + 1
\]\)
\(\[
= \frac{1}{6} + \frac{2}{6} + \frac{6}{6}
\]\)
\(\[
= \frac{1 + 2 + 6}{6}
\]\)
\(\[
= \frac{9}{6} = 1.5
\]\)
Thus, there are 1.5 boos in one bippity, one bop, and one boo combined.
