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