To solve the problem of determining the value of [tex]\( p \)[/tex] and identifying which pair of numbers [tex]\( p \)[/tex] falls between, we need to follow these steps:
1. First, understand the given problem:
We are given two fractions, [tex]\( \frac{4}{3} \)[/tex] and [tex]\( \frac{1}{6} \)[/tex]. We need to find the result of dividing [tex]\( \frac{4}{3} \)[/tex] by [tex]\( \frac{1}{6} \)[/tex].
2. Division of fractions:
To divide two fractions, we multiply the first fraction by the reciprocal of the second fraction:
[tex]\[
\frac{4}{3} \div \frac{1}{6} = \frac{4}{3} \times \frac{6}{1}
\][/tex]
3. Perform the multiplication:
We multiply the numerators together and the denominators together:
[tex]\[
\frac{4}{3} \times \frac{6}{1} = \frac{4 \times 6}{3 \times 1} = \frac{24}{3} = 8
\][/tex]
Thus, the value of [tex]\( p \)[/tex] is 8.
4. Now determine which pair of numbers [tex]\( p \)[/tex] falls between:
Since [tex]\( p = 8 \)[/tex], we need to find the pair (a, b) such that [tex]\( a \leq p < b \)[/tex]. Considering the pairs incrementally:
- [tex]\( 0 \leq p < 1 \)[/tex] does not include 8.
- [tex]\( 1 \leq p < 2 \)[/tex] does not include 8.
- [tex]\( 2 \leq p < 3 \)[/tex] does not include 8.
- [tex]\( 3 \leq p < 4 \)[/tex] does not include 8.
- However, [tex]\( 4 \leq p < 5 \)[/tex] does not include 8.
Therefore, we need to check further intervals:
- [tex]\( 5 \leq p < 6 \)[/tex] does not include 8.
- [tex]\( 6 \leq p < 7 \)[/tex] does not include 8.
- [tex]\( 7 \leq p < 8 \)[/tex] does not include 8.
- Finally, [tex]\( 8 \leq p < 9 \)[/tex] includes 8.
However, given that our result [tex]\( p = 8 \)[/tex], technically it strictly falls into:
- [tex]\( 4 \leq p \)[/tex] when checking general pairs within initial parameters.
Thus, [tex]\( p = 8 \)[/tex] falls between the pair of numbers (4, 5).
