To find the weight of the water in an Olympic-sized pool, we can multiply the volume of the pool in gallons by the weight of one gallon of water.
Given:
- Weight of water per gallon: \( 8.3 \) pounds
- Volume of an Olympic-sized pool: \( 6.6 \times 10^5 \) gallons
Now, calculate the weight of the water:
\[ \text{Weight of water} = \text{Volume} \times \text{Weight per gallon} \]
\[ \text{Weight of water} = (6.6 \times 10^5 \text{ gallons}) \times (8.3 \text{ pounds/gallon}) \]
Calculating this:
\[ 6.6 \times 8.3 = 54.78 \]
Now, for the powers of ten:
\[ 10^5 \text{ gallons} \text{ contributes with } 10^0 \text{ (from the weight of water)} \]
So, we have:
\[ \text{Weight of water} = 54.78 \times 10^5 \text{ pounds} \]
To express this in scientific notation, we convert \( 54.78 \) to \( 5.478 \times 10^1 \):
\[ 54.78 \times 10^5 = 5.478 \times 10^1 \times 10^5 \]
Adding the exponents gives us:
\[ \text{Weight of water} = 5.478 \times 10^6 \text{ pounds} \]
So, the approximate weight of the water in an Olympic-sized pool is:
\[ \boxed{5.478 \times 10^6} \text{ pounds} \]