Question

a gallon of water weighs about 8.3 pounds. An Olympic size pool hold 6.6×10 to the fifth power gallons what is the approximate weight of the water of an Olympic size pool right answer in scientific notation

Answer 1

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:

\[
\boxed{5.478 \times 10^6} \text{ pounds}
\]