Question

A gallon of water weighs about 8.3 lb. An Olympic-sized swimming pool holds 6.6×105
gallons. What is the approximate weight of the water in an Olympic-sized swimming pool? Write the answer in scientific notation.(1 point)
Responses

5.478×105
lb.
5.478 times 10 superscript 5 baseline lb.

54.78×105
lb.
54.78 times 10 superscript 5 baseline lb.

54.78×106
lb.
54.78 times 10 superscript 6 baseline lb.

5.478×106
lb.

Answer 1

To find the approximate weight of the water in an Olympic-sized swimming pool, we can use the following calculation:

1. The volume of the swimming pool is \( 6.6 \times 10^5 \) gallons.
2. Each gallon of water weighs about 8.3 lbs.

Now, we calculate the total weight:

\[
\text{Total weight} = \text{Volume} \times \text{Weight per gallon}
\]

\[
\text{Total weight} = (6.6 \times 10^5 \text{ gallons}) \times (8.3 \text{ lb/gallon})
\]

Performing the multiplication:

\[
\text{Total weight} = 6.6 \times 8.3 \times 10^5
\]

Calculating \( 6.6 \times 8.3 \):

\[
6.6 \times 8.3 = 54.78
\]

Thus, we have:

\[
\text{Total weight} = 54.78 \times 10^5 \text{ lb}
\]

In scientific notation, we should express this with a single digit before the decimal place:

\[
54.78 \times 10^5 \text{ lb} = 5.478 \times 10^6 \text{ lb}
\]

So the approximate weight of the water in an Olympic-sized swimming pool is:

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