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
5.478×106
lb.
5.478 times 10 superscript 6 baseline lb.
5.478×105
lb.
5.478 times 10 superscript 5 baseline lb.
54.78×106
lb.
54.78 times 10 superscript 6 baseline lb.
54.78×105
lb.
gallons. What is the approximate weight of the water in an Olympic-sized swimming pool? Write the answer in scientific notation.(1 point)
Responses
5.478×106
lb.
5.478 times 10 superscript 6 baseline lb.
5.478×105
lb.
5.478 times 10 superscript 5 baseline lb.
54.78×106
lb.
54.78 times 10 superscript 6 baseline lb.
54.78×105
lb.
Answers
GPT-4o mini
To find the weight of the water in an Olympic-sized swimming pool, you can multiply the volume of the pool in gallons by the weight of a gallon of water.
1. The number of gallons in an Olympic-sized swimming pool is approximately \(6.6 \times 10^5\) gallons.
2. The weight of a gallon of water is about 8.3 pounds.
Now, you can calculate the total weight:
\[
\text{Total weight} = \text{Volume in gallons} \times \text{Weight per gallon}
\]
\[
\text{Total weight} = (6.6 \times 10^5 \, \text{gallons}) \times (8.3 \, \text{lb/gallon})
\]
Calculating that:
\[
6.6 \times 8.3 = 54.78
\]
Then, you multiply the powers of ten:
\[
\text{Total weight} = 54.78 \times 10^5 \, \text{lb}
\]
To express this in scientific notation, we need it to be in the form of \(a \times 10^b\) where \(1 \leq a < 10\). We can convert \(54.78 \times 10^5\) as follows:
\[
54.78 = 5.478 \times 10^1
\]
So,
\[
54.78 \times 10^5 = 5.478 \times 10^1 \times 10^5 = 5.478 \times 10^{6}
\]
Thus, the approximate weight of the water in an Olympic-sized swimming pool is:
\[
\boxed{5.478 \times 10^6 \, \text{lb}}
\]
1. The number of gallons in an Olympic-sized swimming pool is approximately \(6.6 \times 10^5\) gallons.
2. The weight of a gallon of water is about 8.3 pounds.
Now, you can calculate the total weight:
\[
\text{Total weight} = \text{Volume in gallons} \times \text{Weight per gallon}
\]
\[
\text{Total weight} = (6.6 \times 10^5 \, \text{gallons}) \times (8.3 \, \text{lb/gallon})
\]
Calculating that:
\[
6.6 \times 8.3 = 54.78
\]
Then, you multiply the powers of ten:
\[
\text{Total weight} = 54.78 \times 10^5 \, \text{lb}
\]
To express this in scientific notation, we need it to be in the form of \(a \times 10^b\) where \(1 \leq a < 10\). We can convert \(54.78 \times 10^5\) as follows:
\[
54.78 = 5.478 \times 10^1
\]
So,
\[
54.78 \times 10^5 = 5.478 \times 10^1 \times 10^5 = 5.478 \times 10^{6}
\]
Thus, the approximate weight of the water in an Olympic-sized swimming pool is:
\[
\boxed{5.478 \times 10^6 \, \text{lb}}
\]