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
54.78×105
lb.
54.78 times 10 superscript 5 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.
5.478×106
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
54.78×105
lb.
54.78 times 10 superscript 5 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.
5.478×106
lb.
Answers
GPT-4o mini
To find the weight of the water in an Olympic-sized swimming pool, we can multiply the weight of one gallon of water by the total number of gallons in the pool:
\[
\text{Weight of water} = \text{Weight per gallon} \times \text{Total gallons}
\]
Given that the weight of one gallon of water is about 8.3 lb and the pool holds \(6.6 \times 10^5\) gallons, we can calculate:
\[
\text{Weight of water} = 8.3 \, \text{lb/gallon} \times 6.6 \times 10^5 \, \text{gallons}
\]
Calculating this step by step:
1. First multiply \(8.3\) by \(6.6\):
\[
8.3 \times 6.6 = 54.78
\]
2. Now, since we are multiplying by \(10^5\), we can express the total weight as:
\[
\text{Weight of water} = 54.78 \times 10^5 \, \text{lb}
\]
3. To express it in proper scientific notation, we note that \(54.78\) can be rewritten as \(5.478 \times 10^1\). Hence:
\[
54.78 \times 10^5 = 5.478 \times 10^1 \times 10^5 = 5.478 \times 10^{6} \, \text{lb}
\]
So, the approximate weight of the water in an Olympic-sized swimming pool is:
\[
\boxed{5.478 \times 10^{6} \, \text{lb}}
\]
\[
\text{Weight of water} = \text{Weight per gallon} \times \text{Total gallons}
\]
Given that the weight of one gallon of water is about 8.3 lb and the pool holds \(6.6 \times 10^5\) gallons, we can calculate:
\[
\text{Weight of water} = 8.3 \, \text{lb/gallon} \times 6.6 \times 10^5 \, \text{gallons}
\]
Calculating this step by step:
1. First multiply \(8.3\) by \(6.6\):
\[
8.3 \times 6.6 = 54.78
\]
2. Now, since we are multiplying by \(10^5\), we can express the total weight as:
\[
\text{Weight of water} = 54.78 \times 10^5 \, \text{lb}
\]
3. To express it in proper scientific notation, we note that \(54.78\) can be rewritten as \(5.478 \times 10^1\). Hence:
\[
54.78 \times 10^5 = 5.478 \times 10^1 \times 10^5 = 5.478 \times 10^{6} \, \text{lb}
\]
So, the approximate weight of the water in an Olympic-sized swimming pool is:
\[
\boxed{5.478 \times 10^{6} \, \text{lb}}
\]