Question

A gallon of water weighs about 8.3 lb. An Olympic-sized swimming pool holds 6.6×105%0D%0A6.6%0D%0A×%0D%0A10%0D%0A5%0D%0A gallons. What is the approximate weight of the water in an Olympic-sized swimming pool? Write the answer in scientific notation.(1 point)%0D%0AResponses%0D%0A%0D%0A5.478×106%0D%0A5.478%0D%0A×%0D%0A10%0D%0A6%0D%0A lb.%0D%0A5 point 4 7 8 times 10 to the 6th power lb.%0D%0A%0D%0A54.78×106%0D%0A54.78%0D%0A×%0D%0A10%0D%0A6%0D%0A lb.%0D%0A54 point 7 8 times 10 to the 6th power lb.%0D%0A%0D%0A54.78×105%0D%0A54.78%0D%0A×%0D%0A10%0D%0A5%0D%0A lb.%0D%0A54 point 7 8 times 10 to the 5th power lb.%0D%0A%0D%0A5.478×105%0D%0A5.478%0D%0A×%0D%0A10%0D%0A5%0D%0A lb.

Answer 1

To find the weight of the water in an Olympic-sized swimming pool, we first need to calculate the volume in gallons and then convert that to weight in pounds.

1. **Volume of the pool in gallons**:
The volume of an Olympic-sized swimming pool is given as \( 6.6 \times 10^5 \) gallons.

2. **Weight per gallon**:
Each gallon of water weighs approximately 8.3 pounds.

3. **Total weight**:
To find the total weight of the water in the pool:
\[
\text{Total weight} = \text{Volume} \times \text{Weight per gallon}
\]
\[
\text{Total weight} = (6.6 \times 10^5 \text{ gallons}) \times (8.3 \text{ lb/gallon})
\]
We can calculate:
\[
6.6 \times 8.3 = 54.78
\]
Therefore,
\[
\text{Total weight} = 54.78 \times 10^5 \text{ lb}
\]

4. **Convert to scientific notation**:
We need to express this value in proper scientific notation:
\[
54.78 \times 10^5 = 5.478 \times 10^6 \text{ lb}
\]

So, the approximate weight of the water in an Olympic-sized swimming pool is:
\[
\boxed{5.478 \times 10^6 \text{ lb}}
\]

Answer 2

u sure

Answer 3

Yes, let's confirm the calculations step by step.

1. **Volume of the Olympic-sized swimming pool**:
- The provided volume is \(6.6 \times 10^5\) gallons.

2. **Weight of water per gallon**:
- Each gallon of water weighs approximately 8.3 pounds.

3. **Total weight calculation**:
- To find the total weight of all the water in the pool, we multiply the volume by the weight per gallon:
\[
\text{Total weight} = \text{Volume} \times \text{Weight per gallon} = (6.6 \times 10^5 \text{ gallons}) \times 8.3 \text{ lb/gallon}
\]
- First, we perform the multiplication of the numerical values:
\[
6.6 \times 8.3 = 54.78
\]

4. **Putting it all together**:
- Now we multiply this by \(10^5\):
\[
\text{Total weight} = 54.78 \times 10^5 \text{ lb}
\]
- In scientific notation, it is proper to express it as:
\[
54.78 \times 10^5 = 5.478 \times 10^6 \text{ lb}
\]

Thus, the calculations are correct, and the final answer, which is the approximate weight of the water in an Olympic-sized swimming pool, is indeed:
\[
\boxed{5.478 \times 10^6 \text{ lb}}
\]

If there's anything specific you'd like me to clarify further, feel free to ask!