An average American household uses about 1.04×10^4 kWh (kilowatt hours) of electricity a year. If a power station generates 2.496×10^10 kWh per year, how many households can it serve? Write your answer in scientific notation, expressed to the exact decimal place.
To find out how many households a power station can serve, we need to divide the total electricity generated by the power station by the average electricity usage per household.
Total electricity generated by power station = 2.496×10^10 kWh
Average electricity usage per household = 1.04×10^4 kWh
Number of households served = Total electricity generated by power station / Average electricity usage per household
Number of households served = (2.496×10^10 kWh) / (1.04×10^4 kWh)
Number of households served = 2.4×10^6
Therefore, the power station can serve 2.4×10^6 households.
There are about 1.6×10^3 meters in a mile. About how many miles are there in 3.52×10^7 meters? Write your answer in scientific notation, expressed to the exact decimal place.
To find out how many miles are in 3.52×107 meters, we need to divide the given number of meters by the number of meters in a mile.
Number of meters in a mile = 1.6×103 meters
Number of meters given = 3.52×107 meters
Number of miles = Number of meters given / Number of meters in a mile
Number of miles = (3.52×107 meters) / (1.6×103 meters)
Number of miles = 2.2×104
Therefore, there are approximately 2.2×104 miles in 3.52×107 meters.
An online data storage facility has 1.326×10^13 bytes of storage available. It has 2.6×10^4 customers. How many bytes of storage can each customer use? Write your answer in scientific notation, expressed to the exact decimal place.
To find out how many bytes of storage each customer can use, we need to divide the total bytes of storage available by the number of customers.
Total bytes of storage available = 1.326×10^13 bytes
Number of customers = 2.6×10^4
Bytes of storage per customer = Total bytes of storage available / Number of customers
Bytes of storage per customer = (1.326×10^13 bytes) / (2.6×10^4 customers)
Bytes of storage per customer = 5.1×10^8
Therefore, each customer can use approximately 5.1×10^8 bytes of storage.
A public aquarium is being built to contain neon tetras. Each neon tetra needs 1.25 gallons of water. The aquarium will contain 1×10^5 gallons of water. How many tetras can it house? Write the answer in scientific notation.
To find out how many tetras the aquarium can house, we need to divide the total gallons of water in the aquarium by the amount of water each neon tetra needs.
Total gallons of water in the aquarium = 1×10^5 gallons
Water required per neon tetra = 1.25 gallons
Number of tetras that can be housed = Total gallons of water in the aquarium / Water required per neon tetra
Number of tetras that can be housed = (1×10^5 gallons) / (1.25 gallons)
Number of tetras that can be housed = 8×10^4
Therefore, the aquarium can house approximately 8×10^4 tetras.
A feather weighs 0.0082 g. How many feathers would you need to make a weight of 4.1×10^3 g? Write the answer in scientific notation.
To find out how many feathers are needed to make a weight of 4.1×10^3 g, we need to divide the total weight we want to achieve by the weight of each feather.
Weight of each feather = 0.0082 g
Total weight desired = 4.1×10^3 g
Number of feathers needed = Total weight desired / Weight of each feather
Number of feathers needed = (4.1×10^3 g) / (0.0082 g)
Number of feathers needed = 5×10^5
Therefore, you would need approximately 5×10^5 feathers to achieve a weight of 4.1×10^3 g.
A rectangular park has an area of 1.36×10^8 ft2 . The length of the park is 1.6×10^4 ft. Divide the area by the length to find the width of the park.(1 point)
Responses
8.5×104 ft.
8.5 times 10 superscript 4 baseline ft.
8.5×103 ft.
8.5 times 10 cubed ft.
8.5×102 ft.
8.5 times 10 squared ft.
8.5×105 ft.