Expressions and Equations: Performing Operations with Scientific Notation (CCSS.8.EE.4)
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Common Core 8th Grade Math › Expressions and Equations: Performing Operations with Scientific Notation (CCSS.8.EE.4)
Each of $2\times 10^{3}$ servers can handle $3\times 10^{4}$ requests per second at peak. About how many requests per second can the data center handle in total? Give your answer in scientific notation.
$5\times 10^{7}$
$6\times 10^{12}$
$6\times 10^{7}$
$6\times 10^{6}$
Explanation
Multiply coefficients: $3\times 2=6$. Add exponents: $10^{4}\times 10^{3}=10^{4+3}=10^{7}$. Total $=6\times 10^{7}$ requests/second.
A shipment has a total mass of $4\times 10^{12}$ grams and is evenly divided into $8\times 10^{3}$ identical crates. What is the mass of one crate, in scientific notation?
$5\times 10^{8}$ g
$5\times 10^{9}$ g
$0.5\times 10^{8}$ g
$4\times 10^{9}$ g
Explanation
Divide coefficients: $4\div 8=0.5$. Subtract exponents: $10^{12}\div 10^{3}=10^{12-3}=10^{9}$. So $0.5\times 10^{9}=5\times 10^{8}$ grams after adjusting the coefficient to be between 1 and 10.
A GPS readout shows a glacier's speed as 3E-3 kilometers per year (technology notation). What is this speed in millimeters per year, written in scientific notation?
$3\times 10^{-9}$ mm/year
$3\times 10^{3}$ mm/year
$3\times 10^{2}$ mm/year
$3\times 10^{9}$ mm/year
Explanation
Interpret 3E-3 as $3\times 10^{-3}$ km/year. Convert km to mm: $1\text{ km}=10^{6}\text{ mm}$. So $3\times 10^{-3}\times 10^{6}=3\times 10^{3}$ mm/year.
A camera sensor records $3 \times 10^{4}$ pixels per row and has $2 \times 10^{3}$ rows. How many pixels are in the whole image?
$6 \times 10^{7}$
$3 \times 10^{7}$
$6 \times 10^{12}$
$6 \times 10^{6}$
Explanation
Multiply coefficients: $3\times 2=6$. Add exponents: $10^{4}\times 10^{3}=10^{4+3}=10^{7}$. So the product is $6\times 10^{7}$.
A transmitter outputs $8 \times 10^{5}$ watts and splits the power evenly among $4 \times 10^{2}$ receivers. How many watts does each receiver get?
$2 \times 10^{8}$
$12 \times 10^{3}$
$2 \times 10^{3}$
$2 \times 10^{-3}$
Explanation
Divide coefficients: $8\div 4=2$. Subtract exponents: $10^{5}\div 10^{2}=10^{5-2}=10^{3}$. Result: $2\times 10^{3}$ watts.
Which quantity is larger?
- Distance A: $5 \times 10^{11}$ meters
- Distance B: 600 billion meters
Quantity A is larger
Quantity B is larger
They are equal
Cannot be determined
Explanation
600 billion meters is $6 \times 10^{11}$ meters. With the same power of 10, compare coefficients: $6>5$, so Distance B is larger.
A seafloor spreads at 3.0E-2 meters per year (calculator display), i.e., $3.0 \times 10^{-2}$ m/year. Express this speed in millimeters per year.
$3 \times 10^{-5}$ mm/year
$3 \times 10^{5}$ mm/year
$3 \times 10^{-2}$ mm/year
$3 \times 10^{1}$ mm/year
Explanation
Convert meters to millimeters by multiplying by $10^{3}$: $(3.0\times 10^{-2})\times (1\times 10^{3})=3.0\times 10^{1}=30$ mm/year.
Each crate holds $2 \times 10^{3}$ screws. A shipment contains $3 \times 10^{4}$ crates. How many screws are in the shipment, written in scientific notation?
$6 \times 10^{7}$
$5 \times 10^{7}$
$6 \times 10^{12}$
$6 \times 10^{1}$
Explanation
Multiply coefficients and add exponents: $(2 \times 10^{3})(3 \times 10^{4}) = (2\cdot 3) \times 10^{3+4} = 6 \times 10^{7}$.
A probe travels $1.2 \times 10^{8}$ km in $3 \times 10^{4}$ s. What is its average speed in km/s written in scientific notation?
$4 \times 10^{4}$ km/s
$4 \times 10^{3}$ km/s
$0.4 \times 10^{4}$ km/s
$0.4 \times 10^{3}$ km/s
Explanation
Divide coefficients and subtract exponents: $\dfrac{1.2 \times 10^{8}}{3 \times 10^{4}} = (\dfrac{1.2}{3}) \times 10^{8-4} = 0.4 \times 10^{4} = 4 \times 10^{3}$ km/s.
Ridge A spreads at $3.2 \times 10^{-2}$ m/year. Ridge B spreads at 25 mm/year. Which ridge is spreading faster?
They are equal
Ridge B
Not enough information
Ridge A
Explanation
Convert Ridge A: $3.2 \times 10^{-2}$ m/year $= 0.032$ m/year $= 32$ mm/year. Compare to 25 mm/year, so Ridge A (32 mm/year) is faster.