Question

Difficulty: MediumDecimals and Scientific Notation

An astronomical digital sensor captures 4.5×1074.5 \times 10^7 raw data packets during an observation run, where each packet has a size of 1.6×1041.6 \times 10^{-4} megabytes. A compression algorithm reduces the total volume of these combined packets by 75%75\%. What is the total volume, in megabytes, of the compressed data expressed in scientific notation?

  1. 1.8×1031.8 \times 10^3Answer
  2. B
    5.4×1035.4 \times 10^3
  3. C
    7.2×1037.2 \times 10^3
  4. D
    1.8×1041.8 \times 10^4
  5. E
    7.2×1037.2 \times 10^{-3}

Answer

1.8×1031.8 \times 10^3 megabytes
Multiplying 4.5×1074.5 \times 10^7 by 1.6×1041.6 \times 10^{-4} gives 7.2×1037.2 \times 10^3 megabytes. Decreasing this total by 75%75\% retains 25%25\% of the volume, which equals 0.25×7.2×103=1.8×1030.25 \times 7.2 \times 10^3 = 1.8 \times 10^3 megabytes.

Step-by-Step Solution

1
Calculate the total uncompressed size by multiplying the packet count by individual packet size.
(4.5×107)×(1.6×104)=(4.5×1.6)×107+(4)=7.2×103(4.5 \times 10^7) \times (1.6 \times 10^{-4}) = (4.5 \times 1.6) \times 10^{7 + (-4)} = 7.2 \times 10^3 megabytes.
Total volume is the product of quantity and unit size.
2
Determine the remaining fraction of data after a 75%75\% size reduction.
100%75%=25%=0.25100\% - 75\% = 25\% = 0.25
A 75%75\% reduction leaves 25%25\% of the original data.
3
Multiply the uncompressed volume by 0.250.25 and write the result in standard scientific notation.
0.25×(7.2×103)=1.8×1030.25 \times (7.2 \times 10^3) = 1.8 \times 10^3 megabytes.
The coefficient 1.81.8 meets the condition 1a<101 \leq a < 10 for scientific notation a×10na \times 10^n.

Key Concept

Decimals and Scientific Notation
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