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mbma526

  • one year ago

Will Fan And Medal Please help in a science lab, a cell is multiplying at a constant rate per second. The scientist recorded the cell's growth for the first five seconds. The data set was 2, 4, 16, 256, 65,536. Describe the growth pattern. How might knowing this pattern help the scientist to predict the cell's growth over a period of 10 seconds?

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  1. mbma526
    • one year ago
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    @heretohelpalways

  2. anonymous
    • one year ago
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    The pattern is squaring 2² = 4 4² = 16 16² = 256, etc

  3. anonymous
    • one year ago
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    1 second = 2^1 = 2 cells 2 seconds = 2^2 = 4 cells 3 seconds = 2^4 = 16 cells 4 seconds = 2^8 = 256 cells 5 seconds = 2^16 = 65,526 cells ---------------------------- N = 2^(2^(s-1)) where N = number of cells and s = seconds ---------------------------- It checks ... 1 second N = 2^(2^(1-1)) = 2^(2^0) = 2^1 = 2 2 seconds N = 2^(2^(2-1)) = 2^(2^1) = 2^2 = 4 3 seconds N = 2^(2^(3-1)) = 2^(2^2) = 2^4 = 16 4 seconds N = 2^(2^(4-1)) = 2^(2^3) = 2^8 = 256 5 seconds N = 2^(2^(5-1)) = 2(2^4) = 2^16 = 65,636 So for 10 seconds we have: N = 2^(2^(10 - 1)) = 2^(2^9) = 2^512 That is approximately equal to 1.3408 x 10^154 NOTE: The question is TOTALLY unrealistic. 1) No cell - not even the faster dividing bacterium - can divide in just 1 second. So it is impossible to even get N = 2^s, let alone anything like N = 2^2^(s-1)). 2) The number of cells after 10 second is impossible. There are only an estimated 10^80 elementary particles in the observable universe. Therefore, it is quite impossible to have anything like 10^154 cells, even if each cell consisted of nothing but a single elementary particle. 3) LONG LONG before a real population could reach a number of cells anywhere even remotely close to 10^80 (let alone 10^154), the population would stop growing: resources would become limited and wastes would accumulate. No population can continue growing at a geometric or exponential rate.

  4. mbma526
    • one year ago
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    thank you @peachpi @emilyalborn12

  5. anonymous
    • one year ago
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    you're welcome

  6. anonymous
    • one year ago
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    https://answers.yahoo.com/question/index?qid=20100606233945AARxvyE @emilyalborn12 give your source credit

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