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24 May, 07:32

Moore's law says that the number of transistors in a dense integrated circuit increases by 41, percent every year. In 1974, a dense integrated circuit was produced with 5000 transistors.

Which expression gives the number of transistors in a dense integrated circuit in 1979?

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Answers (2)
  1. 24 May, 07:56
    0
    5000 (1+0.41) ^5

    Step-by-step explanation:

    did the khan academy
  2. 24 May, 11:05
    0
    5000 (1+0.41) ^5

    Step-by-step explanation:

    Growth at a rate of 41% means the number of transistors keeps its 100%, and adds 41%, percent more. So each year, the number is multiplied by 100% + 41%, which is the same as a factor of (1+0.41).

    After 0 years (i. e., in 1974), the number of transistors in a dense integrated circuit was 5000.

    After 1 year, that number was multiplied by 1 + 0.41, so it was 5000 (1+0.41)

    After 2 years, the number was again multiplied by 1+0.41, so 5000⋅ (1+0.41) ⋅ (1+0.41), which is 5000 (1+0.41) ^2

    We can keep doing this until we get to 5 years after 1974, which is 1979.

    In conclusion, this is an expression for the number of transistors in a dense integrated circuit in 1979:

    5000 (1+0.41) ^5
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