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tensa zangetsu [6.8K]
3 years ago
6

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

nse integrated circuit was produced with 5000 transistors.
Which expression gives the number of transistors in a dense integrated circuit in 1979?
Mathematics
2 answers:
Veseljchak [2.6K]3 years ago
8 0

Answer:

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

Lilit [14]3 years ago
7 0

Answer:

5000(1+0.41)^5

Step-by-step explanation:

did the khan academy

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