Moore's law says that the number of transistors in a dense integrated circuit increases by $41 \%$ every year. In $1974$ , a dense integrated circuit was produced with $5000$ transistors. $\newline$ Which expression gives the number of transistors in a dense integrated circuit in $1979 ?$ $\newline$ Choose $1$ answer: $\newline$ (A) $5000 \cdot 0.41^{5}$ $\newline$ (B) $5000+(1+0.41)^{5}$ $\newline$ (C) $5000(1+0.41)^{5}$ $\newline$ (D) $5000+0.41^{5}$

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Grade 8

Divide numbers written in scientific notation

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Q. Moore's law says that the number of transistors in a dense integrated circuit increases by

41 \%

every year. In

1974

, a dense integrated circuit was produced with

5000

transistors.

\newline

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

1979 ?

\newline

Choose

1

answer:

\newline

(A)

5000 \cdot 0.41^{5}

\newline

(B)

5000+(1+0.41)^{5}

\newline

(C)

5000(1+0.41)^{5}

\newline

(D)

5000+0.41^{5}

Understand Moore's Law: Understand Moore's Law in the context of the problem. $\newline$ Moore's Law suggests that the number of transistors on a dense integrated circuit doubles approximately every two years. However, in this problem, it is stated that the number of transistors increases by $41\%$ every year. This means that each year, the number of transistors is $141\%$ (or $1.41$ times) of the previous year's number.
Determine Initial Transistors: Determine the initial number of transistors and the number of years that have passed. $\newline$ The initial number of transistors in $1974$ is given as $5000$ . The number of years from $1974$ to $1979$ is $5$ years.
Apply Percentage Increase: Apply the percentage increase to find the expression for the number of transistors in $1979$ . $\newline$ Since the number of transistors increases by $41\%$ each year, we multiply the initial number by $(1 + 0.41)$ for each year that has passed. This is equivalent to raising $(1 + 0.41)$ to the power of the number of years, which is $5$ in this case.
Write Expression: Write the expression using the information from the previous steps. $\newline$ The expression for the number of transistors in $1979$ is the initial number of transistors multiplied by $(1 + 0.41)$ raised to the power of $5$ . This gives us: $\newline$ $5000 \times (1 + 0.41)^5$
Match with Choices: Match the expression with the given choices. $\newline$ The correct expression that matches our calculation is: $\newline$ $5000 \times (1.41)^5$ $\newline$ This corresponds to choice (C) $5000(1+0.41)^{5}$ .