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. Which expression gives the number of transistors in a dense integrated circuit in 1979? Choose 1 answer: (A) 5000+0.41^(5) (B) 5000*0.41^(5) (C) 5000(1+0.41)^(5) (D) 5000+(1+0.41)^(5)

Understanding Moore's Law: Understand Moore's Law. Moore's Law suggests that the number of transistors on a dense integrated circuit doubles approximately every two years. However, in this problem, we are given a yearly increase of 41%. This means that each year, the number of transistors is 141% of the previous year's number.Converting the percentage increase: Convert the percentage increase to a decimal multiplier. An increase of 41% per year means that the number of transistors is multiplied by 1.41 (which is 1 + 0.41) each year.Determining the expression for 1979: Determine the expression for the number of transistors in 1979. Since the increase is compounded annually, we need to use an exponential growth formula. The initial number of transistors is 5000, and the growth happens over 5 years (from 1974 to 1979). The expression for the number of transistors in 1979 is therefore 5000 * (1.41)^5.Matching the expression with options: Match the expression with the given options. The correct expression from Step 3 is 5000 * (1.41)^5, which is equivalent to 5000 * (1 + 0.41)^5. This matches option (C) 5000(1+0.41)^(5).

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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) $5000+0.41^{5}$ $\newline$ (B) $5000 \cdot 0.41^{5}$ $\newline$ (C) $5000(1+0.41)^{5}$ $\newline$ (D) $5000+(1+0.41)^{5}$

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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+0.41^{5}

\newline

(B)

5000 \cdot 0.41^{5}

\newline

(C)

5000(1+0.41)^{5}

\newline

(D)

5000+(1+0.41)^{5}

Understanding Moore's Law: Understand Moore's Law. $\newline$ Moore's Law suggests that the number of transistors on a dense integrated circuit doubles approximately every two years. However, in this problem, we are given a yearly increase of $41\%$ . This means that each year, the number of transistors is $141\%$ of the previous year's number.
Converting the percentage increase: Convert the percentage increase to a decimal multiplier. $\newline$ An increase of $41\%$ per year means that the number of transistors is multiplied by $1.41$ (which is $1 + 0.41$ ) each year.
Determining the expression for $1979$ : Determine the expression for the number of transistors in $1979$ . $\newline$ Since the increase is compounded annually, we need to use an exponential growth formula. The initial number of transistors is $5000$ , and the growth happens over $5$ years (from $1974$ to $1979$ ). The expression for the number of transistors in $1979$ is therefore $5000 \times (1.41)^5$ .
Matching the expression with options: Match the expression with the given options. $\newline$ The correct expression from Step $3$ is $5000 \times (1.41)^5$ , which is equivalent to $5000 \times (1 + 0.41)^5$ . This matches option (C) $5000(1+0.41)^{5}$ .