Question

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:

5000 - 0.415

5000(1 +0.41)

5000+ 0.41

5000 + (1 +0.41)5

Answer 1

432.34.234.32

Explanation:

Answer 2

The expression gives the number of transistors in a dense integrated circuit in 1979 is:

`y = 5000(1+0.41)^5\\\\or\\\\y = 5000(1.41)^5`

Solution:

The increasing function is given as:

`y = a(1+r)^t`

Where,

y is the future value

a is the initial value

r is the growth rate

t is the number of years

From given,

a = 5000

`r = 41 \% = (41)/(100) = 0.41`

t = 1974 to 1979 = 5 years

Substituting the values in formula,

`y = 5000(1+0.41)^5\\\\y = 5000(1.41)^5`

Thus the expression gives the number of transistors in a dense integrated circuit in 1979 is found