QUestion is attached httpsgyazocom37a05a41e43bb9fe84d69d23d6

QUestion is attached

https://gyazo.com/37a05a41e43bb9fe84d69d23d666abbc

Solution

The growth function G (t) is given by

G(t) = mG₀ / [ G₀ + ( m – G₀) e^-kmt } where G₀ = 2 million is the initial number of internet users in 1990 when t = 0, k = 0.0018, t is the number of years since 1990, and m = 350 million is the maximum number at which the total internet users are expected to level out.

In 1995, t = 5 so the number of internet users is 350*2/ [ 2 + ( 350 – 2)e^{-0.0018*350*5} ] = 700/ [ 2 + 348e^-3.15 ] = 700/ ( 2 + 348 * 0.04285212686) = 700/ ( 2 + 14.91) = 700/ 16.91 = 41.40 million ( approximately)

In 2000, t = 10 , so that the number of internet users is 350*2/ [ 2 + ( 350 – 2)e^{-0.0018*350*10} ] =                            700 / ( 2 + 348e^{- 6.30} ) = 700 / ( 2 + 348*0.00183630477) = 700/ ( 2 + 0.64) = 700/2.64 = 265.15 million (approx).

In 2010, t = 20, so that the number of internet users is 350*2/ [ 2 + ( 350 – 2)e^{-0.0018*350*20} ] =                            700 / ( 2 + 348e^{- 12.60} ) = 700 / ( 2 + 348*0.00000337201) = 700/ ( 2 + 0.0012) = 700/2.0012 = 349.79 million ( approx)

During 1990-1995, the growth in the number of internet users is (41.2- 2)* 100/ 2 * 5 = 3920/10 = 392 % p.a.

During 1995- 2010, the rate of growth in the number of internet users is ( 265.15 – 41.20)* 100 / 41.20 * 15 =

22395/ 618 = 36.24 % ( approx) p.a.