Answer
$\mathrm{v}(10) =58.4 \quad$ MHz
In 1990 the processor speed was $ 58.4 \quad$ MHz
$\mathrm{v}(16) =200 \quad$ MHz
In 1996 the processor speed was $ 200 \quad$ MHz
$\mathrm{v}(28)=3800 \quad$ MHz
In 2008 the processor speed was $ 3800 \quad$ MHz
Work Step by Step
t=10 represents the year 1980+10=1990.
$\mathrm{t}=10$ belongs to the first interval, we use $\mathrm{v}(\mathrm{t})=8(1.22)^{10}$
$\mathrm{v}(10)=8(1.22)^{10}=58.4 \quad$ MHz
In 1990 the processor speed was $ 58.4 \quad$ MHz
t=16 represents the year 1980+16=1996.
$\mathrm{t}=16$ belongs to the second interval, we use $\mathrm{v}(\mathrm{t})=400\mathrm{t}-6200$
$\mathrm{v}(16)=400(16)-6200=200 \quad$ MHz
In 1996 the processor speed was $ 200 \quad$ MHz
t=28 represents the year 1980+28=2008.
$\mathrm{t}=28$ belongs to the third interval, we use $\mathrm{v}(\mathrm{t})=3800$
$\mathrm{v}(28)=3800 \quad$ MHz
In 2008 the processor speed was $ 3800 \quad$ MHz
