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T3=7.94

by applying formula $T(n+1)=\alpha (Tn)+(1-\alpha )\tau n 0<=\alpha <=1$

T0=20, where B.T=(t0,t1,t2)=(6,2,8)

$T1=\alpha (t0)+(1-\alpha )T0$

      =(0.99)(6)+(1-0.99)(20)

       =6.14

T2=(0.99)(2)+(1-0.99)(6.14)

    =2.04

T3=(0.99)(8)+(1-0.99)(2.04)

   =7.94

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