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Question Number 201972 by Lekhraj last updated on 17/Dec/23
Answered by mr W last updated on 18/Dec/23
Ξ¦β©½x=12[1+erf(xβΞΌ2Ο)]witherf(x)=2Οβ«0xeβt2dt(i)(a)12[1+erf(45β60102)]=12(1β0.886)β6%(i)(b)12[erf(75β60102)βerf(50β60102)]=12(0.866β(β0.683))β77%(ii)12[1+erf(xβ60102)]=0.9erf(xβ60102)=0.8xβ60102=erfβ1(0.8)=0.9062βx=60+0.9062Γ102β73(%)
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