Let a is a constant, thenf(t)=a[(t+10)−2]=a(t+8),0<t<40a∫040t+8dt=1a[t22+8t]040=1a[4022+8(40)]=1a(1120)=1∴a=11120P(t<10)=11120∫010t+8dt=11120[t22+8t]040=11120[1022+8(10)]=11120×130=13112\text {Let a is a constant, then}\\ f(t)=a[(t+10)-2]=a(t+8), 0<t<40\\ a\int _0^{40}t+8dt=1\\ a\left[\frac{t^2}{2}+8t\right]^{40}_0=1\\ a[\frac{40^2}{2}+8(40)]=1 a(1120)=1\\ \therefore a=\frac{1}{1120}\\ P(t<10)=\frac{1}{1120}\int _0^{10}t+8dt\\ =\frac{1}{1120}\left[\frac{t^2}{2}+8t\right]^{40}_0\\ =\frac{1}{1120}[\frac{10^2}{2}+8(10)]\\ =\frac{1}{1120}\times 130\\ =\frac{13}{112}Let a is a constant, thenf(t)=a[(t+10)−2]=a(t+8),0<t<40a∫040t+8dt=1a[2t2+8t]040=1a[2402+8(40)]=1a(1120)=1∴a=11201P(t<10)=11201∫010t+8dt=11201[2t2+8t]040=11201[2102+8(10)]=11201×130=11213
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Thank you for a feedback. The question was solved using those conditions which were initially described. Though (t + 10)^-2 can be considered in this question.
I think this (t + 10)−2 is supposed to be (t + 10)^-2 instead
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Thank you for a feedback. The question was solved using those conditions which were initially described. Though (t + 10)^-2 can be considered in this question.
I think this (t + 10)−2 is supposed to be (t + 10)^-2 instead