Let k is a constant, thenf(t)=k[(t+10)−2]=k(t+8),0<t<40k∫040(t+8)dt=1k[t22+8t]040=1k[4022+8(40)]=1k(1120)=1∴k=11120P(t<10)=11120∫010(t+8)dt=11120[t22+8t]010=11120[1022+8(10)]=11120×130=13112\text {Let k is a constant, then}\\ f(t)=k[(t+10)-2]=k(t+8), 0<t<40\\ k\int _0^{40}(t+8)dt=1\\ k\left[\frac{t^2}{2}+8t\right]^{40}_0=1\\ k[\frac{40^2}{2}+8(40)]=1\\ k(1120)=1\\ \therefore k=\frac{1}{1120}\\ P(t<10)=\frac{1}{1120}\int _0^{10}(t+8)dt\\ =\frac{1}{1120}\left[\frac{t^2}{2}+8t\right]^{10}_0\\ =\frac{1}{1120}[\frac{10^2}{2}+8(10)]\\ =\frac{1}{1120}\times 130\\ =\frac{13}{112}Let k is a constant, thenf(t)=k[(t+10)−2]=k(t+8),0<t<40k∫040(t+8)dt=1k[2t2+8t]040=1k[2402+8(40)]=1k(1120)=1∴k=11201P(t<10)=11201∫010(t+8)dt=11201[2t2+8t]010=11201[2102+8(10)]=11201×130=11213
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