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Question Number 111745 by mohammad17 last updated on 04/Sep/20

Commented by mohammad17 last updated on 04/Sep/20

$$helpmesir$$

Commented by Aziztisffola last updated on 05/Sep/20

$${\int }_{100}^{\mathrm{\infty }}0.{002\mathrm{e}}^{-0.002\mathrm{x}}\mathrm{dx}={[-{\mathrm{e}}^{-0.002\mathrm{x}}]}_{100}^{\mathrm{\infty }}$$$$={\mathrm{e}}^{-0.002\times 100}={\mathrm{e}}^{-0.2}=...$$

Commented by Aziztisffola last updated on 04/Sep/20

$$1.{\int }_0^{+\mathrm{\infty }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=1$$$$2.\mathrm{a})\mathrm{F}\left(\mathrm{x}\right)=\mathrm{p}\left(\right]-\mathrm{\infty };\mathrm{x}])=1-{\mathrm{e}}^{-\lambda \mathrm{x}}$$$$\mathrm{b}){\int }_{25}^{\mathrm{\infty }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=0.9512$$$$\Rightarrow \lambda =0.002$$

Commented by mohammad17 last updated on 04/Sep/20

$$thankyousircanyoucompletethesolutionsofallquestionplease?$$

Commented by Aziztisffola last updated on 04/Sep/20

$$2\mathrm{b}){\int }_{25}^{\mathrm{\infty }}\lambda {\mathrm{e}}^{-\lambda \mathrm{x}}\mathrm{dx}=0.9512$$$$\Rightarrow {[-{\mathrm{e}}^{-\lambda \mathrm{x}}]}_{25}^{\mathrm{\infty }}=0.9512$$$$\Rightarrow {\mathrm{e}}^{-25\lambda }=0.9512$$$$\Rightarrow -25\lambda =\ln (0.9512)$$$$\lambda =\frac{\ln (0.9512)}{-25}=0.002$$$$3.\mathrm{a})\mathrm{p}(\mathrm{X}⩾100)={\int }_{100}^{\mathrm{\infty }}0.{002\mathrm{e}}^{-0.002\mathrm{x}}\mathrm{dx}=...$$$$\mathrm{b}){\mathrm{p}}_{(\mathrm{X}⩾100)}(\mathrm{X}⩾150)=\frac{\mathrm{p}(\mathrm{X}⩾150)}{\mathrm{p}(\mathrm{X}⩾100)}=...$$

Commented by mohammad17 last updated on 05/Sep/20

$$sir3.a)and3.b)canyoucomplete$$

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