Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

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

Commented by mohammad17 last updated on 04/Sep/20

help me sir

$${help}\:{me}\:{sir} \\ $$

Commented by Aziztisffola last updated on 05/Sep/20

∫_(100) ^( ∞) 0.002e^(−0.002x) dx=[−e^(−0.002x) ]_(100) ^∞   =e^(−0.002×100) =e^(−0.2) =...

$$\int_{\mathrm{100}} ^{\:\infty} \mathrm{0}.\mathrm{002e}^{−\mathrm{0}.\mathrm{002x}} \mathrm{dx}=\left[−\mathrm{e}^{−\mathrm{0}.\mathrm{002x}} \right]_{\mathrm{100}} ^{\infty} \\ $$$$=\mathrm{e}^{−\mathrm{0}.\mathrm{002}×\mathrm{100}} =\mathrm{e}^{−\mathrm{0}.\mathrm{2}} =... \\ $$

Commented by Aziztisffola last updated on 04/Sep/20

1 .∫_0 ^( +∞) f(x)dx=1  2.  a) F(x)=p(]−∞;x])=1−e^(−λx)   b) ∫_(25) ^( ∞) f(x)dx=0.9512  ⇒λ=0.002

$$\mathrm{1}\:.\int_{\mathrm{0}} ^{\:+\infty} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\mathrm{1} \\ $$$$\left.\mathrm{2}\left..\left.\:\:\mathrm{a}\right)\:\mathrm{F}\left(\mathrm{x}\right)=\mathrm{p}\left(\right]−\infty;\mathrm{x}\right]\right)=\mathrm{1}−\mathrm{e}^{−\lambda\mathrm{x}} \\ $$$$\left.\mathrm{b}\right)\:\int_{\mathrm{25}} ^{\:\infty} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\mathrm{0}.\mathrm{9512} \\ $$$$\Rightarrow\lambda=\mathrm{0}.\mathrm{002} \\ $$

Commented by mohammad17 last updated on 04/Sep/20

thank you sir can you complete the solutions of all question please ?

$${thank}\:{you}\:{sir}\:{can}\:{you}\:{complete}\:{the}\:{solutions}\:{of}\:{all}\:{question}\:{please}\:? \\ $$

Commented by Aziztisffola last updated on 04/Sep/20

2 b) ∫_(25) ^∞ λe^(−λx) dx=0.9512  ⇒[−e^(−λx) ]_(25) ^∞ =0.9512  ⇒e^(−25λ) =0.9512  ⇒−25λ=ln(0.9512)  λ=((ln(0.9512))/(−25))=0.002  3.a) p(X≥100)=∫_(100) ^∞ 0.002e^(−0.002x) dx=...  b) p_((X≥100)) (X≥150)=((p(X≥150))/(p(X≥100)))=...

$$\left.\mathrm{2}\:\mathrm{b}\right)\:\int_{\mathrm{25}} ^{\infty} \lambda\mathrm{e}^{−\lambda\mathrm{x}} \mathrm{dx}=\mathrm{0}.\mathrm{9512} \\ $$$$\Rightarrow\left[−\mathrm{e}^{−\lambda\mathrm{x}} \right]_{\mathrm{25}} ^{\infty} =\mathrm{0}.\mathrm{9512} \\ $$$$\Rightarrow\mathrm{e}^{−\mathrm{25}\lambda} =\mathrm{0}.\mathrm{9512} \\ $$$$\Rightarrow−\mathrm{25}\lambda=\mathrm{ln}\left(\mathrm{0}.\mathrm{9512}\right) \\ $$$$\lambda=\frac{\mathrm{ln}\left(\mathrm{0}.\mathrm{9512}\right)}{−\mathrm{25}}=\mathrm{0}.\mathrm{002} \\ $$$$\left.\mathrm{3}.\mathrm{a}\right)\:\mathrm{p}\left(\mathrm{X}\geqslant\mathrm{100}\right)=\int_{\mathrm{100}} ^{\infty} \mathrm{0}.\mathrm{002e}^{−\mathrm{0}.\mathrm{002x}} \mathrm{dx}=... \\ $$$$\left.\mathrm{b}\right)\:\mathrm{p}_{\left(\mathrm{X}\geqslant\mathrm{100}\right)} \left(\mathrm{X}\geqslant\mathrm{150}\right)=\frac{\mathrm{p}\left(\mathrm{X}\geqslant\mathrm{150}\right)}{\mathrm{p}\left(\mathrm{X}\geqslant\mathrm{100}\right)}=... \\ $$

Commented by mohammad17 last updated on 05/Sep/20

sir 3.a)and 3.b) can you complete

$$\left.{s}\left.{ir}\:\mathrm{3}.{a}\right){and}\:\mathrm{3}.{b}\right)\:{can}\:{you}\:{complete} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com