Question Number 132515 by mohammad17 last updated on 14/Feb/21
Commented by mohammad17 last updated on 14/Feb/21
$${help}\:{me}\:{sir} \\ $$
Answered by EDWIN88 last updated on 14/Feb/21
![Q1. ∫_0 ^∞ c.e^(−2x) dx = 1 −(1/2)c.e^(−2x) ]_0 ^∞ = 1 (1/2)c = 1 ⇒ c = 2](Q132522.png)
$$\mathrm{Q1}.\:\int_{\mathrm{0}} ^{\infty} \mathrm{c}.\mathrm{e}^{−\mathrm{2x}} \:\mathrm{dx}\:=\:\mathrm{1}\: \\ $$$$\left.\:−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{c}.\mathrm{e}^{−\mathrm{2x}} \:\right]_{\mathrm{0}} ^{\infty} \:=\:\mathrm{1}\: \\ $$$$\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{c}\:=\:\mathrm{1}\:\Rightarrow\:\mathrm{c}\:=\:\mathrm{2}\: \\ $$
Answered by EDWIN88 last updated on 14/Feb/21
$$\mathrm{Q1C}.\: \\ $$$$\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{disjoint}\:\mathrm{even}\:\Rightarrow\mathrm{P}\left(\mathrm{A}\cap\mathrm{B}\right)=\:\mathrm{0} \\ $$$$\mathrm{n}\left(\mathrm{B}\cap\mathrm{A}^{\mathrm{c}} \right)\:=\:\mathrm{n}\left(\mathrm{B}\right) \\ $$$$\:\frac{\mathrm{n}\left(\mathrm{B}\cap\mathrm{A}^{\mathrm{c}} \right)}{\mathrm{n}\left(\mathrm{S}\right)}\:=\:\frac{\mathrm{n}\left(\mathrm{B}\right)}{\mathrm{n}\left(\mathrm{S}\right)}\:\Rightarrow\mathrm{P}\left(\mathrm{B}\cap\mathrm{A}^{\mathrm{c}} \right)\:=\:\mathrm{P}\left(\mathrm{B}\right)=\frac{\mathrm{1}}{\mathrm{3}} \\ $$
Answered by guyyy last updated on 15/Feb/21
Commented by guyyy last updated on 15/Feb/21
Commented by mohammad17 last updated on 16/Feb/21
$${sir}\:{can}\:{you}\:{help}\:{me}\:{in}\:{this}\:{question}\:{pls} \\ $$$${prove}\:{that}\:{Log}\left({z}_{\mathrm{1}} {z}_{\mathrm{2}} \right)={Log}\left({z}_{\mathrm{1}} \right)+{Log}\left({z}_{\mathrm{2}} \right) \\ $$$${with}\:{condition}\left(−\pi<{Argz}_{\mathrm{1}} +{Argz}_{\mathrm{2}} <\pi\right) \\ $$
Commented by guyyy last updated on 19/Feb/21
