- #1

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- Homework Statement
- see attached

- Relevant Equations
- understanding of continous and discrete distribution

I am refreshing on this; ..after a long time...

Note that i do not have the solution to this problem.

I will start with part (a).

##f(u)= 3u-\dfrac{3u^2}{2k}## with limits ##0≤u≤k##

it follows that,

##3k - \dfrac{3k}{2}=1##

##\dfrac{3k}{2}=1##

##k=\dfrac {2}{3}##

For part (b),

##E(T)=\int_0^{\frac{2}{3}} u⋅(3-\dfrac{9}{2}u )du=\left[\dfrac{3}{2}×\dfrac{4}{9}-\dfrac{3}{2}×\dfrac{8}{27}\right]=\dfrac{6-4}{9}= \dfrac{2}{9}##

Ok let me know if that's correct...

Note that i do not have the solution to this problem.

I will start with part (a).

##f(u)= 3u-\dfrac{3u^2}{2k}## with limits ##0≤u≤k##

it follows that,

##3k - \dfrac{3k}{2}=1##

##\dfrac{3k}{2}=1##

##k=\dfrac {2}{3}##

For part (b),

##E(T)=\int_0^{\frac{2}{3}} u⋅(3-\dfrac{9}{2}u )du=\left[\dfrac{3}{2}×\dfrac{4}{9}-\dfrac{3}{2}×\dfrac{8}{27}\right]=\dfrac{6-4}{9}= \dfrac{2}{9}##

Ok let me know if that's correct...

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