+0  
 
0
54
3
avatar+454 

i dont get question D please explain thanks

YEEEEEET  Nov 19, 2018
 #1
avatar+701 
+5

If you did not know already, "μ" is the greek letter "mu", and it stands for micro, or \(10^-6\) (10^-6).

 

The "η" is the greek letter "eta", and it stands for the greek letter "eta", and it stands for the "Dirichlet eta function".  In mathematics, in the area of analytic number theory, the Dirichlet eta function is defined by the following Dirichlet series, which converges for any complex number having real part > 0: 

\({\displaystyle \eta (s)=\sum _{n=1}^{\infty }{(-1)^{n-1} \over n^{s}}={\frac {1}{1^{s}}}-{\frac {1}{2^{s}}}+{\frac {1}{3^{s}}}-{\frac {1}{4^{s}}}+\cdots }\)

 

...Except this is not analytic number theory, and "μ" simply represents tthe mean of X and "η" simply represents the median of X. Questions b) and c) tell us that "μ"  = 1 and "η" = \(\dfrac{13}{12}\). (The top part was a joke...cool)

\(3\cdot μ = 3\) and \(3 - \dfrac{13}{12} = 1\dfrac{11}{12}\). Question d) asks us to find the probabliltiy that \(X\lt 1\dfrac{11}{12}\)\(X\) can only be greater than \(1\dfrac{11}{12}\) when \(f(x) = \dfrac{3-x}{4}\). I believe you can go on from here and find the probabililty that \(X\lt 1\dfrac{11}{12}\). If you need more help, you could re-state the question or ask for more information in forums. smiley

 

- PartialMathematician

PartialMathematician  Nov 19, 2018
 #2
avatar+3197 
+2

what are you going on about partialmathematician?

 

\(\text{for (d) you've already calculated }\mu=\dfrac{13}{12} \text{ and }\eta=1\\ \text{so you are asked to find }P\left[X < 3 \cdot \dfrac{13}{12} - 1\right] =\\ P\left[X <\dfrac 9 4\right]\)

 

\(\text{this will be}\\ \displaystyle \int \limits_0^1~\dfrac 1 2~dx + \int_1^{\frac 9 4}\dfrac{3-x}{4}~dx = \dfrac{119}{128}\\ \)

Rom  Nov 20, 2018
 #3
avatar+94126 
+1

Partial Mathematician (PM) is an enthusiastic kid Rom. 

He is a little more enthusiastic than his knowledge can support but he'll settle in here ok I think.

 

I like this place to be full of people who are enthusiastic about mathematics and I have already seen evidence that PM is learning.

That is right isn't it PM  ?

 

 

You could back off on questions that you do not understand though PM  wink

 

Did you copy this stuff from somewhere and just hope it was relevant? 

You know that is plagerism and illegal.  You must at least say where you copy your info from. 

Melody  Nov 20, 2018

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