Results 1 to 5 of 5

Math Help - Can anyone help?

  1. #1
    Newbie
    Joined
    Aug 2012
    From
    uk
    Posts
    24

    Can anyone help?

    k/32 (x2) 0<x>2
    fx 3/32 (6-x) 2<x>6
    0 otherwise

    Find what K is?


    Can anyone help with this question?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,616
    Thanks
    1579
    Awards
    1

    Re: Can anyone help?

    Quote Originally Posted by wrexlive1 View Post
    k/32 (x2) 0<x>2
    fx 3/32 (6-x) 2<x>6
    0 otherwise

    Find what K is?
    The question makes no sense.
    It is never the case that 0<x>2 NOR 2<x>6.

    What does K have to do with anything?

    Please correct the notation!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Aug 2012
    From
    uk
    Posts
    24

    Re: Can anyone help?

    k/32 (x^2) x is less than or equal to 0 and x<2
    3/32 (6-x) x is less than or equal to 2 and x is greater or equal to 6.

    The question is what is the value of k?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,572
    Thanks
    1412

    Re: Can anyone help?

    If you are serious then you have real problems. There is no reason why "k" cannot have any value at all or even "no" value.


    Now if the problem were to "find the value of k such that the function defined is continuous" then there would be one unique value for k.

    Does the problem say anything about "cotinuous"? Do you know what "continuous" means?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Aug 2012
    From
    uk
    Posts
    24

    Re: Can anyone help?

    yes this is a continuous random variable
    Follow Math Help Forum on Facebook and Google+


/mathhelpforum @mathhelpforum