# Thread: Can anyone help?

1. ## Can anyone help?

 k/32 (x2) 02 fx 3/32 (6-x) 26 0 otherwise

Find what K is?

Can anyone help with this question?

2. ## Re: Can anyone help?

Originally Posted by wrexlive1
 k/32 (x2) 02 fx 3/32 (6-x) 26 0 otherwise

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

What does K have to do with anything?

Please correct the notation!

3. ## 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?

4. ## 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?

5. ## Re: Can anyone help?

yes this is a continuous random variable