Determine the value of c such that the function f(x, y) = cxy for 0 < x < 3 and 0 < y < 3 satisfies the properties of a joint probability density function. determine the following. round your answers to four decimal places (e. g. 98.7654).

Thus the value of c such that the function f(x,y) = cxy for 0 < x < 3 and 0 < y < 3 satisfies the properties of a joint probability density function is:

c=0.0494

Step-by-step explanation:

Lets start by understanding what we are looking for, thus what is a Joint Probability Density Function.

In principle, and considering a given Probability Space, if two variables exist, say and , then the Joint Probability defined as , essentially denotes the probability that each and exist in a discrete value set,  particularly specified for these variables.

If the given variables, (here and ) are continuous, then three properties must be met by . However in this question we are only interested in the first two properties, defined as:

Now lets look at our case here. We know the following:

Thus we can say that:

,    otherwise

,    ,

With respect to that, then we can write our integral for solution as follow:

      Eqn(1).

Having obtained Eqn(1) we can start solving for our constant as follow:

           Take constant outside of the Integrals

            Compute first integral for

      Apply Boundary Conditions

               Compute and Simplify

                        Compute second integral for  

            Apply Boundary Conditions

                         Simplify and solve for c

                       Answer rounded to 4 d.p.

Thus the value of c such that the function f(x,y) = cxy for 0 < x < 3 and 0 < y < 3 satisfies the properties of a joint probability density function is:

f(x) has value 150 at x= 0

and 0 at x= 3

so d(f(x))/dt = slope = -150/3 = -50

sina-sinb

cosb+cosa

tana-tanb

cos0+cos x pi/3

20tan

step-by-step explanation: