Mathematics, 30.10.2019 19:31, mng32

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).

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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 $f_{XY}(x,y)$ , 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 $f_{XY}(x,y)$ . However in this question we are only interested in the first two properties, defined as:

$f_{XY}(x,y)\geq 0$ for all x,y

$\int\limits^\infty_{-\infty} \int\limits^\infty_{-\infty}{} \,f_{XY}(x,y) dydx=1$

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

$f_{XY}(x,y)=cxy$

Thus we can say that:

$f_{XY}(x,y)=0$ , otherwise

$f_{XY}(x,y)=cxy$ , ,

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

$\int\limits^3_b {}\int\limits^3_b {cxy} \, dydx =1$ Eqn(1).

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

$c\int\limits^3_b {}\int\limits^3_b {xy} \, dydx =1$ Take constant outside of the Integrals

$c\int\limits^3_b{x\frac{y^2}{2} }|^3_0 \, dx =1$ Compute first integral for

$c\int\limits^3_b{x[\frac{3^2}{2}-\frac{0^2}{2}] } \, dx =1$ Apply Boundary Conditions

$c\int\limits^3_b{\frac{9}{2}x } \, dx =1$ Compute and Simplify

$\frac{9}{2}c[ \frac{x^2}{2}]|^3_0=1$ Compute second integral for

$\frac{9}{2}c[ \frac{3^2}{2}- \frac{0^2}{2}]=1$ Apply Boundary Conditions

$\frac{81}{4}c=1$ Simplify and solve for c

$81c=4\\ c=\frac{4}{81}\\ c=0.04938\\$

c=0.0494 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:

c=0.0494

answered: Guest

Answer: d. 48 step-by-step explanation: i used a calculator so i am 100% sure these are right mark as brainliest?

answered: Guest

i think there was a typo. you wrote pat brings 3 quarts of ice-cream to the party. each serving of ice-cream is 1 6 quart. how many total servings of ice-cream does pat bring to the party? i think the bold is supposed to be 1/6. so 1 quart of ice cream = 6 servings. 6 * 3 = 18.

the answer is 18.