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3 years ago

Which statement best describes the relation (3, 4), (4, 3), (6, 3), (7, 8), (5, 4)?

Mathematics

1 answer:

snow_lady [41]3 years ago

3 0

Answer:

This relation represent y as a function of x, because each value of x is associated with a single value of y

Step-by-step explanation:

we know that

A function is a relation from a set of inputs (x values) to a set of possible outputs (y values) where each input is related to exactly one output

we have the relation

$(3, 4), (4, 3), (6, 3), (7, 8), (5, 4)$

This relation represent y as a function of x, because each value of x is associated with a single value of y

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$f_{X,Y,Z}(x,y,z)=\begin{cases}Ce^{-(0.5x+0.2y+0.1z)}&\text{for }x\ge0,y\ge0,z\ge0\\0&\text{otherwise}\end{cases}$

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$\displaystyle\int_0^\infty\int_0^\infty\int_0^\infty f_{X,Y,Z}(x,y,z)\,\mathrm dx\,\mathrm dy\,\mathrm dz=100C=1\implies \boxed{C=0.01}$

b. Find the marginal joint density of $X$ and $Y$ by integrating the joint density with respect to $z$ :

$f_{X,Y}(x,y)=\displaystyle\int_0^\infty f_{X,Y,Z}(x,y,z)\,\mathrm dz=0.01e^{-(0.5x+0.2y)}\int_0^\infty e^{-0.1z}\,\mathrm dz$

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Then

$\displaystyle P(X\le1.375,Y\le1.5)=\int_0^{1.5}\int_0^{1.375}f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy$

$\approx\boxed{0.12886}$

c. This probability can be found by simply integrating the joint density:

$\displaystyle P(X\le1.375,Y\le1.5,Z\le1)=\int_0^1\int_0^{1.5}\int_0^{1.375}f_{X,Y,Z}(x,y,z)\,\mathrm dx\,\mathrm dy\,\mathrm dz$

$\approx\boxed{0.012262}$

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