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2 years ago

Factorise 8x+12 ffffffffffdfffdfdfd

Mathematics

2 answers:

bagirrra123 [75]2 years ago

7 0

Answer:
Factor out 4 out 8x-12
= 4(2x-3)

Misha Larkins [42]2 years ago

6 0

You have to find the Greatest common factor (GCF) to factor this binomial. First you should see what goes into both 8 and 12... the greatest number that goes into both is: 4... so basically you would have to divide 4 into both 8 and 12. You will get 2 and 6. Once you finish getting the GCF of both numbers, then you would put both the number in parenthesis like this (2+6). But there’s still a variable, which is x. So you would put that in the parenthesis with its coefficient. It will look like, (2x+6). Once you get this, get your GCF of 4 and put it on the outside of your parenthesis. It should look like, 4(2x+6)

Your answer should be —> 4(2x+6)

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Suppose X, Y, and Z are random variables with the joint density function f(x, y, z) = Ce−(0.5x + 0.2y + 0.1z) if x ≥ 0, y ≥ 0, z

kompoz [17]

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

is a proper joint density function if, over its support, $f$ is non-negative and the integral of $f$ is 1. The first condition is easily met as long as $C\ge0$ . To meet the second condition, we require

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

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

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

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

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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3 years ago

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