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jek_recluse [69]
3 years ago
8

Help me! Sorry for the other question.... let’s try it again! 1/2 x + 3/5 x = 5/4

Mathematics
2 answers:
Ann [662]3 years ago
8 0

Answer:

x = 1 3/22

Step-by-step explanation:

1/2 x + 3/5 x = 5/4

We need to get rid of the fractions by multiplying by 20 on each side

20 (1/2 x + 3/5 x) = 20 * 5/4

Distribute

10x + 12x = 25

Combine like terms

22x = 25

Divide each side by 22

22x/22 = 25/22

x = 22/22 + 3/22

x = 1 3/22

Volgvan3 years ago
4 0

Answer:

x = 1\frac{3}{22}

Step-by-step explanation:

=> \frac{1}{2} x + \frac{3}{5} x = \frac{5}{4}

LCM = 20

So, Multiplying both sides by 20

=> 20 (\frac{x}{2} + \frac{3x}{5}) = 5 * 5

10 * x + 4*3x = 25\\10x+12x = 25\\22x = 25

Dividing both sides by 22

x = \frac{25}{22}

x = 1\frac{3}{22}

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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
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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}

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}

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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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\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

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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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