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Juli2301 [7.4K]
3 years ago
7

1. Check the divisibility of the following numbers by 2, 3, 9 and 11 a) 76543 b) 98765436

Mathematics
1 answer:
Alla [95]3 years ago
6 0

Answer:

1. ( I didnt understand the question but I divided them using the calculator.)

A.) 76543

÷ 2= 38,271.5

÷ 3= 25,514.333...

÷ 9= 8,504.777...

÷ 11= 6,958.454545...

B.) 98765436

÷ 2= 49,382,718

÷ 3= 32,921,812

÷ 9= 10,973,937.333...

÷ 11= 8,978,676

2. B

A.) 67894

÷ 4= 16,973.5

÷ 8= 8,486.75

B.) 9685048

÷ 4= 2,421,262

÷ 8= 1,210,631

Explanation:

I used the calculator to divide.

I hope this helps! I'm sorry if it's wrong.

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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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a.

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}

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

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\approx\boxed{0.012262}

7 0
3 years ago
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