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3 years ago

Help plz show work

Mathematics

1 answer:

Sidana [21]3 years ago

7 0

Answer: see below

<u>Step-by-step explanation:</u>

7) Algebraic Inequality: x ≠ 3

--> -∞ < x < 3 and 3 < x < ∞

Graph: ←-----------o-----------→

Interval Notation: (-∞,3) ∪ (3, ∞)

8) This is a duplicate question of #7

9) Algebraic Inequality: x ≠ 2 and x ≠ 7

--> -∞ < x < 2 and 2 < x < 7 and 7 < x < ∞

Graph: ←-------o-------o-------→

2 7

Interval Notation: (-∞, 2) ∪ (2, 7) ∪ (7, ∞)

10) Algebraic Inequality: x ≠ -4 and x ≠ 0

--> -∞ < x < -4 and -4 < x < 0 and 0 < x < ∞

Graph: ←-------o-------o-------→

-4 0

Interval Notation: (-∞, -4) ∪ (-4, 0) ∪ (0, ∞)

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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

dexar [7]

Answer:

The value of the constant C is 0.01 .

Step-by-step explanation:

Given:

Suppose X, Y, and Z are random variables with the joint density function,

$f(x,y,z) = \left \{ {{Ce^{-(0.5x + 0.2y + 0.1z)}; x,y,z\geq0 } \atop {0}; Otherwise} \right.$

The value of constant C can be obtained as:

$\int_x( {\int_y( {\int_z {f(x,y,z)} \, dz }) \, dy }) \, dx = 1$

$\int\limits^\infty_0 ({\int\limits^\infty_0 ({\int\limits^\infty_0 {Ce^{-(0.5x + 0.2y + 0.1z)} } \, dz }) \, dy } )\, dx = 1$

$C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0 {e^{-0.2y }(\int\limits^\infty_0 {e^{-0.1z} } \, dz }) \, dy }) \, dx = 1$

$C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0{e^{-0.2y}([\frac{-e^{-0.1z} }{0.1} ]\limits^\infty__0 }) \, dy }) \, dx = 1$

$C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0 {e^{-0.2y}([\frac{-e^{-0.1(\infty)} }{0.1}+\frac{e^{-0.1(0)} }{0.1} ]) } \, dy }) \, dx = 1$

$C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0 {e^{-0.2y}[0+\frac{1}{0.1}] } \, dy }) \, dx =1$

$10C\int\limits^\infty_0 {e^{-0.5x}([\frac{-e^{-0.2y} }{0.2}]^\infty__0 }) \, dx = 1$

$10C\int\limits^\infty_0 {e^{-0.5x}([\frac{-e^{-0.2(\infty)} }{0.2}+\frac{e^{-0.2(0)} }{0.2}] } \, dx = 1$

$10C\int\limits^\infty_0 {e^{-0.5x}[0+\frac{1}{0.2}] } \, dx = 1$

$50C([\frac{-e^{-0.5x} }{0.5}]^\infty__0}) = 1$

$50C[\frac{-e^{-0.5(\infty)} }{0.5} + \frac{-0.5(0)}{0.5}] =1$

$50C[0+\frac{1}{0.5} ] =1$

$100C = 1$ ⇒ $C = \frac{1}{100}$

C = 0.01

3 0

2 years ago

2x+9=1x-5 what is x?

Vlada [557]

Answer:

2x+9=1x-5

2x-1x=-5-9

x=-14

4 0

1 year ago

Read 2 more answers

The maximum height of a vehicle that can safely pass under a bridge is 12 feet 5 inches. A truck measures 162 inches in height.

Leni [432]

1 foot = 12 inches. If you multiply 12 by 12 and then add 5, you get 149. 149 is less than 162 so it will be able to fit!

8 0

2 years ago

Divided by 85 show me the work

ivanzaharov [21]

Answer:

Step-by-step explanation:

What divided by 85?

3 0

3 years ago

Plz help me!!!!!!!!!

Over [174]

Answer: $\bold{\dfrac{4\pm \sqrt{6}}{2}}$

<u>Step-by-step explanation:</u>

$\dfrac{3}{y-2}-2=\dfrac{1}{y-1}\\\\\\\text{Multiply by the LCD (y-2)(y-1) to clear the denominator:}\\\\\dfrac{3}{y-2}(y-2)(y-1)-2(y-2)(y-1)=\dfrac{1}{y-1}(y-2)(y-1)\\\\\\3(y-1)-2(y-2)(y-1)=1(y-2)\\\\3y-3-2(y^2-3y+2)=y-2\\\\3y-3-2y^2+6y-4=y-2\\\\-2y^2+9y-7=y-2\\\\0=2y^2-8y+5\quad \rightarrow \quad a=2,\ b=-8,\ c=5\\\\\\\text{Quadratic formula is: }x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\\x=\dfrac{-(-8)\pm \sqrt{(-8)^2-4(2)(5)}}{2(2)}\\\\\\.\ =\dfrac{8\pm \sqrt{64-40}}{2(2)}$

$.\ =\dfrac{8\pm \sqrt{24}}{2(2)}\\\\\\.\ =\dfrac{8\pm 2\sqrt{6}}{2(2)}\\\\\\.\ =\dfrac{4\pm \sqrt{6}}{2}$

7 0

3 years ago