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Pepsi [2]
3 years ago
9

The probability density function f(x) for a uniform random variable x defined over the interval [2, 10] is

Mathematics
1 answer:
stellarik [79]3 years ago
3 0
F(x) = {1/8, 2 ≤ x ≤ 10}
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Perform the indicated operation. Be sure the answer is reduced.
avanturin [10]
<h3>Given Equation:-</h3>

\boxed{ \rm  \frac{4x^{2}y^{3}z}{9} \times  \frac{45y}{8 {x}^{5} {z}^{5} }}

<h3>Step by step expansion:</h3>

\dashrightarrow \sf\dfrac{4x^{2}y^{3}z}{9} \times  \dfrac{45y}{8 {x}^{5} {z}^{3} }

\\  \\

\dashrightarrow \sf\dfrac{ \cancel4x^{2}y^{3}z}{9} \times  \dfrac{45y}{ \cancel8 {x}^{5} {z}^{3} }

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\dashrightarrow \sf\dfrac{x^{2}y^{3}z}{9} \times  \dfrac{45y}{2{x}^{5} {z}^{3} }

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\dashrightarrow \sf\dfrac{x^{2}y^{3}z}{ \cancel9} \times  \dfrac{ \cancel{45}y}{2{x}^{5} {z}^{3} }

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\dashrightarrow \sf\dfrac{x^{2}y^{3}z}{1} \times  \dfrac{5y}{2{x}^{5} {z}^{3} }

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\dashrightarrow \sf\dfrac{x^{0}y^{3}z}{1} \times  \dfrac{5y}{2{x}^{5 - 2} {z}^{3} }

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\dashrightarrow \sf\dfrac{y^{3}z}{1} \times  \dfrac{5y}{2{x}^{3} {z}^{3} }

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\dashrightarrow \sf\dfrac{y^{3}z {}^{0} }{1} \times  \dfrac{5y}{2{x}^{3} {z}^{3 - 1} }

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\dashrightarrow \sf\dfrac{y^{3}}{1} \times  \dfrac{5y}{2{x}^{3} {z}^{2} }

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\dashrightarrow \sf  \dfrac{5y \times  {y}^{3} }{2{x}^{3} {z}^{2} }

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\dashrightarrow \sf  \dfrac{5y {}^{0}  \times  {y}^{3 + 1} }{2{x}^{3} {z}^{2} }

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\dashrightarrow \sf  \dfrac{5 \times  {y}^{4} }{2{x}^{3} {z}^{2} }

\\  \\

\dashrightarrow \bf  \dfrac{5 {y}^{4} }{2{x}^{3} {z}^{2} }

\\  \\

\therefore \underline{ \textbf{ \textsf{option \red c \: is \: correct}}}

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Answer:

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How do you find the bisector of an angle
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