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Gwar [14]
2 years ago
12

8x = 96 How would I solve for x

Mathematics
2 answers:
Vaselesa [24]2 years ago
6 0

How to solve your problem

fomenos2 years ago
5 0

x = 12

Step-by-step explanation:

8x = 96 \\  \\ divide \:  \: both \:  \: sides

8x \div 8 = 96 \div 8 \\  \\ divide \\  \\  = x = 12 \\  \\ hope \:  \: it \:  \: helps \:

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1/6 * 1/2 = 1/12

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The Weibull distribution is widely used in statistical problems relating to aging of solid insulating materials subjected to agi
Tresset [83]

Answer:

a

  P(X \le 250 ) =  0.7564 [/tex]  ,  P(X <  250 )  =  0.7564   ,

    P(X <  300 )  =  0.09922

b

P(100 <  X  < 250 ) =0.644

c

 x  = 192.1

Step-by-step explanation:

From the question we are told that

   The value for \alpha  =  2.6

    The value for \beta = 220

Generally the  Weibull distribution function is mathematically represented as

      F( x , \alpha ,  \beta ) =  \left \{  0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  x < 0} \atop { 1- e^{-(\frac{x}{\beta } )^{\alpha } }}\ \ \ \ \ \ x \ge 0} \right

Generally the probability that a specimen's lifetime is at most 250 is mathematically represented as

      P(X \le  250 ) =  F(250, 2.7 , 220 )

      P(X \le 250 )=1 - e^{- (\frac{250}{220} )^{2.7}}

      P(X \le 250 ) =  1 - 0.2436

      P(X \le 250 ) =  0.7564

Generally the probability that a specimen's lifetime is less than 250

      P(X <  250 ) =  F(250, 2.7 , 220 )

      [texP(X <  250 ) =1 - e^{- (\frac{250}{220} )^{2.7}}[/tex]

      P(X <  250 )  =  1 - 0.2436

      P(X <  250 )  =  0.7564    

Generally the probability that a specimen's lifetime is more than 300

     P(X >  300 ) = 1- p(X \le 300 )

      P(X >  300 ) = 1-  F(300, 2.7 , 220 )

      [texP(X <  300) =1- [1 - e^{- (\frac{300}{220} )^{2.7}}][/tex]

      P(X <  300 )  =  0.09922

Generally the probability that a specimen's lifetime is between 100 and 250 is

     P(100 <  X  < 250 ) =  P(X < 250) - P(X < 100)

=>  P(100 <  X  < 250 ) =F(250 , 2.7 , 220 ) - F(100 , 2.7 , 220 )

=>  P(100 <  X  < 250 ) =(1 - e^{-(\frac{250}{220})^{2.7}}) - (1 - e^{-(\frac{100}{220})^{2.7}})

=>  P(100 <  X  < 250 ) = (1 - 0.244 ) - (1- 0.888)

=>  P(100 <  X  < 250 ) =0.644

Generally the value  such that exactly 50% of all specimens

    P(X > x) = 1-P(X <  x) = 0.50

=>  P(X > x) = 1- (1 - e^{- (\frac{x}{220}) ^{2.7}}) = 0.50

=>  P(X> x ) = e^(- \frac{x}{20})^{2.7}  = 0.50

=>  P(X> x ) = (- \frac{x}{20})^{2.7}  = ln0.50

=>   P(X> x ) =  \frac{x}{20}  =[ -ln0.50 ] ^{frac{1}{2.7}}

=>   x  = 220[ -ln0.50 ] ^{frac{1}{2.7}}

=>   x  = 192.1

     

8 0
3 years ago
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