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Anarel [89]
3 years ago
15

If F(x)=9x which of the following is the inverse of F(x)

Mathematics
2 answers:
user100 [1]3 years ago
8 0

Answer:

Inverse of F(x)=9x = 1/9x

Step-by-step explanation:

If F(x)=9x

which of the following is the inverse of F(x)

When talking about inverse of a function, we are talking about reversal. Therefore, an inverse function is the inverse of another function. In order to get the inverse of a particular function, we reverse the denominator to become the numerator and the numerator to become the denominator.

So for the function,

F(x)=9x, the numerator is 9x and the denominator is 1

Reversing the numerator and denominator, .

Inverse of F(x)=9x = 1/9x

Dennis_Churaev [7]3 years ago
6 0

Answer:

F^{-1}(x)=\frac{1}{9} x

Step-by-step explanation:

In order to find the inverse of a function, you can follow the next steps:

Step 1: Write f(x)=y

Step 2: Solve for x

Step 3: Replace every x with a y and replace every y with an x

Step 4: Replace y with f^{-1}(x)

Step 5: Verify the result by checking that:

f(a)=b \rightarrow f^{-1}(b)=a

Therefore:

Step 1:

y=9x

Step 2:

x=\frac{y}{9}

Step 3:

y=\frac{x}{9}

Step 4

y=f^{-1}(x)=F^{-1}(x)=\frac{x}{9}

Step 5:

Let:

a=5

F(a)=F(5)=9*5=45\\\\Hence\\\\b=45

Now:

F^{-1}(b)=F^{-1}(45)=\frac{45}{9} =5

Since:

f(a)=b \rightarrow f^{-1}(b)=a

We have verified the result.

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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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The amount of time required for an oil and filter change on an automobile is normally distributed with a mean of 45 minutes and
Firdavs [7]

Answer:

1a

  P(39 <  X < 48  ) = 0.8767

1b

    95% of all sample means will fall between 40.1  <  \mu < 49.9

1c

    \= x = 41. 795

2

   n =  25

Step-by-step explanation:

From the question we are told that

   The mean is n   =  45

   The population standard deviation is  \sigma =  10

   The sample size is n  =  16

Generally the standard error of the mean is mathematically represented as

       \sigma_{x} =  \frac{ \sigma}{\sqrt{n} }

=>    \sigma_{x} =  \frac{ 10 }{\sqrt{16 } }

=>    \sigma_{x} = 2.5

Generally the probability that the sample mean will be between 39 and 48 minutes is

    P(39 <  X < 48  ) =  P( \frac{ 39 - 45}{ 2.5} <  \frac{X - \mu }{\sigma } <  \frac{ 48 - 45}{ 2.5} )

=> P(39 <  X < 48  ) =  P(-2.4 < Z< 1.2 )

=> P(39 <  X < 48  ) =  P( Z< 1.2 ) - P(Z <  -2.4)

From the z table  the area under the normal curve to the left corresponding to  1.2  and  -2.4  is

=> P( Z< 1.2 ) = 0.88493

and  

    P( Z< - 2.4 ) = 0.0081975

So

   P(39 <  X < 48  ) = 0.88493 -0.0081975

=> P(39 <  X < 48  ) = 0.8767

From the question we are told the confidence level is  95% , hence the level of significance is    

      \alpha = (100 - 95 ) \%

=>   \alpha = 0.05

Generally from the normal distribution table the critical value  of   is  

   Z_{\frac{\alpha }{2} } =  1.96

Generally the margin of error is mathematically represented as  

      E = Z_{\frac{\alpha }{2} } *  \frac{\sigma }{\sqrt{n} }

=>   E = 1.96 * 2.5  

=>   E =4.9  

Generally the  95% of all sample means will fall between

      \mu  -E <  and   \mu   +E

=>   45  -4.9\   and \  45  + 4.9

Generally the value which  90% of sample means is  greater than is mathematically represented

      P( \= X >  \= x  ) = 0.90

=>   P( \= X >  \= x  ) =  P( \frac{\= X  - \mu }{ \sigma_x} >  \frac{\= x  -45 }{ 2.5}  ) = 0.90

=>  P( \= X >  \= x  ) =  P( Z >  z  ) = 0.90

Generally from the z-table  the critical  value  of  0.90  is  

      z = -1.282

      \frac{\= x  -45 }{ 2.5}  = -1.282

=>   \= x = 41. 795

Considering question 2

 Generally we are told that the standard deviation of the mean to be one fifth of the population standard deviation, this is mathematically represented as

         s = \frac{1}{5} \sigma

  Generally the standard deviation of the sample mean is mathematically  represented as

          s = \frac{\sigma }{ \sqrt{n} }

=>       \frac{1}{5} \sigma = \frac{\sigma }{ \sqrt{n} }

=>       n =  5^2

=>       n =  25

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3 years ago
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