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vampirchik [111]
3 years ago
5

Consider the following statement: The square of a prime number is not prime. (a) Write this as an if-then statement, using caref

ul mathematical language and notation. (b) Prove or disprove the statement.
Mathematics
1 answer:
Hitman42 [59]3 years ago
7 0

The square of a prime number is not prime.

a) let x ∈ R, If x  ∈ {prime numbers}, then x^{2}∉{prime numbers}

there says that if x is a real and x is in the set of the prime numbers, then the square of x isn't in the set of prime numbers.

b) Prove or disprove the statement.

ok, if x is a prime number, then x only can be divided by himself. Now is easy to see that x^{2} = x*x can be divided by himself and x, then x*x is not a prime number, because can be divided by another number different than himself

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The area of a rectangle is represented by x^2- 7x + 10. What are the missing side lengths of the rectangle?
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Length = x - 2
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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
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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

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=>   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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