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earnstyle [38]
3 years ago
12

Describe how (2^3)(2^-4) can be simplified.

Mathematics
2 answers:
Morgarella [4.7K]3 years ago
5 0

Answer:

2^ (-1)

1/2

Step-by-step explanation:

Add the exponents which have common bases.

3 + (-4) = -1

2^ (-1)

1 / 2

Please mark for Brainliest!! :D Thanks!!

For more information, please comment below and I'll respond ASAP!

lukranit [14]3 years ago
3 0

Answer:

The simplified form of the provided expression is 2^{-1}\ or\ \frac{1}{2}

Step-by-step explanation:

Consider the provided expression:

(2^3)(2^{-4})

Use the product rule of exponent:

a^m \cdot a^n=a^{m+n}

Now use the above formula to simplify the provided expression.

(2^3)(2^{-4})=2^{3+(-4)}

(2^3)(2^{-4})=2^{3-4}

(2^3)(2^{-4})=2^{-1}

Hence, the simplified form of the provided expression is 2^{-1}\ or\ \frac{1}{2}

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Answer:

1 a  p -value  =   0.030054

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1c  p -value  = 0.0039768  

2a  p-value  =   0.00099966

2b  p-value  =  0.00999706

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Step-by-step explanation:

Considering question a

  The alternative hypothesis is H1:μ>μ0

   The test statistics is  z =1.88

Generally from the z-table  the  probability of   z =1.88 for a right tailed test is

    p -value  =  P(Z > 1.88) = 0.030054

Considering question b

  The alternative hypothesis is H1:μ<μ0

   The test statistics is  z=−2.75

Generally from the z-table  the  probability of   z=−2.75 for a left tailed test is

    p -value  =  P(Z < -2.75) = 0.0029798

Considering question c

  The alternative hypothesis is H1:μ≠μ0

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Generally from the z-table  the  probability of  z=2.88 for a right  tailed test is

    p -value  = P(Z >2.88) =  0.0019884    

Generally the p-value for the two-tailed test is

    p -value  = 2 *  P(Z >2.88) =  2 * 0.0019884    

=> p -value  = 0.0039768  

Considering question 2a

    The alternative hypothesis is H1:μ>μ0

     The sample size is  n=16

     The  test statistic is  t =  3.733

Generally the degree of freedom is mathematically represented as

        df =  n - 1

=>     df =  16 - 1

=>     df =  15

Generally from the t distribution table  the probability of   t =  3.733 at a degree of freedom of  df =  15 for a right tailed test is  

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Considering question 2b

    The alternative hypothesis is H1:μ<μ0

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     The  test statistic is ,t= −2.500

Generally from the t distribution table  the probability of   t= −2.500 at a degree of freedom of  df=23 for a left  tailed test is  

       p-value  =  t_{-2.500 ,  23} = 0.00999706

Considering question 2c

    The alternative hypothesis is H1:μ≠μ0

     The sample size is  n= 7

     The  test statistic is ,t= −2.2500

Generally the degree of freedom is mathematically represented as

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Generally from the t distribution table  the probability of   t= −2.2500 at a degree of freedom of  df =  6 for a left   tailed test is  

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Generally the p-value  for t= −2.2500 for a two tailed test is

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