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Ivan
1 year ago
6

Use completing the square to solve for x in the equation (x 7) (x minus 9) = 25.

Mathematics
1 answer:
gavmur [86]1 year ago
5 0

The values of x are 1+\sqrt{89} and 1-\sqrt{89}.

To find the values of x:

Given equation: (x+7)(x-9)=25

Then: x(x-9)+7(x-9)=25

Using the distributive property: a.(b+c)=a.b+a.c

x^{2} -9x+7x-63=25

Combine like terms:

x^{2} -2x-63=25

Subtract 25 from both sides and obtain:

x^{2} -2x-88=0

Using completing square form:

Add and subtract (\frac{2}{2} )^{2} =1 we have:

x^{2} -2x-88+1-1=0\\(x-1)^{2} -89=0

Add 89 to both sides we have:

(x-1)^{2} =89

Taking square roots on both sides, obtain:

x-1= ± \sqrt{89}

Add 1 to both sides we have:

x=1±\sqrt{89}

Therefore, the values of x are 1+\sqrt{89} and 1-\sqrt{89}.

Know more about square roots here:

brainly.com/question/428672

#SPJ4

The complete question is given below:

Use completing the square to solve (x + 7)(x – 9) = 25 for x.

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The true average diameter of ball bearings of a certain type is supposed to be 0.5 in. A one-sample t test will be carried out t
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Complete Question

The complete question is shown on the first uploaded image

Answer:

a

   C

b

    C

c

    C

d  

     A

Step-by-step explanation:

From the question we are told that

    The population mean is  \mu  =  0.5 \  in

     

Generally the Null hypothesis is  H_o  :  \mu = 0. 5 \ in

                The Alternative hypothesis is  H_a  :  \mu  \ne  0.5 \ in

Considering the parameter given for part a  

       The sample size is  n =  15  

        The  test statistics is  t =  1.66

        The level of significance \alpha  =  0.05

The degree of freedom is evaluated as

            df =  n-  1

           df =  15-  1

           df =  14

Using the critical value calculator at (social science statistics web site )  

           t_{\frac{\alpha}{2} ,df } =  t_{\frac{0.05 }{2} ,14} =  2.145

We are making use of this  t_{\frac{\alpha }{2} } because it is a one-tail test

Looking at the value of  t and t_{\frac{\alpha }{2} } the we see that  t < t_{\frac{\alpha }{2}  } so the null hypothesis would not be rejected

Considering the parameter given for part b  

       The sample size is  n =  15  

        The  test statistics is  t =  -1.66

        The level of significance \alpha  =  0.05

The degree of freedom is evaluated as

            df =  n-  1

           df =  15-  1

           df =  14

Using the critical value calculator at (social science statistics web site )  

           t_{\frac{\alpha}{2} ,df } =  t_{\frac{0.05 }{2} ,14} =  -2.145

Looking at the value of  t and t_{\frac{\alpha}{2} ,df } the we see that t does not lie in the area covered by  t_{\frac{\alpha}{2}  , df } (i.e the area from -2.145 downwards on the normal distribution curve ) hence we fail to reject the null hypothesis

 

Considering the parameter given for part  c

       The sample size is  n =  26  

        The  test statistics is  t =  -2.55

        The level of significance \alpha  =  0.01

The degree of freedom is evaluated as

            df =  n-  1

           df =  26-  1

           df =  25

Using the critical value calculator at (social science statistics web site )  

           t_{\frac{\alpha}{2} ,df } =  t_{\frac{0.01 }{2} ,25} =  2.787

Looking at the value of  t and t_{\frac{\alpha }{2} } the we see that t does not lie in the area covered by  t_{\alpha , df } (i.e the area from -2.787 downwards on the normal distribution curve ) hence we fail to reject the null hypothesis

Considering the parameter given for part  d

       The sample size is  n =  26  

        The  test statistics is  t =  -3.95

        The level of significance \alpha  =  0.01

The degree of freedom is evaluated as

            df =  n-  1

           df =  26-  1

           df =  25

Using the critical value calculator at (social science statistics web site )  

           t_{\frac{\alpha}{2} ,df } =  t_{\frac{0.01 }{2} ,25} =  -2.787

Looking at the value of  t and t_{\frac{\alpha}{2}  } the we see that  t  lies in the area covered by  t_{\alpha , df } (i.e the area from -2.787 downwards on the normal distribution curve ) hence we  reject the null hypothesis

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3 years ago
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