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Alika [10]
3 years ago
14

-6(3-6r)=-18+4r solve

Mathematics
2 answers:
Darya [45]3 years ago
8 0

Answer:

r = 0

Step-by-step explanation:

-6 (3 - 6r) = -18 + 4r

-18 + 36r = -18 + 4r

-18 (+18) + 36r = -18 (+18) + 4r

36r = 0 + 4r (-4r)

(-4r) + 36r = 32r

32r = 0

0/32 = 0

8090 [49]3 years ago
5 0

Answer:

r = 0

Step-by-step explanation:

-6(3 - 6r) = -18 + 4r //Distribute -6

-18 + 36r = -18 + 4r // Add 18 on both sides

36r = 4r //Subtract 4r on both sides

36r - 4r = 0 //Combine like terms

32r = 0 //Divide 32 on both sides

r = 0/32

r = 0

Check:

-6(3 -6r) = -18 + 4r //Replace 0 in place of r, since r=0

-6(3 - 6(0)) = -18 + 4(0) //Multiple

-6(3) = -18

-18 = -18 //True, therefore r=0

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Step-by-step explanation:

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Step-by-step explanation:

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For f(x)=2x+1 and g(x)=x^2-7, find (f.g)(x)
hram777 [196]

Option C (in the figure) is correct.

Step-by-step explanation:

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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
yulyashka [42]

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

6 0
3 years ago
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