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disa [49]
3 years ago
8

The length width and height of a rectangular solid are 8,4,1 respectively what is the length of the longest line segment whose e

ndpoints are two vertices of this solid
Mathematics
1 answer:
LUCKY_DIMON [66]3 years ago
3 0

Answer:

9 units

Step-by-step explanation:

Given

Length = 8

Width = 4

Height = 1

Required

Determine the length of the longest segment

To do this, we use:

Longest = \sqrt{Length^2 + Width^2 + Height^2}

So, we have:

Longest = \sqrt{8^2 + 4^2 + 1^2}

Longest = \sqrt{64 + 16 + 1}

Longest = \sqrt{81}

Take positive square root

Longest = 9

<em>Hence, the length of the longest segment is 9 units</em>

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Answer:

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Step-by-step explanation:

Shawn is placing a fence around a circular garden with diameter of 15 foot

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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 )  

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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

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        The  test statistics is  t =  -3.95

        The level of significance \alpha  =  0.01

The degree of freedom is evaluated as

            df =  n-  1

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Answer:

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Step-by-step explanation:

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Answer:

Step-by-step explanation:

You cannot solve the problem be cause you dont know how many trees there are.

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