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Naddika [18.5K]
3 years ago
15

Pls help I will mark brainliest and 50 pts

Mathematics
1 answer:
const2013 [10]3 years ago
5 0

Answer:

5) Slope is 3/2 or 1.5

6) Slope is -1.

Step-by-step explanation:

To find the slope of any line given two points, we can use the slope-formula:

m=\frac{y_2-y_1}{x_2-x_1}

Let's use the slope formula for each of our tables.

Table 5)

From this table, let's pick any two points. It really doesn't matter which we pick. I'm going to pick the first two columns.

So, our points are (0, -4) and (2, -1).

Now, we can use the slope formula. Let's let (0, -4) be (x₁, y₁) and let's let (2, -1) be (x₂, y₂). Substitute:

m=\frac{-1-(-4)}{2-0}

Simplify:

m=\frac{-1+4}{2}

Add:

m=\frac{3}{2}=1.5

So, the slope of the line from Table 5 is 3/2 or 1.5.

Table 6:

Again, we can pick any two points. I'm going to use the first two columns.

So, our points are (-4, 7) and (-1, 4).

Let's let (-4, 7) be (x₁, y₁) and let's let (-1, 4) be (x₂, y₂). Substitute:

m=\frac{4-7}{-1-(-4)}

Simplify:

m=\frac{4-7}{-1+4}

Add or subtract:

m=\frac{-3}{3}

Divide:

m=-1

So, the slope of the line from Table 6 is -1.

And we're done!

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Complete Question

The complete question is shown on the first uploaded image

Answer:

a

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b

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c

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d  

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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

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Considering the parameter given for part a  

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        The  test statistics is  t =  1.66

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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

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

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

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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

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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