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sveticcg [70]
3 years ago
14

(−4 + 5) − (6 + 7) = x x 0

Mathematics
1 answer:
weeeeeb [17]3 years ago
3 0

1

Add the numbers

(

−

4

+

5

)

−

(

6

+

7

)

=

0

({\color{#c92786}{-4}}+{\color{#c92786}{5}})-(6+7)=xx^{0}

(−4+5)−(6+7)=xx0

(

1

)

−

(

6

+

7

)

=

0

({\color{#c92786}{1}})-(6+7)=xx^{0}

(1)−(6+7)=xx0

2

Add the numbers

1

−

(

6

+

7

)

=

0

1-({\color{#c92786}{6}}+{\color{#c92786}{7}})=xx^{0}

1−(6+7)=xx0

1

−

(

1

3

)

=

0

1-({\color{#c92786}{13}})=xx^{0}

1−(13)=xx0

3

Multiply the numbers

1

−

1

⋅

1

3

=

0

1{\color{#c92786}{-1}} \cdot {\color{#c92786}{13}}=xx^{0}

1−1⋅13=xx0

1

−

1

3

=

0

1{\color{#c92786}{-13}}=xx^{0}

1−13=xx0

4

Subtract the numbers

1

−

1

3

=

0

{\color{#c92786}{1-13}}=xx^{0}

1−13=xx0

−

1

2

=

0

{\color{#c92786}{-12}}=xx^{0}

−12=xx0

5

Combine exponents

−

1

2

=

0

-12={\color{#c92786}{xx^{0}}}

−12=xx0

−

1

2

=

1

-12={\color{#c92786}{x^{1}}}

−12=x1

Show less

Solution

−

1

2

=

1

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An article reported on the results of an experiment in which half of the individuals in a group of 60 postmenopausal overweight
slavikrds [6]

Answer:

Step-by-step explanation:

From the given information:

Let's first compute the null and alternative hypothesis

Null hypothesis:

\mathbf{H_o: \mu_1-\mu_2=1}

Alternative hypothesis:

H_a :\mu_1 -\mu_2 > 1

The number of samples is half in a group of 60

i.e

n_1=n_2 = 30

the sample mean for sample 1 \overline x_1 = 5.7

the standard deviation for sample 1 s_1 = 3.1

the sample mean for sample 2 \overline x_2 = 3.9

the standard  deviation for sample 2 s_2 = 2.7

degree of freedom  for this test can be computed by using the formula:

df = \dfrac{\begin {pmatrix} \dfrac{s_1^2}{n_1} + \dfrac{s^2_2}{n_2}   \end {pmatrix}^2  }  {\dfrac{ (\dfrac{s_1^2}{n_1}^2)}{ n_1-1}  + \dfrac{ (\dfrac{s_2^2}{n_2}^2)}{ n_2-1}  }

df = \dfrac{\begin {pmatrix} \dfrac{3.1^2}{30} + \dfrac{2.7^2}{30}   \end {pmatrix}^2  }  {\dfrac{ \begin {pmatrix} \dfrac{3.1^2}{30} \end {pmatrix}^2}{ 30-1}  + \dfrac{ \begin {pmatrix} \dfrac{2.7^2}{30} \end {pmatrix}^2}{ 30-1}  }}

df = 114.68

The test statistics can be computed as follows:

t = \dfrac{ \overline x_1 -\overline x_2- (\mu_1 -\mu_2)}{\sqrt{ \dfrac{s_1^2}{n_1} +\dfrac{s_2^2}{n_2} }}

t = \dfrac{ 5.7-3.9- (1)}{\sqrt{ \dfrac{3.1^2}{30} +\dfrac{2.7^2}{30} }}

t = 1.07

Using the level of significance of 0.1, the P-value for the test statistics at the df of 114.68 is:

P-value = 0.143

Decision rule: To reject the null hypothesis if the level of significance is greater than the p-value.

Conclusion: We fail to reject the null hypothesis because the p-value is greater than the level of significance at 0.1.

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8 0
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Fynjy0 [20]
Hey there,

The answer is true

Hope it helped!
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Vika [28.1K]

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