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KATRIN_1 [288]
2 years ago
5

If H is the midpoint of GE and J is the midpoint of FE, determine the following lengths.

Mathematics
2 answers:
Alinara [238K]2 years ago
3 0

Answer: The value of x is 5, the length of HJ is 8 and the length of 4.

Explanation:

It is given that H is the midpoint of GE and J is the midpoint of FE.

According to midpoint theorem the line segment connecting the midpoint of two sides is parallel to the three side and its length is half of the third side.

Since JH is connecting the midpoints.

HJ=\frac{1}{2} (GF)

x+3=\frac{1}{2} (4x-4)

x+3=2x-2

x=5

The value of x is 5.

HJ=x+3=5+3=8

JE=x-1=5-1=4

Therefore HJ=8 and JE=4.

Usimov [2.4K]2 years ago
3 0
HJ is a mid- segment,  GF=2HJ
so 
4x-4 =2(x+3)
4x-4=2x+6
2x=10
x=5
HJ= x+3=5+3=8
JE=x-1=5-1=4
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Answer:

(1) The degrees of freedom for unequal variance test is (14, 11).

(2) The decision rule for the 0.01 significance level is;

  • If the value of our test statistics is less than the critical values of F at 0.01 level of significance, then we have insufficient evidence to reject our null hypothesis.      
  • If the value of our test statistics is more than the critical values of F at 0.01 level of significance, then we have sufficient evidence to reject our null hypothesis.  

(3) The value of the test statistic is 0.3796.

Step-by-step explanation:

We are given that you are an expert on the fashion industry and wish to gather information to compare the amount earned per month by models featuring Liz Claiborne's attire with those of Calvin Klein.

The following is the amount ($000) earned per month by a sample of 15 Claiborne models;

$3.5, $5.1, $5.2, $3.6, $5.0, $3.4, $5.3, $6.5, $4.8, $6.3, $5.8, $4.5, $6.3, $4.9, $4.2 .

The following is the amount ($000) earned by a sample of 12 Klein models;

$4.1, $2.5, $1.2, $3.5, $5.1, $2.3, $6.1, $1.2, $1.5, $1.3, $1.8, $2.1.

(1) As we know that for the unequal variance test, we use F-test. The degrees of freedom for the F-test is given by;

\text{F}_(_n__1-1, n_2-1_)

Here, n_1 = sample of 15 Claiborne models

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So, the degrees of freedom = (n_1-1, n_2-1) = (15 - 1, 12 - 1) = (14, 11)

(2) The decision rule for 0.01 significance level is given by;

  • If the value of our test statistics is less than the critical values of F at 0.01 level of significance, then we have insufficient evidence to reject our null hypothesis.      
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                          T.S. = \frac{s_1^{2} }{s_2^{2} } \times \frac{\sigma_2^{2} }{\sigma_1^{2} }  ~ \text{F}_(_n__1-1, n_2-1_)

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So, the test statistics =  \frac{1.007}{2.653 } \times 1  ~ \text{F}_(_1_4,_1_1_)

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Hence, the value of the test statistic is 0.3796.

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