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adell [148]
2 years ago
10

Geomtry plzz help 15 points

Mathematics
2 answers:
luda_lava [24]2 years ago
8 0

Answer:

False.

Step-by-step explanation:

The sides of a triangle rule asserts that the sum of the lengths of any two sides of a triangle has to be greater than the length of the third side. However, 8+6=14 is less than 15. Hence, the triangle cannot be formed.

Dafna1 [17]2 years ago
6 0

Answer:

False

Step-by-step explanation:

Sum of two sides ( lengths ) must be greater than the third side ( length ), then the sides will form a triangle.

8 + 15 = 23 > 6

15 + 6 = 21 > 8

8 + 6 = 14 < 15

Since,

14 < 15

So,

Three segments with length 8, 15 and 6 cannot make a triangle.

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Answer:

p_v= 0.01

Since the significance level is 0.05 we see that pv so we have enough evidence to reject the null hypothesis. And the best conclusion for this case would be:

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Step-by-step explanation:

Previous concepts

Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".  

The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"  

Solution to the problem

The hypothesis for this case are:

Null hypothesis: \mu_{A}=\mu_{B}=\mu_{C}= \mu_D

Alternative hypothesis: Not all the means are equal \mu_{i}\neq \mu_{j}, i,j=A,B,C,D

In order to find the mean square between treatments (MSTR), we need to find first the sum of squares and the degrees of freedom.

If we assume that we have p=4 groups and on each group from j=1,\dots,p we have n_j individuals on each group we can define the following formulas of variation:  

SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2  

SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2  

SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2  

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SST=SS_{between}+SS_{within}  

And in order to test this hypothesis we need to ue an F statistic and for this case the p value calculated is

p_v= 0.01

Since the significance level is 0.05 we see that pv so we have enough evidence to reject the null hypothesis. And the best conclusion for this case would be:

b. at least some, but not all, of the gestation periods across all four species are the same

Because is only to identify if AT LEAST one mean is different NOT to conclude that the all the means are different.

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