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a_sh-v [17]
3 years ago
7

PLS HELP 15POINTS 2/3x2 - 4 = 2

Mathematics
1 answer:
Mandarinka [93]3 years ago
4 0

Answer:

x1 = -3 / x2 = 3

Step-by-step explanation:

First multiply both sides of the equation by three getting you to 2x^ - 12 =6

Then, move the content to the right-hand side and change its sign so you will ADD 12 to the other side  you 2x^ = 6+12

6 + 12 = 18

Divide both sides of the equation by 2 so you get x^ = 9

Now simplify by taking the square root to get positive 3giving

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Vsevolod [243]
X + y = 2
Substitute x + 8 into y
x + x + 8 = 2
2x + 8 =2
(Subtract 8 from both sides)
2x = -6
(Divide both sides by 2)
x = -3

Subsitute x = -3 into either equation to find y
x + y = 2
-3 + y = 2
(Add 3 to both sides)
y = 2 + 3
y = 5

Or y can be solved for using:
y = x + 8
y = -3 + 8
y = 5
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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:

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

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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  

And we have this property  

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

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