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cricket20 [7]
2 years ago
6

What is the domain (in interval notation) of the following functions?

Mathematics
1 answer:
alexgriva [62]2 years ago
4 0
1.\\g(x)=\frac{3}{5x-4}\\\\D:5x-4\neq0\to5x\neq4\ \ \ /:5\to x\neq\frac{4}{5}\to x\in\mathbb{R}\ \backslash\ \{\frac{4}{5}\}\\\\2.\\h(x)=\frac{\sqrt{x}}{x-5}\\\\D:x\geq0\ \wedge\ x-5\neq0\to x\geq0\ \wedge\ x\neq5\to x\in\left

3.\\f(x)=\frac{\sqrt{x}}{x^2-5x}\\\\D:x\geq0\ \wedge\ x^2-5x\neq0\to x\geq0\ \wedge\ x(x-5)\neq0\\\\\to x\geq0\ \wedge\ x\neq0\ \wedge\ x\neq5\to x\in\mathbb{R^+}\ \backslash\ \{5\}\\\\4.\\g(x)=\frac{\sqrt{x}+5}{x^2-x-20}\\\\D:x\geq0\ \wedge\ x^2-x-20\neq0\to x\geq0\ \wedge\ (x+4)(x-5)\neq0\\\\\to x\geq0\ \wedge\ x\neq-4\ \wedge\ x\neq5\to x\in\left

5.\\h(x)=\frac{3}{x^2+1}\\\\D:x^2+1\neq0\to x^2\neq-1\to x\in\mathbb{R}\\\\6.\\f(x)=\frac{\sqrt{x-2}}{x+1}\\\\D:x-2\geq0\ \wedge\ x+1\neq0\to x\geq2\ \wedge\ x\neq-1\to x\in\left

7.\\g(x)=\frac{x^2}{3x^2-x-2}\\\\D:3x^2-x-2\neq0\to (3x+2)(x-1)\neq0\to x\neq-\frac{2}{3}\ \wedge\ x\neq1\\\\\to x\in\mathbb{R}\ \backslash\ \{-\frac{2}{3};\ 1\}\\\\8.\\h(x)=3(x-4)^2-7\\\\D:x\in\mathbb{R}
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In a one-way ANOVA, what is the difference between the among-groups variance MSA and the within-groups variance MSW?
LuckyWell [14K]

Answer:

MSR=MSA=\frac{SSR}{k-1}

And MSE=MSW=\frac{SSE}{N-k}

The difference is that MSA takes incount the variation between the groups and the grand mean, and the MSW takes in count the variation within groups respect to the mean of each group .

Step-by-step explanation:

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"  

If we assume that we have j groups and on each group from j=1,\dots,j 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=Treatment}=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}=6750+8000=14750  

The degrees of freedom for the numerator on this case is given by df_{num}=k-1 where k  represent the number of groups.  

The degrees of freedom for the denominator on this case is given by df_{den}=df_{between}=N-k.  

And the total degrees of freedom would be df=N-1  

We can find the MSR=MSA=\frac{SSR}{k-1}

And MSE=MSW=\frac{SSE}{N-k}

The difference is that MSA takes incount the variation between the groups and the grand mean, and the MSW takes in count the variation within groups respect to the mean of each group .

And the we can find the F statistic F=\frac{MSR}{MSE}=

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What is the value of x
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We know that the right triangle has one 90 degree angle and two acute (< 90 degree) angles. Since the sum of the angles of a triangle is always 180 degrees.

<u>To find x:</u>

(10x - 20) + (6x + 8) =  180

16x - 12 = 180

      +12     +12

16x = 192

-----    -----

16        16

x = 12

<u>Check:</u>

(10(12) - 20) + (6(12) + 8) = 180

(120 -  20) + (72 + 8) = 180

100 + 80 = 180

180 = 180

<u>Angle (10x - 20):</u>

(10(12) - 20)

(120 - 20)

100

<u>Angle (6x + 8):</u>

(6(12) + 8)

(72 + 8)

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Answer:

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