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3 years ago

In the diagram AB bisects FAE. BF=5x and BE=x2+6. Solve for x.

1 answer:

5 0

$If\ AB^{\to}\ bisects\ \angle FAE\ then\ m\angle BAF=m\angle BAE\\\\5x=x^2+6\\\\x^2-5x+6=0\\\\x^2-2x-3x+6=0\\\\x(x-2)-3(x-2)=0\\\\(x-2)(x-3)=0\iff x-2=0\ or\ x-3=0\\\\\boxed{x=2\ or\ x=3}$

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Read 2 more answers

To test whether the mean time needed to mix a batch of material is the same for machines produced by three manufacturers, Jacob'

TiliK225 [7]

Answer:

a) Independent variable: The manufacturer. It's a categorical and qualitative variable.

b) Dependent variable: time needed to mix the material, and is a quantitative variable

c) On this case since we want to compare the means of different groups of manufacturers on specific 3 manufacturers (A,B and C) the best way to analyze this is with the ANOVA procedure.

d) ANOVA is the correct procedure since this procedure is used to analyze the differences among group means in a sample, and that's what we want to analyze.

The hypothesis for this case are given by

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

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

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"

Part A

Independent variable: The manufacturer. It's a categorical and qualitative variable.

Part B

Dependent variable: time needed to mix the material, and is a quantitative variable

Part C

On this case since we want to compare the means of different groups of manufacturers on specific 3 manufacturers (A,B and C) the best way to analyze this is with the ANOVA procedure.

Part D

ANOVA is the correct procedure since this procedure is used to analyze the differences among group means in a sample, and that's what we want to analyze.

The hypothesis for this case are given by

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

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

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 $p$ 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

$SST=SS_{between}+SS_{within}$

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3 years ago