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2 years ago

Factor completly x^2 -64

Mathematics

1 answer:

Hoochie [10]2 years ago

7 0

∑ Hey, jillianwagler ⊃

Answer:

$\left(x+8\right)\left(x-8\right)$

Step-by-step explanation:

⇒ Given info:

Factor completely $x^2 -64$

⇒ Solution:

Rewrite 64 as 8 ² because 8² is equal to 64. 8 ² = 8 × 8 = 64.

$=x^2-8^2$

Applying difference of two squares formula: $x^2-y^2=\left(x+y\right)\left(x-y\right)$

$=x^2-8^2=\left(x+8\right)\left(x-8\right)\\$

$=\left(x+8\right)\left(x-8\right)$

Answer~: $\left(x+8\right)\left(x-8\right)$

xcookiex12

8/19/2022

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ruslelena [56]

Answer:

$f^{-1}(x)=-\frac{5}{3} x+10$

Step-by-step explanation:

Steps to find the inverse function:

1) replace f(x) with "y"

$y=-\frac{3}{5} x+6$

2) solve for x:

$y-6=-\frac{3}{5} x\\x=-\frac{5}{3}(y-6)\\ x=\frac{5}{3} (6-y)\\x=-\frac{5}{3} y+10$

3) replace "x" with $f^{-1}(x)$ , and y with x:

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3 years ago

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Yakvenalex [24]

Answer:

a) (iii) ANOVA

b) The ANOVA test is more powerful than the t test when we want to compare group of means.

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"

If we assume that we have $p=10$ groups and on each group from $j=1,\dots,p=10$ 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

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

Solution to the problem

Part a

(i) confidence interval

False since the confidence interval work just when we have just on parameter of interest, but for this case we have more than 1.

(ii) t-test

Can be a possibility but is not the best method since every time that we conduct a t-test we have a chance that we commit a Type I error.

(iii) ANOVA

This one is the best method when we want to compare more than 1 group of means.

(iv) Chi square

False for this case we don't want to analyze independence or goodness of fit, so this one is not the correct test.

Part b

The ANOVA test is more powerful than the t test when we want to compare group of means.

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4 years ago

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4 years ago

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Finally,, add the two fractions together to get your final answer
$\frac{7}{9}$
This means that the correct answer to your question is $\frac{7}{9}$ .
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4 years ago