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kondor19780726 [428]
3 years ago
3

PLEASEEE HELPPPPPP!!!!!!!The tile along the edge of a triangular community pool needs to be replaced. A right triangles, where t

he hypotenuse is labeled 8 x squared, the height is labeled 4 x squared + 15, and the base is labeled 8 x + 10. Which expression represents the total perimeter of the pool edge?
Mathematics
1 answer:
creativ13 [48]3 years ago
4 0

Answer:

the answer is b

Step-by-step explanation:

i did the test

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Joyce baked a cake using 2 3/4 cups of brown sugar and 1 3/4 cups of powdered sugar. How many cups of sugar did she use in her c
BartSMP [9]
So she has used two kinds of sugar, brown and powdered.

She used 2 \frac{3}{4} cups of brown sugar and  1 \frac{3}{4} cups of powdered sugar: we need to add the two numbers:

2 \frac{3}{4}+1 \frac{3}{4}= 2+1+\frac{3}{4}+ \frac{3}{4}=3\frac{6}{4}=3+1+ \frac{2}{4}  =4 \frac{2}{4}=4 \frac{1}{2}=4.5

so the correct answer is 4.5
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Which of the following is equivalent to 7a^4 + 3a^4?
nasty-shy [4]
(7+3)a^4

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A study of 10 different weight loss programs involved 500 subjects. Each of the 10 programs had 50 subjects in it. The subjects
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.

8 0
4 years ago
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