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meriva
4 years ago
14

Find solution and show work for: (3.4^17)4

Mathematics
1 answer:
stira [4]4 years ago
7 0

Answer:

231.2

Step-by-step explanation:

first multiply 34 and 17 which will give you 578. Then put the decimal after the 7 because you only have one decimal in the tenths place: 57.8. Then multiply 578 by 4: 2312. Then put the decimal after the 1 to get 231.2

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140 is decreased to 273
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Is there more to the question,

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4 years ago
Solve for the equation below<br><br> 4x^n + x^4n = ?
slavikrds [6]
4x + 6 = -10<span>Simplifying 4x + 6 = -10 Reorder the terms: 6 + 4x = -10 Solving 6 + 4x = -10 Solving for variable 'x'. Move all terms containing x to the left, all other terms to the right. Add '-6' to each side of the equation. 6 + -6 + 4x = -10 + -6 Combine like terms: 6 + -6 = 0 0 + 4x = -10 + -6 4x = -10 + -6 Combine like terms: -10 + -6 = -16 4x = -16 Divide each side by '4'. x = -4 Simplifying x = -4</span>
7 0
3 years ago
Which of the fractions 13/20, 3/5, 3/4 and 7/10 is greatest​
Andrei [34K]

Answer:

3/4

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
You are dealt one card from a 52 card deck. Then the card is replaced in the deck, the deck is shuffled, and you draw again. Fin
lora16 [44]

Answer:

3/52

Step-by-step explanation:

There are 13 types of cards (A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K), out of which 3 (J, Q, K) are face cards. This means that the probability for the first half is 3/13. There are 4 suites (Club, Spade, Hearts, and Diamonds). This means the probability of attaining a club is 1/4. 3/13 * 1/4 = 3/52

4 0
3 years ago
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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