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4 years ago

What is the lowest term of 10 / 12

2 answers:

3 0

If you want to simplify it, 10/12 simplifies to 5/6. You divide both top and bottom by two to get the simplest form of this fraction.

Alex Ar [27]4 years ago

3 0

Two two twotwo two two two two

You might be interested in

How to solve for 7/8 X c for c=8

zvonat [6]

Simple 7/8 is being multiplied by 8 or 8/1. When that is multiplied together 7/8*8/1, the 8's cancel each other out, and the answer is 7.

7 0

4 years ago

Read 2 more answers

The inverse of the function f(x) = x + 10 is shown.

Vadim26 [7]

we have

$f(x) = \frac{1}{2}x + 10$

Step $1$

Let

y=f(x)

$y = \frac{1}{2}x + 10$

Exchange the variable x for y and variable y for x

$x = \frac{1}{2}y + 10$

Clear variable y

Multiply by $2$ both sides

$2x = y + 20$

$y = 2x-20$

Let

$f(x)^{-1} =y$

$f(x)^{-1}=2x-20$

therefore

the answer is

the inverse of the function is $f(x)^{-1}=2x-20$

the missing value is $20$

5 0

3 years ago

Read 2 more answers

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

Your are part of a team studying an endangered species of land snail in upstate New York. You and your colleagues capture and ma

ELEN [110]

Answer:

720 individuals

Step-by-step explanation:

According to my research, I can say that based on the information provided within the question you can conclude that the population of land snails has about 720 individuals. We can calculate this by using the Rule of Three formula that is attached below.

x ⇒ 80 marked

45 ⇒ 5 marked

$\frac{45*80}{5} = x$