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Alenkasestr [34]
3 years ago
13

The sum of twice a number and 18 less than the number is the same as the difference between -30 and the number

Mathematics
2 answers:
maksim [4K]3 years ago
6 0
2x + x - 18 = -30 - x
Monica [59]3 years ago
6 0
Lets say the number is x. The formula would be:
2x + x - 18 = x -30
To find x, you have to get rid of everything els, so minus x from both sides.
2x - 18 = -30
Since -30 is the smallest number, you get rid of that next. Add 30 to both sides
2x + 12 = 0
Now get rid of +12 by doing -12.
2x = -12
divide both sides by 2 to get x by itself
x = -6.

To double check your answer, do the reverse. Substitute -6 iinto the equation wherever it says x.

2x + x - 18 = x -30 becomes (2 x -6) + (-6 - 18) = -6 - 30.
Use BIDMAS (Brackets Indices Division Multiplication Addition Subtraction) to get your answer. Brackets are first
-12 + -24 = -6 - 30
-36 = -6 - 30 which is basically -36 = -36
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To test whether or not there is a difference between treatments A, B, and C, a sample of 12 observations has been randomly assig
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Answer:

1. 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

2. D. 36

3. C. 34

4. B. 1.059

5. B. 8.02

Step-by-step explanation:

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 1

The hypothesis for this case are:

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

Part 2

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}

We need to find the mean for each group first and the grand mean.

\bar X =\frac{\sum_{i=1}^n x_i}{n}

If we apply the before formula we can find the mean for each group

\bar X_A = 27, \bar X_B = 24, \bar X_C = 30. And the grand mean \bar X = 27

Now we can find the sum of squares between:

SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2

Each group have a sample size of 4 so then n_j =4

SS_{between}=SS_{model}=4(27-27)^2 +4(24-27)^2 +4(30-27)^2=72

The degrees of freedom for the variation Between is given by df_{between}=k-1=3-1=2, Where  k the number of groups k=3.

Now we can find the mean square between treatments (MSTR) we just need to use this formula:

MSTR=\frac{SS_{between}}{k-1}=\frac{72}{2}=36

D. 36

Part 3

For the mean square within treatments value first we need to find the sum of squares within and the degrees of freedom.

SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2

SS_{error}=(20-27)^2 +(30-27)^2 +(25-27)^2 +(33-27)^2 +(22-24)^2 +(26-24)^2 +(20-24)^2 +(28-24)^2 +(40-30)^2 +(30-30)^2 +(28-30)^2 +(22-30)^2 =306

And the degrees of freedom are given by:

df_{within}=N-k =3*4 -3 = 12-3=9. N represent the total number of individuals we have 3 groups each one with a size of 4 individuals. And k the number of groups k=3.

And now we can find the mean square within treatments:

MSE=\frac{SS_{within}}{N-k}=\frac{306}{9}=34

C. 34

Part 4

The test statistic F is given by this formula:

F=\frac{MSTR}{MSE}=\frac{36}{34}=1.059

B. 1.059

Part 5

The critical value is from a F distribution with degrees of freedom in the numerator of 2 and on the denominator of 9 such that we have 0.01 of the area in the distribution on the right.

And we can use excel to find this critical value with this function:

"=F.INV(1-0.01,2,9)"

And we will see that the critical value is F_{crit}=8.02

B. 8.02

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