Answer:
Type the correct answer in the box. Use numerals instead of words. If necessary, use / for the fraction bar. Write the expressio
n in simplest form. 11 I + 3
2 answers:
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Answer:
The reduced row-echelon form of the linear system is
Step-by-step explanation:
We will solve the original system of linear equations by performing a sequence of the following elementary row operations on the augmented matrix:
- Interchange two rows
- Multiply one row by a nonzero number
- Add a multiple of one row to a different row
To find the reduced row-echelon form of this augmented matrix
You need to follow these steps:
- Divide row 1 by 2
- Subtract row 1 from row 2
- Subtract row 1 multiplied by 5 from row 3
- Subtract row 2 multiplied by 3 from row 1
- Subtract row 2 multiplied by 3 from row 3
- Multiply row 2 by 2
- Divide row 3 by −19
- Subtract row 3 multiplied by 16 from row 1
- Add row 3 multiplied by 6 to row 2
Answer:
B) 4.07
Step-by-step explanation:
First we need to calculate the mean of all the data, which is the same as the mean of the means of each grade of gasoline:
Regular BelowRegular Premium SuperPremium
39.31 36.69 38.99 40.04
39.87 40.00 40.02 39.89
39.87 41.01 39.99 39.93
X1⁻=39.68 X2⁻= 39.23 X3⁻= 39.66 X4⁻= 39.95
Xgrand⁻ = (39.68+39.23+39.66+39.95)/4 = 39.63
Next we need to calculate the sum of squares within the group (SSW) and the sum of squares between the groups (SSB), and the respective degrees of freedom):
SSW = [ (39.31-39.68)² + (39.87-39.68)² + (39.87-39.68)² ] + [ (36.69-39.23)² + (40.00-39.23)² + (41.01-39.23)² ] + [ (38.99-39.66)² + (40.02-39.66)² + (39.99-39.66)² ] + [ (40.04-39.95)² + (39.89-39.95)² + (39.93-39.95)² ] = [0.2091] + [10.2129] + [0.6874] + [0.0121] = 11.12
SSW = 11.12
Degrees of freedom in this case is calculated by m(n-1), with m being the number of grades of gasoline (4) and n being the number of trial results for each one (3), so we would have 4(3-1) = 8 degrees of freedom
SSB = [ (39.68-39.63)² + (39.68-39.63)² + (39.68-39.63)²] + [ (39.23-39.63)² + (39.23-39.63)² + (39.23-39.63)² ] + [ (39.66-39.63)² + (39.66-39.63)² + (39.66-39.63)² ] + [ (39.95-39.63)² + (39.95-39.63)² +(39.95-39.63)² ] = [0.0075] + [0.48] + [0.0027] + [0.3072] = 0.7974
SSB = 0.80
For this case, the degrees of freedom are m-1, so we would have 4-1 = 3 degrees of freedom
Now we can establish the hypothesis for the test:
H0: μ1 = μ2 = μ3 = μ4
The null hypothesis states that the means of miles per gallon for each fuel are the same, indicating that the drade of gasoline does not make a difference, therefore our alternative hypothesis will be:
H1: the grade of gasoline does makes a difference
We will use the F statistic to test the hypothesis, which is calculated like follows:
F - statistic = (SSB/m-1) / (SSW/m(n-1)) = (0.80/3) / (11.12/8) = 0.19
We know that the level of significance we are using is α = 0.05, so to find the critical value F we need to look at some table of critical values for the F distribution for the 0.05 significance level (like the attached image). Then we just need to look fot the value that is located in the intersection between the degrees of freedom we have in the numerator (horizontal) and the denominator (vertical) of the statistic (3 and 8). That critical value is:
Fc = 4.07
