Solve 47.5 + 4.5x ≤ 65.
1 answer:
You might be interested in
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
Answer:
The values of variables x and m are 11 and 17
Step-by-step explanation:
The question has missing details as the diagram of the trapezoid isn't attached.
(See attachment).
Given that trapezoid CHLE is isosceles then the angles at the base area equal (4x)
And
The angles at the top are also equal
8m = 11x + 15
At this point, the four angles in the trapezoid are 8m, 11x + 15, 4x and 4x..
The sum of interior= 360
So,
11x + 15 + 8m + 4x + 4x = 360
Collect like terms
11x + 4x + 4x + 8m = 360 - 15
19x + 8m = 345
Substitute 11x + 15 for 8m
19x + 11x + 15 = 345
30x + 15 = 345
30x = 345 - 15
30x = 330
Divide through by 30
30x/30 = 330/30
x = 11
Recall that 8m = 11x + 15;
8m = 11(11) + 15
8m = 121 + 15
8m = 136
Divide through by 8
8m/8 = 136/8
m = 17
Hence, the values of variables x and m are 11 and 17
