Mona wrote 7 tests, and her test average is 93. There will be one more test before the end of the year. What is the lowest grade
1 answer:
For this case, the first thing we are going to do is assume that all the tests are worth the same.
Then, we define a variable:
x: score of Mona's last test
We write now the inequality that models the problem:
From here, we clear the value of x:
Answer:
the lowest grade that Mona can get for her last test so that her test average is 90 or more is:
x = 87
Then, we define a variable:
x: score of Mona's last test
We write now the inequality that models the problem:
From here, we clear the value of x:
Answer:
the lowest grade that Mona can get for her last test so that her test average is 90 or more is:
x = 87
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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
