We can start by setting up an equation using weighted averages:
0.76(100 - 40 - 15) + x(40) + 0(15) = 0.8(100)
Here, x represents the percentage Sarah needs to score on her test after completing corrections. We subtract the weights of the test and quiz from 100 since they cover the remaining percentage of her grade.
Simplifying the equation:
0.76(45) + 40x = 80
34.2 + 40x = 80
40x = 45.8
x = 1.145
To convert this to a percentage, we can multiply by 100:
x = 114.5%
This means that Sarah would need to score 114.5% on her test after corrections to bring her grade up to an 80%. This is not realistically possible, so we can conclude that she cannot achieve this given the current weights and grades.
Sarahs grade is at a 76%
her test is worth 40% of her grade
her quiz is worth 15%
how many test corrections will she need to do to bring her grade up to a 80%
6 answers
er test is worth 40% of her grade
her quiz is worth 15%
On test she scores about 66% and Quizes 89%
how many test and quiz corrections will she need to do to bring her grade up to a 80%
her quiz is worth 15%
On test she scores about 66% and Quizes 89%
how many test and quiz corrections will she need to do to bring her grade up to a 80%
We can set up an equation to solve for the number of test and quiz corrections she needs to do:
0.66(40) + 0.89(15) + x(40) + y(15) = 0.8(100)
Here, x represents the percentage increase she needs to achieve on her test, and y represents the percentage increase she needs to achieve on her quiz. We multiply each weight by the percentage she has currently scored in each category, and then add x and y as the weights for the percentages she needs to increase to reach an 80% overall grade.
Simplifying the equation:
26.4 + 13.35 + 40x + 15y = 80
39.75 + 40x + 15y = 80
40x + 15y = 40.25
Dividing by 5:
8x + 3y = 8.05
We have one equation with two unknowns, so we cannot solve for x and y uniquely. However, we do know that Sarah needs to do both test and quiz corrections, as there are values of x and y that satisfy this equation. To find specific values of x and y, Sarah can talk to her teacher and see what options are available for test and quiz corrections, and how many points she can earn through each correction. She can then solve for x and y based on these options.
0.66(40) + 0.89(15) + x(40) + y(15) = 0.8(100)
Here, x represents the percentage increase she needs to achieve on her test, and y represents the percentage increase she needs to achieve on her quiz. We multiply each weight by the percentage she has currently scored in each category, and then add x and y as the weights for the percentages she needs to increase to reach an 80% overall grade.
Simplifying the equation:
26.4 + 13.35 + 40x + 15y = 80
39.75 + 40x + 15y = 80
40x + 15y = 40.25
Dividing by 5:
8x + 3y = 8.05
We have one equation with two unknowns, so we cannot solve for x and y uniquely. However, we do know that Sarah needs to do both test and quiz corrections, as there are values of x and y that satisfy this equation. To find specific values of x and y, Sarah can talk to her teacher and see what options are available for test and quiz corrections, and how many points she can earn through each correction. She can then solve for x and y based on these options.
whats the answer?
So she'd need to do 8 corrections?
No, we cannot determine the specific number of test and quiz corrections that Sarah needs to do based on the given information. The equation we derived only tells us that there are values of x and y that satisfy it, but we need additional information about the available options for test and quiz corrections to solve for x and y.
Sarah should speak to her teacher to gather more information about her options for test and quiz corrections, and how many points she can earn through each correction. Once she has this information, she can solve for x and y using the equation 8x + 3y = 8.05 to determine the specific number of test and quiz corrections she needs to do.
Sarah should speak to her teacher to gather more information about her options for test and quiz corrections, and how many points she can earn through each correction. Once she has this information, she can solve for x and y using the equation 8x + 3y = 8.05 to determine the specific number of test and quiz corrections she needs to do.
1 answer