Professor Davis counts his midterm as 2/3 of the grade, and his final as 1/3 of the grade. Jason only scored 64 on the midterm. What range of scores on the final exam would put Jason's average between 70 and 79 inclusive?
2007-01-13 02:50:45 · 11 answers · asked by salarian2001 2 in Science & Mathematics ➔ Mathematics
Here's how you would have to set this problem up to solve it.
Jason already has one fixed grade, 64, which is 2/3 of his semester grade. He needs 1/3 of some unknown value, x, such that when you add it to (2/3)(64) the sum lies somewhere between 70 and 79, inclusive. Mathematically, we would write it as an inequality like this:
70 < or = (2/3)(64) + (x/3) < or = 79.
Multiply by 3 to clear fractions.
210 < or = 128 + x < or = 237
Subtract 128 from both sides to isolate x.
210 - 128 < or = x < or = 137 - 128
82 < or = x < or = 109.
So, Jason needs a score ranging from 82 to 109, inclusive, in order stay in the 'C' range for his final grade.
If Jason made 100 on his final exam, the maximum average grade possible is:
(2/3)(64) + (1/3)(100) = 42.67 + 33.33 = 76.00
Jason's only hope for attaining any higher grade than 76 is that the professor puts bonus questions on the final exam.
2007-01-13 04:49:08 · answer #1 · answered by MathBioMajor 7 · 0⤊ 0⤋
You'd have to get three numbers to sum up all the grades--each of which is "one third" of the semester's total. That total of all of three of them, then divided by three, then gives you the "average" for Jason's three grades.
You say Jason wants to get a "70 to 79" average score?
He already has two 64s, earned by his 2/3 of the total score that he achieved on the midterm. So, he's starting with 128
(2 x 64) and needs 3 x 70 or better, 210 or more, as a score on the final exam in order to get a grade in the range he wants.
Therefore, 210 minus 128=82. Jason has to get an 82 or better on the final to raise his grade to a 70; otherwise according to Professor Davis's grading curve (probably a bad one anyway), he's toast.
(The highest grade he could get by getting 100 on the final is
64 + 64 + 100-228 or a 76 average).
2007-01-13 03:03:11 · answer #2 · answered by Robert David M 7 · 0⤊ 1⤋
Jason need a total of 210 (70x3), he has 128 so far so he need at least an 82 on the final to average 70. To get to 79 he would need a total of 237 so he would need a 109 on the final which is not possible. If he got 100 on the final he would have a total of 228 which is an average of 76. Thats the best he can do.
2007-01-13 02:54:39 · answer #3 · answered by John G 4 · 1⤊ 0⤋
an 82 would give him a 70 average
a 100 would give him a 76 average
a 110 would give him a 79 average
~~so i guess the range is 82<>110~~
2007-01-13 02:57:51 · answer #4 · answered by bjd72003 3 · 0⤊ 1⤋
If some thing is 40% smaller, you purely subtract 40%. So 40% of 40 energy is sixteen: you are trying this by multiplying 40 by 0.4. the respond could be 40 - sixteen = 24 A shorter way could merely be to multiply by 40 by 0.6, on the grounds that's a million-0.4.
2016-10-19 22:25:18 · answer #5 · answered by swindler 4 · 0⤊ 0⤋
81 to 85
2007-01-13 02:55:28 · answer #6 · answered by Anonymous · 0⤊ 1⤋
82 and 109
2007-01-13 02:55:05 · answer #7 · answered by Anonymous · 1⤊ 0⤋
81-95
2007-01-13 03:33:32 · answer #8 · answered by Anonymous · 0⤊ 1⤋
hmmm 82-91?
2007-01-13 02:55:38 · answer #9 · answered by Anonymous · 0⤊ 1⤋
do you own homework....lazy bugger!
2007-01-13 03:00:50 · answer #10 · answered by Dave B 3 · 0⤊ 0⤋