Answer:
Line B? Professional guesser please dont get me fired.
This problem deals with a weighted average. Since the AP classes count twice as much as regular classes, their grades must be counted twice. It's as if for each AP class he's taking, he was taking two classes. The points of each AP class grade is added twice, and the each AP class counts as 2 classes in the number of classes.
Each AP class counts twice and counts as 2 classes.
Class Ben's grade Points Number of Classes
AP English B 3 + 3 2
AP Government B 3 + 3 2
AP Algebra II A 4 + 4 2
Spanish B 3 1
Physics D 1 1
TOTALS 24 8
GPA = (total points)/(number of classes) = 24/8 = 3
<span>Answer: B 3.0</span>
the number of elements in the union of the A sets is:5(30)−rAwhere r is the number of repeats.Likewise the number of elements in the B sets is:3n−rB
Each element in the union (in S) is repeated 10 times in A, which means if x was the real number of elements in A (not counting repeats) then 9 out of those 10 should be thrown away, or 9x. Likewise on the B side, 8x of those elements should be thrown away. so now we have:150−9x=3n−8x⟺150−x=3n⟺50−x3=n
Now, to figure out what x is, we need to use the fact that the union of a group of sets contains every member of each set. if every element in S is repeated 10 times, that means every element in the union of the A's is repeated 10 times. This means that:150 /10=15is the number of elements in the the A's without repeats counted (same for the Bs as well).So now we have:50−15 /3=n⟺n=45
Hey.. don’t click on that link or any links on here !!! but i jus wanna say you didn’t attach an image of your question :).
Answer:
You are correct! It is the answer you have selected in the picture!
Step-by-step explanation:
Hope this helps :)
4x4=16
-4x-4=16 too
