How many students had 3 or more absences? Add together {45%, 21%, 23%} and subtract that sum from 100%: 11%.
How many students had two or more absences? Add together this 11% and 21%: 32% had two or more absences. Let the total number of students be n. Then 32% of these n students add up to 96 students. Thus, 0.32n = 96, and n= 300.
How many students had one absence? The chart shows that that is 45%, and we know that there are 300 students total.
Thus, the number of students who had one absence was 0.45(300) = 135.
Answer:
There are 3,659,040 ways he can choose the books to put on the list.
Step-by-step explanation:
There are
12 novels
8 plays
12 nonfiction.
He wants to include
5 novels
4 plays
2 nonfiction
The order in which the novels, plays and nonfictions are chosen is not important. So we use the combinations formula to solve this problem.
is the number of different combinations of x objects from a set of n elements, given by the following formula.

How many ways can he choose the books to put on the list?
Novels:
5 from a set of 12. So

Plays:
4 from a set of 8. So

Nonfiction:
2 from a set of 12

Total:
Multiplication of novels, plays and nonfiction.

There are 3,659,040 ways he can choose the books to put on the list.
Answer:
f(x)/g(x) = 4x^2+6x-3 remainder -12
or
f(x)/g(x) = (4x^2+6x-3)/(2x+1) + (-12)/(2x+1)
or
((4x²+6x-3)-12)/(2x+1)
Step-by-step explanation:
f(x)= 8x³ + 16x² -15
g(x) = 2x+1.
Using long division,
f(x)/g(x) = 4x^2+6x-3 remainder -12