Answer:
2:3
3:2
Step-by-step explanation:
The fraction is 156/468 and the percentage is 33 percent
Answer: 175
Step-by-step explanation: total = x + 5y dollars and dollars = 300 so
x + 5y = 300 then since we know y = 200 we can substitute that
x + 5(200) = 300 then subtract x from 200
x + 5(200 - x) = 300 now solve for x
x + 5(200) + 5(-x) = 300 we know 5 times 200 is 1000
x + 1000 - 5x = 300 then subtract 5x - x = -4x
-4x + 1000 = 300 switch around 1000 and 300
-4x = 300 - 1000 subtract 300 - 1000 = -700
-4x = -700 multiply
x = -700/(-4) then finish up and solve
x = 175
hope this helps please mark brainliest if you can
Wow !
OK. The line-up on the bench has two "zones" ...
-- One zone, consisting of exactly two people, the teacher and the difficult student.
Their identities don't change, and their arrangement doesn't change.
-- The other zone, consisting of the other 9 students.
They can line up in any possible way.
How many ways can you line up 9 students ?
The first one can be any one of 9. For each of these . . .
The second one can be any one of the remaining 8. For each of these . . .
The third one can be any one of the remaining 7. For each of these . . .
The fourth one can be any one of the remaining 6. For each of these . . .
The fifth one can be any one of the remaining 5. For each of these . . .
The sixth one can be any one of the remaining 4. For each of these . . .
The seventh one can be any one of the remaining 3. For each of these . . .
The eighth one can be either of the remaining 2. For each of these . . .
The ninth one must be the only one remaining student.
The total number of possible line-ups is
(9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) = 9! = 362,880 .
But wait ! We're not done yet !
For each possible line-up, the teacher and the difficult student can sit
-- On the left end,
-- Between the 1st and 2nd students in the lineup,
-- Between the 2nd and 3rd students in the lineup,
-- Between the 3rd and 4th students in the lineup,
-- Between the 4th and 5th students in the lineup,
-- Between the 5th and 6th students in the lineup,
-- Between the 6th and 7th students in the lineup,
-- Between the 7th and 8th students in the lineup,
-- Between the 8th and 9th students in the lineup,
-- On the right end.
That's 10 different places to put the teacher and the difficult student,
in EACH possible line-up of the other 9 .
So the total total number of ways to do this is
(362,880) x (10) = 3,628,800 ways.
If they sit a different way at every game, the class can see a bunch of games
without duplicating their seating arrangement !
Answer
Sorry thats not out topic yet
Step-by-step explanation:
Sorry :c
