The mean is 10,724.28.
Explanation:
The mean is all of the values divided by the number of values there are.
So what you have to do is add all of the numbers and divide it by the amount of numbers there are.
10,150+10,211+10,424+10,769+10,884+11,155+11,477= 75,070.
Since there are 7 numbers, divide 75,070/7 and you get your mean, which is 10,724.28 rounded.
Answer:
3, 4, Point R to Point Q, and depression from Point Q to Point R
Step-by-step explanation:
......
Answer:
7
Step-by-step explanation:
The text of this expression is:
"If 3 is added to twice a number, the result is 17. Find the number"
Let's call 
the number we want to find.
Twice the number is equal to
So, if we add 3 to 2x, we get
And we know that this is equal to 17:
Now we can solve the equation to find x: first, we subtract 3 from both sides
Now we divide both sides by 2:
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 !
