Answer:
it is 2 1/2
Step-by-step explanation:
I hope this helps
If the ratio of red marbles to blue marbles in a jar is 3 to 8, the meaning is for every 3 red marbles there are 8 blue marbles.
3 red marbles------------------8 blue marbles.
ratio=3/8
we have 40 blue marbles; we have to compute the number of red marbles.
1) Method 1; by the rule three.
3 red marbles-----------------8 blue marbles
x----------------------------------40 blue marbles
x=(3 red marbles * 40 blue marbles) / 8 blue marbles=15 red marbles.
answer: B. 15
Method 2: the ratio of red marbles to blue marbles is
ratio=number of red marbles / number of blue marbles
ratio=3/8
if we want to compute the number of red marbles we have to multiply the number of blue marbles by this ratio.
number of red marbles=ratio (red/blue)* number of blue marbles
number of red marbles=(3/8)*40=15
Answer: B.15
Answer:
hi
Step-by-step explanation:
lol ahwvanbw wiwjvwnsiwbwbwnjsbwb
Answer:
The mean is also increased by the constant k.
Step-by-step explanation:
Suppose that we have the set of N elements
{x₁, x₂, x₃, ..., xₙ}
The mean of this set is:
M = (x₁ + x₂ + x₃ + ... + xₙ)/N
Now if we increase each element of our set by a constant K, then our new set is:
{ (x₁ + k), (x₂ + k), ..., (xₙ + k)}
The mean of this set is:
M' = ( (x₁ + k) + (x₂ + k) + ... + (xₙ + k))/N
M' = (x₁ + x₂ + ... + xₙ + N*k)/N
We can rewrite this as:
M' = (x₁ + x₂ + ... + xₙ)/N + (k*N)/N
and (x₁ + x₂ + ... + xₙ)/N was the original mean, then:
M' = M + (k*N)/N
M' = M + k
Then if we increase all the elements by a constant k, the mean is also increased by the same constant k.
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 !
