Answer:
5) 15120
6) 11880
7) 336
Step-by-step explanation:
The formula for permutation where mPn is m!/(m-n)!
Applying this to question 5, we get 9!/4!, which is 15120.
For question 6, we get 12!/8!, which is 11880.
For question 7, we get 8!/5!, which is 336.
Answer:-30
Step-by-step explanation:
Third term=a3
an=a(n-1) x (-9)
a3=5/3 x (3-1) x (-9)
a3=5/3 x 2 x -9
a3=(5x2x-9)/3
a3=-90/3
a3=-30
Well, we could try adding up odd numbers, and look to see when we reach 400. But I'm hoping to find an easier way.
First of all ... I'm not sure this will help, but let's stop and notice it anyway ...
An odd number of odd numbers (like 1, 3, 5) add up to an odd number, but
an even number of odd numbers (like 1,3,5,7) add up to an even number.
So if the sum is going to be exactly 400, then there will have to be an even
number of items in the set.
Now, let's put down an even number of odd numbers to work with,and see
what we can notice about them:
1, 3, 5, 7, 9, 11, 13, 15 .
Number of items in the set . . . 8
Sum of all the items in the set . . . 64
Hmmm. That's interesting. 64 happens to be the square of 8 .
Do you think that might be all there is to it ?
Let's check it out:
Even-numbered lists of odd numbers:
1, 3 Items = 2, Sum = 4
1, 3, 5, 7 Items = 4, Sum = 16
1, 3, 5, 7, 9, 11 Items = 6, Sum = 36
1, 3, 5, 7, 9, 11, 13, 15 . . Items = 8, Sum = 64 .
Amazing ! The sum is always the square of the number of items in the set !
For a sum of 400 ... which just happens to be the square of 20,
we just need the <em><u>first 20 consecutive odd numbers</u></em>.
I slogged through it on my calculator, and it's true.
I never knew this before. It seems to be something valuable
to keep in my tool-box (and cherish always).
From the given data sets we shall obtain the answers as follows:
<span>301 345 305 394 328 317 361 325 310 362
Re-arrange the data in ascending order:
301, 305, 310, 317,325, 328, 345, 361, 362, 394
from the above we see that:
Minimum value=301
Q1=310
median=(325+328)/2=326.5
Q3=361
Maximum value=394
Thus the box plot will be as follows:
</span>
