Answer:
25
Step-by-step explanation:
2 + <em>3</em> + <em>3</em> + 4 + 6 + 7 = 25
Alright, so the first thing I would do is find the LCM of the dividends, which is 20. So we have 10 15/20 and 6 16/20. You can either leave it there and just carry the one (20 in this case) for 3 19/20.
If you multiply 10 by 20 and add 15, or 6 by 20 and add 16 (which I think is easier) you get 215/20 - 136/20 = 79/20. It can be simplified from here if you teacher wants it in the future.
Answer:
129.4
Step-by-step explanation:
Calculation for what's the total length from the front of the first bus to the end of the last
First step is to multiply the given number of buses by the length of each one
Hence,
12.6 meters x 9 = 113.4 meters
Second step
Since there are 8 spaces in between we would multiply the 8 spaces by 2 meters
8 x 2 meters = 16
Now let calculate total length from the front of the first bus to the end of the last
16 + 113.4 meters =129.4
Therefore total length from the front of the first bus to the end of the last will be 129.4
Answer:
There are 67626 ways of distributing the chairs.
Step-by-step explanation:
This is a combinatorial problem of balls and sticks. In order to represent a way of distributing n identical chairs to k classrooms we can align n balls and k-1 sticks. The first classroom will receive as many chairs as the amount of balls before the first stick. The second one will receive as many chairs as the amount of balls between the first and the second stick, the third classroom will receive the amount between the second and third stick and so on (if 2 sticks are one next to the other, then the respective classroom receives 0 chairs).
The total amount of ways to distribute n chairs to k classrooms as a result, is the total amount of ways to put k-1 sticks and n balls in a line. This can be represented by picking k-1 places for the sticks from n+k-1 places available; thus the cardinality will be the combinatorial number of n+k-1 with k-1,
.
For the 2 largest classrooms we distribute n = 50 chairs. Here k = 2, thus the total amount of ways to distribute them is
.
For the 3 remaining classrooms (k=3) we need to distribute the remaining 50 chairs, here we have
ways of making the distribution.
As a result, the total amount of possibilities for the chairs to be distributed is 51*1326 = 67626.