9+y=6 help me please
2 answers:
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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.
Answer:
There is no statistical evidence at 1% level to accept that the mean net contents exceeds 12 oz.
Step-by-step explanation:
Given that a random sample of ten containers is selected, and the net contents (oz) are as follows: 12.03, 12.01, 12.04, 12.02, 12.05, 11.98, 11.96, 12.02, 12.05, and 11.99.
We find mean = 11.015
Sample std deviation = 3.157
a)
(Right tailed test)
Mean difference /std error = test statistic
p value =0.174
Since p >0.01, our alpha, fail to reject H0
Conclusion:
There is no statistical evidence at 1% level to accept that the mean net contents exceeds 12 oz.