Answer:
Step-by-step explanation: if Peytons base pay= $60 plus she earns commission on $x (computers sold)
P= $60+ (0.025x)
You would multiply by 0.025 as this equals 2.5%
Remember PEMDAS- parentheses, exponents, multiplications, division, addition, subtraction
<span>Assuming that "pair up students" means "divide up all 20 of the students into groups of two," and we regard two pairings as the same if and only if, in each pairing, each student has the same buddy, then I believe that your answer of 20! / [(2!)^10 * (10!)] is correct. (And I also believe that this is the best interpretation of the problem as you've stated it.)
There are at least two ways to see this (possibly more).
One way is to note that, first, we have to select 2 students for the first pair; that's C(20, 2) (where by C(20, 2) I mean "20 choose 2"; that is, 20! / (18! * 2!). )
Then, for each way of selecting 2 students for the first pair, I have to select 2 of the remaining 18 students for the second pair, so I multiply by C(18, 2).
Continuing in this manner, I get C(20, 2) * C(18, 2) * ... * C(2, 2).
But it doesn't matter in this situation the order in which I pick the pairs of students. Since there are 10! different orders in which I could pick the individual pairs, then I want to divide the above by 10!, giving me the answer
[C(20, 2) * C(18, 2) * ... * C(2, 2)] / 10!.
This is the same as your answer, because C(n , 2) = n(n - 1) / 2, so we can simplify the above as
[(20 * 19) / 2 * (18 * 17) / 2 * ... * (2 * 1) / 2] / 10!
= 20! / [2^10 * 10!]
= 20! / [(2!)^10 * (10!)].
Another way is to reason as follows:
1. First, arrange the 20 students in a line; there are 20! ways to do this
2. We can get a pairing by pairing the 1st and 2nd students in line together, the 3rd and 4th students together, etc.
3. But if I switch the order of the 1st and 2nd student, then this doesn't give a different pairing. I don't want to count both orderings separately, so I divide by 2!
4. The same argument from step 3 holds for the 3rd and 4th student, the 5th and 6th student, etc., so I need to divide by 2! nine more times
5. Finally, the particular order in which I selected the ten pairings are unimportant--for example, the following orderings don't produce different pairings:
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20
3, 4, 1, 2, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20
So I need to further divide by the number of ways I can arrange the ten pairs, which is 10!.
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1) 26 different outcomes are in the sample space.
2) 1 / 26 is the probability that the computer produces the first letter of your first name.
<u>Step-by-step explanation:</u>
<u>1) You have to find out the different outcomes in the sample space :</u>
- A "Sample space" is defined as the set of all the possible outcomes of an event.
- Here, the given event is randomly selecting a letter from the alphabets.
Therefore, the sample space must contain all the possible alphabets that can be chosen randomly.
The sample space is the set of all the 26 alphabets in English language.
⇒ Sample space = {A,B,C,D...........,Y,Z}
⇒ 26 different outcomes.
<u>2) The probability the computer produces the first letter of your first name :</u>
Here, the required outcome is getting the first letter of your first name.
Probability = No. of required outcomes / total no. of outcomes.
For example, The name Alex Davis has the first letter of the fist name as alphabet 'A'.
∴ Probability = 1 / 26
Similarly, for any first name there is going to be any one alphabet from the 26 alphabets, thus the probability to get the first letter will be always 1 / 26.
