Divide. (6x^2+36x+35)/(x+5) Your answer should give the quotient and the remainder.
1 answer:
Answer:
Step-by-step explanation:
When we divide polynomials by polynomials with more than one term, we use long division or factorization to simplify.
This polynomial isn't easily divided after factorization so we will use long division. And we will use the remainder theorem to write any remainder.
We start long division by finding what multiplies with x+5 to get . This is 6x.
So . We have 6x left as a remainder from 36x.
We now divide x+5 into 6x+35. What multiplies with x+5 to get 6x+35? 6.
So we have 6x+6 as our answer so far and after we multiply 6(x+5)=6x+30 we will have a final remainder of 5.
We write our answer as .
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The correct question statement is:
A probability experiment is conducted in which the sample space of the experiment is S = {4,5,6,7,8,9,10,11,12,13,14,15}. Let event E={7,8,9,10,11,12,13,14,15}. Assume each outcome is equally likely. List the outcomes in
. Find 
.
Solution:
Part 1:
means compliment of the set E. A compliment of a set can be obtained by finding the difference of the set from the universal set. The universal set is the set which contains all the possible outcomes of the events which is S in this case.
So, compliment of E will be equal to S - E. S - E will result in all those elements of S which are not present in E. So, we can write:
Thus the set compliment of E will contain the elements {4,5,6}.So
= {4,5,6}
Part 2)
means probability that if we select any number from the Sample Space S, it will belong the set E compliment.
= (Number of Elements in 
)/Number of elements in S
Number of elements in set S = n(S) = 12
Number of elements in set = 
=3
So,
A probability experiment is conducted in which the sample space of the experiment is S = {4,5,6,7,8,9,10,11,12,13,14,15}. Let event E={7,8,9,10,11,12,13,14,15}. Assume each outcome is equally likely. List the outcomes in
Solution:
Part 1:
So, compliment of E will be equal to S - E. S - E will result in all those elements of S which are not present in E. So, we can write:
Thus the set compliment of E will contain the elements {4,5,6}.So
Part 2)
Number of elements in set S = n(S) = 12
Number of elements in set
So,