What is the midpoint of the segment shown below
2 answers:
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Answer: A. 85.7
Step-by-step explanation:
Given : Two sections of a class took the same quiz.
Section A had 15 students who had a mean score of 80, and Section B had 20 students who had a mean score of 90.
We know that ,
Then , for section A :
Similarly in Section B,
Total scores = Sum of scores in sec A+Sum of scores in sec B
=1200+1800=3000
Total students = Students in sec A +Students in sec B
=15+20=35
Now , the mean score for all of the students on the quiz =
Hence, the approximate mean score for all of the students on the quiz = 85.7
Thus , the correct answer is option A. 85.7.
Answer:
f(r) = (x -1)(x -4+√7)(x -4-√7)
Step-by-step explanation:
The signs of the terms are + - + -. There are 3 changes in sign, so Descartes' rule of signs tells you there are 3 or 1 positive real roots.
The rational roots, if any, will be factors of 9, the constant term. The sum of coefficients is 1 -9 +17 -9 = 0, so you know that r=1 is one solution to f(r) = 0. That means (r -1) is a factor of the function.
Using polynomial long division, synthetic division (2nd attachment), or other means, you can find the remaining quadratic factor to be r^2 -8r +9. The roots of this can be found by various means, including completing the square:
r^2 -8r +9 = (r^2 -8r +16) +9 -16 = (r -4)^2 -7
This is zero when ...
(r -4)^2 = 7
r -4 = ±√7
r = 4±√7
Now, we know the zeros are {1, 4+√7, 4-√7), so we can write the linear factorization as ...
f(r) = (r -1)(r -4 -√7)(r -4 +√7)
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<em>Comment on the graph</em>
I like to find the roots of higher-degree polynomials using a graphing calculator. The red curve is the cubic. Its only rational root is r=1. By dividing the function by the known factor, we have a quadratic. The graphing calculator shows its vertex, so we know immediately what the vertex form of the quadratic factor is. The linear factors are easily found from that, as we show above. (This is the "other means" we used to find the quadratic roots.)
To demonstrate a method for computing the limit itself, let's pick a small value of n. If n = 3, then our limit is
Let a = 1 and b the cosine product, and write them as
with
Now we use the identity
to rationalize the numerator. This gives
As x approaches 0, both a and b approach 1, so the polynomial in a and b in the denominator approaches 6, and our original limit reduces to
For the remaining limit, use the Taylor expansion for cos(x) :
where essentially means that all the other terms in the expansion grow as quickly as or faster than x⁴; in other words, the expansion behaves asymptotically like x⁴. As x approaches 0, all these terms go to 0 as well.
Then
so in our limit, the constant terms cancel, and the asymptotic terms go to 0, and we end up with
Unfortunately, this doesn't agree with the limit we want, so n ≠ 3. But you can try applying this method for larger n, or computing a more general result.
Edit: some scratch work suggests the limit is 10 for n = 6.