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Scilla [17]
3 years ago
12

Let H be an upper Hessenberg matrix. Show that the flop count for computing the QR decomposition of H is O(n2), assuming that th

e factor Q is not assembled but left as a product of rotators.
Mathematics
1 answer:
aleksley [76]3 years ago
8 0

Answer:

Answer is explained in the attached document

Step-by-step explanation:

Hessenberg matrix- it a special type of square matrix,there there are two subtypes of hessenberg matrix that is upper Hessenberg matrix and lower Hessenberg matrix.

upper Hessenberg matrix:- in this type of matrix  zero entries below the first subdiagonal or in another words square matrix of n\times n is said to be in upper Hessenberg form  if ai,j=0

for all i,j with i>j+1.and upper Hessenberg matrix is called unreduced if all subdiagonal entries are nonzero

lower Hessenberg matrix:-  in this type of matrix  zero entries upper the first subdiagonal,square matrix of n\times n is said to be in lower Hessenberg form  if ai,j=0  for all i,j with j>i+1.and lower Hessenberg matrix is called unreduced if all subdiagonal entries are nonzero.

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Simplify to create an equivalent expression <br> −5(3p+3)+9(−7+p)
son4ous [18]

Answer: -6p - 78

Step-by-step explanation:

-5 (3p + 3) + 9(-7 + p)

-15p -15 - 63 + 9p

-6p-78

4 0
3 years ago
Read 2 more answers
A professor at a local community college noted that the grades of his students were normally distributed with a mean of 84 and a
creativ13 [48]

Answer:

A. P(x>91.71)=0.10, so the minimum grade is 91.71

B. P(x<72.24)=0.025 so the maximum grade could be 72.24

C. By rule of three, 200 students took the course

Step-by-step explanation:

The problem says that the grades are normally distributed with mean 84 and STD 6, and we are asked some probabilities. We can´t find those probabilities directly only knowing the mean and STD (In that distribution), At first we need to transfer our problem to a Standard Normal Distribution and there is where we find those probabilities. We can do this by a process called "normalize".

P(x<a) = P( (x-μ)/σ < (a-μ)/σ ) = P(z<b)

Where x,a are data from the original normal distribution, μ is the mean, σ is the STD and z,b are data in the Standard Normal Distribution.

There´s almost no tools to calculate probabilities in other normal distributions. My favorite tool to find probabilities in a Standard Normal Distribution is a chart (attached to this answer) that works like this:

P(x<c=a.bd)=(a.b , d)

Where "a.b" are the whole part and the first decimal of "c" and "d" the second decimal of "c", (a.b,d) are the coordinates of the result in the table, we will be using this to answer these questions. Notice the table only works with the probability under a value (P(z>b) is not directly shown by the chart)

A. We are asked for the minimum value needed to make an "A", in other words, which value "a" give us the following:

P(x>a)=0.10

Knowing that 10% of the students are above that grade "a"

What we are doing to solve it, as I said before, is to transfer information from a Standard Normal Distribution to the distribution we are talking about. We are going to look for a value "b" that gives us 0.10, and then we "normalize backwards".

P(x>b)=0.10

Thus the chart only works with probabilities UNDER a value, we need to use this property of probabilities to help us out:

P(x>b)=1 - P(x<b)=0.10

P(x<b)=0.9

And now, we are able to look "b" in the chart.

P(x<1.28)=0.8997

If we take b=1.285

P(x<1.285)≈0.9

Then

P(x>1.285)≈0.1

Now that we know the value that works in the Standard Normal Distribution, we "normalize backwards" as follows:

P(x<a) = P( (x-μ)/σ < (a-μ)/σ ) = P(z<b)

If we take b=(a+μ)/σ, then a=σb+μ.

a=6(1.285)+84

a=91.71

And because P(x<a)=P(z<b), we have P(x>a)=P(z>b), and our answer will be 91.71 because:

P(x>91.71) = 0.1

B. We use the same trick looking for a value in the Standard Normal Distribution that gives us the probability that we want and then we "normalize backwards"

The maximum score among the students who failed, would be the value that fills:

P(x<a)=0.025

because those who failed were the 2.5% and they were under the grade "a".

We look for a value that gives us:

P(z<b)=0.025 (in the Standard Normal Distribution)

P(z<-1.96)=0.025

And now, we do the same as before

a=bσ+μ

a=6(-1.96)+84

a=72.24

So, we conclude that the maximum grade is 72.24 because

P(x<72.24)=0.025

C. if 5 students did not pass the course, then (Total)2.5%=5

So we have:

2.5%⇒5

100%⇒?

?=5*100/2.5

?=200

There were 200 students taking that course

6 0
3 years ago
Simplify the following expression. <br><br> (65x+50y)-(16x-21y)
GarryVolchara [31]
Your going to want to first distribute the negative out to the second equation
65x+50y-16x+21y
49x+71y
5 0
3 years ago
Read 2 more answers
Pls solve if u can &lt;3 much appreciated
Scorpion4ik [409]

Step-by-step explanation:

that is only possible, if we don't need to use a 2 and a 5 (but especially the 2) to create the 1 in the middle (log(10) = log(2×5) or log (5×2) = 1).

in other words, if the 1 in the middle is just a given, and does not use any of our digits for is creation, then a solution is possible.

first of all, the upper left corner has to be log(0⁰). we cannot use the digit 0 for anything else. and 0^n (except for n = 0) is not a valid argument for a logarithm.

log(0⁰) = 0 log(2×1) = 0.3... log(9/4) = 0.35...

log(1×2) = 0.3... 1 log(3×6) = 1.26...

log(7/3) = 0.37... log(4×5) = 1.3... log(6⁹) = 7.0...

but as soon as we need to build the center 1 by log(10), which takes then away one of the 2s (as the only way to build 10 by multiplication is 2×5 or 5×2), there is no possible solution.

7 0
2 years ago
I need help with this problem
gregori [183]

Answer:

[g+2/3g-1] ÷ [g^2+2g/6g+2]

g+2/3g-1 × 6g+2/g^2+2g

g+2/3g-1 × 2(3g +1)/g(g+2)

1/ 3g-1 × 2(3g+1)/g

2(3g +1)÷ g(3g-1)

6g+2 ÷[3g^2-g]

4 0
2 years ago
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