Question
In: Advanced Math
A= 1 0 -7 7 0 1 0 0 2 -2 10 -7 2 -2 2...
A=
| 1 | 0 | -7 | 7 |
| 0 | 1 | 0 | 0 |
| 2 | -2 | 10 | -7 |
| 2 | -2 | 2 | 1 |
Diagonalize the matrix above. That is, find matrix D and a nonsingular matrix P such that A = PDP-1 . Use the representation to find the entries of An as a function of n.
Solutions
Colby Messinger answered 2 years agoRelated Solutions
A = （1 −7 5 0 0 10 8 2 2 4 10 3 −4 8...
A =
（1 −7 5 0
0 10 8 2
2 4 10 3
−4 8 −9 6）
(1) Count the number of rows that contain negative
components.
(2) Obtain the inverse of A and count the number of columns that
contain even number of positive components.
(3) Assign column names (a,b,c,d) to the columns of A.
(4) Transform the matrix A into a vector object a by stacking
rows.
(5) Replace the diagonal components of A with (0,0,2,3). Hint:...
0 0 0 0 0 0 0 0 0 0 0 6 7 10 10 10...
0
0
0
0
0
0
0
0
0
0
0
6
7
10
10
10
15
15
15
15
15
15
15
20
20
20
20
20
20
20
20
20
20
20
20
25
25
25
25
26
26
27
30
30
30
30
30
30
30
35
35
35
35
35
36
36
36
36
40
40
40
40
40
40
40
40
40
40
45
45
50
50
60
60
80
INTERVAL ESTIMATE
Confidence Level
Sample...
DATA 3 8 2 15 2 2 0 0 4 5 2 7 0 1 5...
DATA
3
8
2
15
2
2
0
0
4
5
2
7
0
1
5
3
0
2
5
4
1
6
9
5
3
1
2
10
6
1
1
2
1
19
6
6
6
7
0
4
1
1
1
0
1
9
2
2
2
1
16
10
10
5
2
3
1
4
4
4
3
6
2
8
5
2
7
1
6
4
0
3
1
1
1
Background:
A group of...
6 5 4 5 0 0 13 48 6 1 0 7 2 0 1 1...
6
5
4
5
0
0
13
48
6
1
0
7
2
0
1
1
0
2
11
5
11
27
4
0
6
Create Standard Deviation Chart (Normal Distribution
Curve)
Given: x y -5 1 -4 5 -3 4 -2 7 -1 10 0 8 1...
Given:
x
y
-5
1
-4
5
-3
4
-2
7
-1
10
0
8
1
9
2
13
3
14
4
13
5
18
What are the confidence limits (alpha = 0.05) for the true mean
value of Y when X = 3?
Mother's age 18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,51 Female 1, 0, 2, 2, 3, 4, 7, 3, 2, 4, 7, 1,...
Mother's age
18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,51
Female 1, 0, 2, 2, 3, 4, 7, 3, 2, 4, 7, 1, 6, 4, 5, 3, 1, 4, 0,
1, 1, 1, 0, 1, 0
Use the stem and leaf plots that you previously created to help
you draw and label histograms on your scratch paper with bin width
of 2 for mothers's age at birth of female students and for mother's
age at birth of male students. Make the lower bound of your first
bin...
f(x,y) = 2/7(2x + 5y) for 0 < x < 1, 0 < y < 1...
f(x,y) = 2/7(2x + 5y) for 0 < x < 1, 0 < y < 1
given X is the number of students who get an A on test 1
given Y is the number of students who get an A on test 2
find the probability that more then 90% students got an A test 2
given that 85 % got an A on test 1
0. 0. 0. 0.0. 0. 0. 0. 0. 1. 1. 1. 1. 1. 1. 2. 2. 2. 3. 4.
0. 0. 0. 0.0. 0. 0. 0. 0. 1. 1. 1. 1. 1. 1. 2. 2. 2.
3. 4.
A.)MEAN –
B.)MEDIAN -
C.)MODE -
D.)STANDARD DEVIATION –
E.)5 NUMBER SUMMARY –
F.)BOX AND WHISKERS PLOT –
G.) OUTLIERS-
0. 0. 0. 0.0. 0. 0. 0. 0. 1. 1. 1. 1. 1. 1. 2. 2. 2. 3. 4.
0. 0. 0. 0.0. 0. 0. 0. 0. 1. 1. 1. 1. 1. 1. 2. 2. 2.
3. 4.
A.)5 NUMBER SUMMARY –
B.)BOX AND WHISKERS PLOT –
C.) OUTLIERS-
Given the following data (2, 5, 1, 0, 5, 0, 7, 2, 3) Calculate the 3rd...
Given the following data (2, 5, 1, 0, 5, 0, 7, 2, 3)
Calculate the 3rd quartile from your "data".
What's the probability that a randomly selected number from
your data is even?
Consider the events A = "a randomly selected number from your
data is even" and B = "a randomly selected number from your data is
above the 3rd quartile". Are these events disjoint, independent,
neither, or both? Show your work.
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