Vendor: HP
Exam Code: HPE2-E67
Exam Name: HPE IT Business Conversations
Version: DEMO
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QUESTION 1
You have a customer who would like to take multiple physical servers and allow
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Exam
: HPE2-E67
Title
: HPE IT Business
Conversations
Version : Demo
1/2
The safer , easier way
Exam:
HPE2-E67
Name:
HPE IT Business Conversations
Version :
V9.0.1
1.What is the common term used for business managers who sponsor non sanctioned IT
projects
A. Nimble IT
B. DevOps IT
C. Shadow IT
D. Converged IT
AnswerA
2.When creating a Business Value
HPE2-E67
QUESTION: 1
Which new IT consumption model is a pay-as-you-deploy model that allows customers to more
efficiently adapt to business changes, and to take a more filexible, scalable approach to
capacity planning?
A. HPE Pre-Provisioning
B. HPE Flex
UNIT 1 LAB PORTFOLIO
PART III
HUMAN TISSUE TYPES
BIOL 2401: HUMAN ANATOMY AND PHYSIOLOGY I
SECTION: _
NAMES:
_
_
_
_
_
HUMAN TISSUE TYPES
DIRECTIONS: Identify the specific name of each tissue and, using the lab review, provide one specific function and on
UNIT 1 LAB PORTFOLIO
PART I
INTRODUCTION TO MICROSCOPY, METRICS, AND THE HUMAN BODY
BIOL 2401: HUMAN ANATOMY AND PHYSIOLOGY I
SECTION: _
NAMES:
_
_
_
_
_
INTRODUCTION TO MICROSCOPY
DIRECTIONS: Match the letter of the definition to the name of the microsco
UNIT 1 LAB PORTFOLIO
PART II
SOMATIC CELLS AND CELL DIVISION
BIOL 2401: HUMAN ANATOMY AND PHYSIOLOGY I
SECTION: _
NAMES:
_
_
_
_
_
INTRODUCTION TO SOMATIC CELLS
TYPICAL SOMATIC CELL
1. Match the number on the figure to the name of the cellular component t
324
C H A P T E R 12
VECTOR GEOMETRY
solution
(LT CHAPTER 13)
See the following three figures:
y
y
y
A
sw
w
w
w
B
v
rv
C
v
sw
rv
x
x
sw
v
x
rv
54. Sketch the parallelogram spanned by v = 1, 4 and w = 5, 2. Add the vector u = 2, 3 to the sketch and express
318
C H A P T E R 12
VECTOR GEOMETRY
(LT CHAPTER 13)
23. Let v = P Q, where P = (2, 5), Q = (1, 2). Which of the following vectors with the given tails and heads are
equivalent to v?
(a) (3, 3), (0, 4)
(b) (0, 0), (3, 7)
(c) (1, 2), (2, 5)
(d) (4, 5), (1,
A wins (Y/N) spinner a outcome
Y
9
Y
9
Y
9
N
5
N
5
Y
5
N
1
N
1
N
1
spinner b outcome
7
6
2
7
6
2
7
6
2
Table 5: The possible values spinners a and b in the game for Problem 2-56.
Nowby symmetry
all of the probabilities in the individual sums are the same
Problem 12 (counting awards)
Part (a): We have 30 students to choose from for the first award, and 30 students to choose
from for the second award, etc. So the total number of different outcomes is given by
305 = 24300000
Part (b): We have 30 students to
Problem 13 (random letter)
The same letter could be chosen if and only if it comes from one of R, E, or V . The
probability of R is chosen from both words is
1
2
2
=
.
7
8
56
The probability of E is chosen from both words is
3
3
1
=
.
7
8
56
Finally t
Chapter 1 (Combinatorial Analysis)
Chapter 1: Problems
Problem 1 (counting license plates)
Part (a): In each of the first two places we can put any of the 26 letters giving 262 possible
letter combinations for the first two characters. Since the five othe
Vectors in the Plane (LT SECTION 13.1)
S E C T I O N 12.1
321
42. Vector v of length 2 making an angle of 30 with the x-axis
solution The desired vector is
v = 2cos 30 , sin 30 = 2
3 1
,
= 3, 1 .
2 2
43. Find all scalars such that 2, 3 has length 1.
s
Problem 10 (committees with a chair)
Part (a): We can select a committee with k members in
son from the k committee members gives
k
n
k
n
k
ways. Selecting a chairper-
possible choices.
n
Part (b): If we choose the non chairperson members first this can b
Vectors in the Plane (LT SECTION 13.1)
S E C T I O N 12.1
323
The length of the vector is:
4i + 3j =
42 + 32 = 5
(b) We use vector algebra and the definition of the standard basis vector to compute the components of the vector 2i 3j:
2i 3j = 21, 0 30, 1 =
Vectors in the Plane (LT SECTION 13.1)
S E C T I O N 12.1
317
y
3v + w
w
v
x
2v 2w
20. Sketch v = 2, 1, w = 2, 2, v + 2w, v 2w.
solution We compute the linear combinations v + 2w and v 2w and then sketch the vectors:
v + 2w = 2, 1 + 22, 2 = 2, 1 + 4, 4 =
the women that get off on floor i. Thus we must have
6
X
i=1
6
X
i=1
mi = 5 mi 0
wi = 3 wi 0 .
The number of solutions to the first equation is given by
10
5+61
= 252 ,
=
5
61
while the number of solutions to the second equation is given by
8
3+61
= 5
316
C H A P T E R 12
VECTOR GEOMETRY
(LT CHAPTER 13)
y
w
x
w
The vector sum v + w is the vector:
v + w = 2, 3 + 4, 1 = 6, 4.
This vector is shown in the following figure:
y
v+w
v
w
x
The vector 2v w is
2v w = 22, 3 4, 1 = 4, 6 4, 1 = 0, 5
It is shown next
A Solution Manual for:
A First Course In Probability
by Sheldon M. Ross.
John L. Weatherwax
December 3, 2012
Introduction
Here youll find some notes that I wrote up as I worked through this excellent book. Ive
worked hard to make these notes as good as I
8 4 + 2 = 6 objects to place. This can be done in 6! way. We can permute the husband
and wife in each pair in 2 2 = 22 ways. Thus we find
P (Ai Aj ) =
22 6!
.
8!
In general, following the same logic we have for r couples
P (Ai Aj Ak ) =
2r (8 r)!
.
8!
Now
Vectors in the Plane (LT SECTION 13.1)
S E C T I O N 12.1
y
(3, 8)
Q
(1, 8)
B, P
(5, 8)
A
x
In Exercises 2932, are AB and P Q parallel? And if so, do they point in the same direction?
29. A = (1, 1),
B = (3, 4), P = (1, 1), Q = (7, 10)
solution We compute
Problem 19 (counting committees with constraints)
6
, but since two men wont serve together
Part (a): We select three men from six in
3
we need to compute the number of these pairings of three men that have the two that wont
serve together. The number of
Part (c): We can have no sophomores and four junior or one sophomore and three juniors
or two sophomores and two juniors or three sophomores and one juniors or four sophomores
and zero juniors. So our probability is given by
4
4
4
4
4
4
4
4
4
4
+
+
+
+
N k 1 middle seat choices.
If we let p(n, 1) be the probability that given one crazy passenger and n total seats to select
from that the last passenger sits in his seat. From the argument above we have a recursive
structure give by
p(N, 1) =
N 1
1
1 X
1
p
332
C H A P T E R 12
VECTOR GEOMETRY
(LT CHAPTER 13)
2. Let v = P0 Q0 , where P0 = (1, 2, 5) and Q0 = (0, 1, 4). Which of the following vectors (with tail P and head
Q) are equivalent to v?
v1
v2
v3
v4
P
(1, 2, 4)
(1, 5, 4)
(0, 0, 0)
(2, 4, 5)
Q
(0, 5, 5)
314
C H A P T E R 12
VECTOR GEOMETRY
(LT CHAPTER 13)
y
Q
a
a0 P
0
x
4. Let v = P Q, where P = (1, 1) and Q = (2, 2). What is the head of the vector v equivalent to v based at (2, 4)?
What is the head of the vector v0 equivalent to v based at the origin? S
328
C H A P T E R 12
VECTOR GEOMETRY
(LT CHAPTER 13)
By the parallelogram law we have
r = OA + AP
(1)
We find the vectors OA and AP :
The vector OA has length L1 and it makes an angle of 90 1 with the x-axis.
The vector AP has length L2 and it makes a
latter number is given by 52!. To compute the number of card ordering that given rise to
event F consider that in selecting the first card we can select any card that is not an ace
and thus have 52 4 = 48 cards to select from. To select the second card we
Problem 36 (choosing two identical cards)
52
2
possible ways to draw two cards from the 52 total. For us to
4
ways. Thus our probability is given by
draw two aces, this can be done in
2
4
2
= 0.00452 .
52
2
Part (a): We have
Part (b): For the two car