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NEW QUESTION: 1
A security analyst is investigating a malware infection that occurred on a Windows system. The system was not connected to a network and had no wireless capability Company policy prohibits using portable media or mobile storage The security analyst is trying to determine which user caused the malware to get onto the system Which of the following registry keys would MOST likely have this information?
A)

B)

C)

D)

A. Option B
B. Option D
C. Option A
D. Option C
Answer: D

NEW QUESTION: 2
An application stores data in an Amazon RDS MySQL DB instance. The database traffic primarily consists of read queries, which are overwhelming the current database. A Solutions Architect wants to scale the database.
What combination of steps will achieve the goal? (Choose two.)
A. Create an ELB Application Load Balancer
B. Create read replicas in different Availability Zones
C. Migrate the MySQL database to a PostgreSQL database
D. Migrate the MySQL database to Amazon Aurora
E. Add the MySQL database instances to an Auto Scaling group
Answer: A,B

NEW QUESTION: 3



R1# show running-config
interface Loopback0
description ***Loopback***
ip address 192.168.1.1 255.255.255.255
ip ospf 1 area 0
!
interface Ethernet0/0
description **Connected to R1-LAN**
ip address 10.10.110.1 255.255.255.0
ip ospf 1 area 0
!
interface Ethernet0/1
description **Connected to L2SW**
ip address 10.10.230.1 255.255.255.0
ip ospf hello-interval 25
ip ospf 1 area 0
!
router ospf 1
log-adjacency-changes

R2# show running-config
R2
!
interface Loopback0
description **Loopback**
ip address 192.168.2.2 255.255.255.255
ip ospf 2 area 0
!
interface Ethernet0/0
description **Connected to R2-LAN**
ip address 10.10.120.1 255.255.255.0
ip ospf 2 area 0
!
interface Ethernet0/1
description **Connected to L2SW**
ip address 10.10.230.2 255.255.255.0
ip ospf 2 area 0
!
router ospf 2
log-adjacency-changes

R3# show running-config
R3
username R6 password CISCO36
!
interface Loopback0
description **Loopback**
ip address 192.168.3.3 255.255.255.255
ip ospf 3 area 0
!
interface Ethernet0/0
description **Connected to L2SW**
ip address 10.10.230.3 255.255.255.0
ip ospf 3 area 0
!
interface Serial1/0
description **Connected to R4-Branch1 office**
ip address 10.10.240.1 255.255.255.252
encapsulation ppp
ip ospf 3 area 0
!
interface Serial1/1
description **Connected to R5-Branch2 office**
ip address 10.10.240.5 255.255.255.252
encapsulation ppp
ip ospf hello-interval 50
ip ospf 3 area 0
!
interface Serial1/2
description **Connected to R6-Branch3 office**
ip address 10.10.240.9 255.255.255.252
encapsulation ppp
ip ospf 3 area 0
ppp authentication chap
!
router ospf 3
router-id 192.168.3.3
!

R4# show running-config
R4
!
interface Loopback0
description **Loopback**
ip address 192.168.4.4 255.255.255.255
ip ospf 4 area 2
!
interface Ethernet0/0
ip address 172.16.113.1 255.255.255.0
ip ospf 4 area 2
!
interface Serial1/0
description **Connected to R3-Main Branch office**
ip address 10.10.240.2 255.255.255.252
encapsulation ppp
ip ospf 4 area 2
!
router ospf 4
log-adjacency-changes

R5# show running-config
R5
!
interface Loopback0
description **Loopback**
ip address 192.168.5.5 255.255.255.255
ip ospf 5 area 0
!
interface Ethernet0/0
ip address 172.16.114.1 255.255.255.0
ip ospf 5 area 0
!
interface Serial1/0
description **Connected to R3-Main Branch office**
ip address 10.10.240.6 255.255.255.252
encapsulation ppp
ip ospf 5 area 0
!
router ospf 5
log-adjacency-changes

R6# show running-config
R6
username R3 password CISCO36
!
interface Loopback0
description **Loopback**
ip address 192.168.6.6 255.255.255.255
ip ospf 6 area 0
!
interface Ethernet0/0
ip address 172.16.115.1 255.255.255.0
ip ospf 6 area 0
!
interface Serial1/0
description **Connected to R3-Main Branch office**
ip address 10.10.240.10 255.255.255.252
encapsulation ppp
ip ospf 6 area 0
ppp authentication chap
!
router ospf 6
router-id 192.168.3.3
!


A. Option B
B. Option C
C. Option D
D. Option A
Answer: A

NEW QUESTION: 4
Which of the following is NOT a property of a public key cryptosystem?
(Let P represent the private key, Q represent the public key and M the
plaintext message.)
A. P[Q(M)] = M
B. P and Q are difficult to generate from a particular key value.
C. Q[P(M)] = M
D. It is computationally infeasible to derive P from Q.
Answer: B
Explanation:
The answer refers to the initial computation wherein the private and public keys are computed. The computation in this direction is relatively straightforward. Answers "Q[P(M)] = M" and "P[Q(M)] = M" state the true property of public key cryptography which is that a plaintext message encrypted with the private key can be decrypted by the public key and vice versa. Answer "It is computationally infeasible to derive P from Q" states that it is computationally infeasible to derive the private key from the public key. Obviously, this is a critical property of public key cryptography.