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NEW QUESTION: 1
Refer to the exhibit. Which two statements are true about interVLAN routing in the topology that is shown in the exhibit? (Choose two.)

A. Router1 needs more LAN interfaces to accommodate the VLANs that are shown in the exhibit.
B. Router1 and Switch2 should be connected via a crossover cable.
C. The FastEthernet 0/0 interface on Router1 and the FastEthernet 0/1 interface on Switch2 trunk ports must be configured using the same encapsulation type.
D. The FastEthernet 0/0 interface on Router1 must be configured with subinterfaces.
E. Router1 will not play a role in communications between host A and host D.
F. Host E and host F use the same IP gateway address.
Answer: C,D
Explanation:
http://www.cisco.com/en/US/tech/tk389/tk815/technologies_configuration_example09186a00800
949fd.shtml

NEW QUESTION: 2
A company has gigabytes of web log files stored in an Amazon S3 bucket. A Solutions Architect wants to copy those files into Amazon Redshift for analysis. The company's security policy mandates that data is encrypted at rest both in the Amazon Redshift cluster and the Amazon S3 bucket.
Which process will fulfill the security requirements?
A. Enable server-side encryption on the Amazon S3 bucket. Launch an encrypted Amazon Redshift cluster. Copy the data into the Amazon Redshift cluster.
B. Enable server-side encryption on the Amazon S3 bucket. Launch an unencrypted Amazon Redshift cluster. Copy the data into the Amazon Redshift cluster.
C. Launch an encrypted Amazon Redshift cluster. Copy the data from the Amazon S3 bucket into the Amazon Redshift cluster. Copy data back to the Amazon S3 bucket in encrypted form.
D. Enable server-side encryption on the Amazon S3 bucket. Copy data from the Amazon S3 bucket into an unencrypted Redshift cluster. Enable encryption on the cluster.
Answer: A
Explanation:
Explanation/Reference:
Reference: https://aws.amazon.com/blogs/big-data/encrypt-your-amazon-redshift-loads-with-amazon-s3- and-aws-kms/

NEW QUESTION: 3
Which statement is NOT correct about the Fitted Line Plot shown here?

A. A reactant value between 6 and 8 units yields around 40 to 60
B. When the reactant increases, the expected yield would increase
C. The independent variable is the reactant
D. If the reactant was 10 units, with 95% confidence we would expect a minimum yield of 148 units
E. With at least 95% confidence, we can expect less than 10 units of Yield when the reactant is at a value of 1
Answer: A

NEW QUESTION: 4
Which single Cisco IOS ACL entry permits IP addresses from 172.16.80.0 to 172.16.87.255?
A. permit 172.16.80.0 0.0.248.255
B. permit 172.16.80.0 255.255.248.0
C. permit 176.16.80.0 255.255.252.0
D. permit 172.16.80.0 255.255.240.0
E. permit 172.16.80.0 0.0.3.255
F. permit 172.16.80.0 0.0.7.255
Answer: F
Explanation:
www.cisco.com/en/US/products/sw/secursw/ps1018/products_tech_note09186a00800a5b 9a.shtml ACL Summarization Note: Subnet masks can also be represented as a fixed length notation. For example, 192.168.10.0/24 represents 192.168.10.0 255.255.255.0. This list describes how to summarize a range of networks into a single network for ACL optimization. Consider these networks. 192.168.32.0/24 192.168.33.0/24 192.168.34.0/24 192.168.35.0/24 192.168.36.0/24 192.168.37.0/24 192.168.38.0/24 192.168.39.0/24
The first two octets and the last octet are the same for each network. This table is an explanation of how to summarize these into a single network.
The third octet for the previous networks can be written as seen in this table, according to the octet bit position and address value for each bit.
Decimal 128 64 32 16 8 4 2 1 32 0 0 1 0 0 0 0 0 33 0 0 1 0 0 0 0 1 34 0 0 1 0 0 0 1 0 35 0 0 1 0 0 0 1 1 36 0 0 1 0 0 1 0 0 37 0 0 1 0 0 1 0 1 38 0 0 1 0 0 1 1 0 39 0 0 1 0 0 1 1 1 M M M M M D D D
Since the first five bits match, the previous eight networks can be summarized into one network (192.168.32.0/21 or 192.168.32.0 255.255.248.0). All eight possible combinations of the three low-order bits are relevant for the network ranges in question. This command defines an ACL that permits this network. If you subtract 255.255.248.0 (normal mask) from 255.255.255.255, it yields 0.0.7.255. access-list acl_permit permit ip 192.168.32.0 0.0.7.255

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