Refer to the exhibit.

Which type of antenna does the radiation pattern represent?

• A.omnidirectional
• B.directional patch
• C.multidirectional
• D.Yagi

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Which method displays text directly into the active console with a synchronous EEM applet policy?

• A.event manager applet boom
  event syslog pattern 'UP'
  action 1.0 gets 'logging directly to console'
• B.event manager applet boom
  event syslog pattern 'UP'
  action 1.0 syslog priority direct msg 'log directly to console'
• C.event manager applet boom
  event syslog pattern 'UP'
  action 1.0 puts 'logging directly to console'
• D.event manager applet boom
  event syslog pattern 'UP'
  action 1.0 string 'logging directly to console'

Correct Answer: B

Explanation
To enable the action of printing data directly to the local tty when an Embedded Event Manager (EEM) applet is triggered, use the action puts command in applet configuration mode.
The following example shows how to print data directly to the local tty:

The action puts command applies to synchronous events. The output of this command for a synchronous applet is directly displayed to the tty, bypassing the syslog.
Reference: https://www.cisco.com/c/en/us/td/docs/ios-xml/ios/eem/command/eem-cr-book/eemcr- a1.html

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Drag and drop the characteristics from the left onto the routing protocols they describe on the right.

Correct Answer:

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Which configuration restricts the amount of SSH that a router accepts 100 kbps?
A)

B)

C)

D)

• A.Option D
• B.Option B
• C.Option A
• D.Option C

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Which standard access control entry permits from odd-numbered hosts in the 10.0.0.0/24 subnet?

• A.Permit 10.0.0.0.0.0.0.1
• B.Permit 10.0.0.1.0.0.0.0
• C.Permit 10.0.0.1.0.0.0.254
• D.Permit 10.0.0.0.255.255.255.254

Correct Answer: C

Remember, for the wildcard mask, 1's are I DON'T CARE, and 0's are I CARE. So now let's analyze a simple ACL:
access-list 1 permit 172.23.16.0 0.0.15.255
Two first octets are all 0's meaning that we care about the network 172.23.x.x. The third octet of the wildcard mask, 15 (0000 1111 in binary), means that we care about first 4 bits but don't care about last 4 bits so we allow the third octet in the form of 0001xxxx (minimum:00010000 = 16; maximum: 0001111 = 31).

The fourth octet is 255 (all 1 bits) that means I don't care.
Therefore network 172.23.16.0 0.0.15.255 ranges from 172.23.16.0 to 172.23.31.255.
Now let's consider the wildcard mask of 0.0.0.254 (four octet: 254 = 1111 1110) which means we only care the last bit. Therefore if the last bit of the IP address is a "1" (0000 0001) then only odd numbers are allowed. If the last bit of the IP address is a "0" (0000 0000) then only even numbers are allowed.
Note: In binary, odd numbers are always end with a "1" while even numbers are always end with a "0".
Therefore in this question, only the statement "permit 10.0.0.1 0.0.0.254" will allow all oddnumbered hosts in the 10.0.0.0/24 subnet.

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