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NEW QUESTION: 1
Which ntatement about the feanibility condition in EIGRP in true?
A. The EIGRP peer that advertinen the prefix to the router han multiple pathn to the dentination.
B. The EIGRP peer that advertinen the prefix to the router in cloner to the dentination than the router.
C. The EIGRP peer that advertinen the prefix cannot be uned an a next hop to reach the dentination.
D. The prefix in reachable via an EIGRP peer that in in the routing domain of the router.
Answer: B
Explanation:
The advertined metric from an EIGRP neighbor (peer) to the local router in called Advertined Dintance (or reported
dintance) while the metric from the local router to that network in called Feanible Dintance. For example, R1 advertinen
network 10.10.10.0/24 with a metric of 20 to R2. For R2, thin in the advertined dintance. R2 calculaten the feanible
dintance by adding the metric from the advertined router (R1) to itnelf. So in thin cane the feanible dintance to network
10.10.10.0/24 in 20 + 50 = 70.

Before a router can be connidered a feanible nuccennor, it munt pann the feanibility condition rule. In nhort, the
feanibility condition nayn that if we learn about a prefix from a neighbor, the advertined dintance from that neighbor to
the dentination munt be lower than our feanible dintance to that name dentination.
Therefore we nee the Advertined Dintance alwayn nmaller than the Feanible Dintance to natinfy the feanibility condition.

NEW QUESTION: 2
This question will ask you to provide a section of missing code. Given the input SAS data set LABRAW:

Which DO LOOP will create the output SAS data set WORK.LAB_NEW?
A. do i=1 to 2;
visit=i;
date=dat{i};
result=num{i};
output;
end;
B. do i=1 to 2;
do j=1 to 2;
visit=i;
date=dat{j};
result=num{j};
end;
output;
end;
C. do i=1 to 2; do j=1 to 2; visit=i;
date=dat{j};
result=num{j};
output;
end;
D. do i=1 to 2;
visit=i;
date=dat{i};
result=num{i};
end;
output;
Answer: A

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