Answer:
b. the same data type
Explanation:
Any number of variables can be declared in a statement as long as the variables have the same data type. For example:
1) int a,b,c,d,e;
Here each of the declared variables a,b,c,d,e have the type int.
2) char p,q,r,s,t,u,v,w;
In this case variables p to w all have the type char.
3) float x,y,z;
x,y and z are all of type float.
Answer:
0.8488
Explanation:
Let E =error found by test 1
Let F=error found by test 2
Let G=error found by test 3
Let H=error found by test 4
Let I= error found by test 5
Given P(E)=0.1, P(F)=0.2, P(G)=0.3, P (H)= 0.4, P (I)=0.5
therefore P(notE)=0.9, P(notF)=0.8, P(notG)=0.7, P(not H)=0.6, P (notI)=0.5
Tests are independent P(not E & not F ¬ G & not H & not I=P(notE)*P(notF)*P(notG)*P (notH)*P (not I) =0.9*0.8*0.7*0.6*0.5 =0.1512
P(found by at least one test)= 1- P(not found by any test)=1-P(not E& not F & not G & not H & not I ) = 1-0.1512 = 0.8488
Answer:
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Over a TCP connection, suppose host A sends two segments to host B, host B sends an acknowledgement for each segment, the first acknowledgement is lost, but the second acknowledgement arrives before the timer for the first segment expires is True.
True
<u>Explanation:</u>
In network packet loss is considered as connectivity loss. In this scenario host A send two segment to host B and acknowledgement from host B Is awaiting at host A.
Since first acknowledgement is lost it is marked as packet lost. Since in network packet waiting for acknowledgement is keep continues process and waiting or trying to accept acknowledgement for certain period of time, once period limits cross then it is declared as packet loss.
Meanwhile second comes acknowledged is success. For end user assumes second segments comes first before first segment. But any how first segment expires.
Answer:
If your using java, then its supposed to be "System.out.print("a")"
Explanation:
its supposed to have quotations
