Continued from previous. Suppose that the number of jobs in a processor may be modeled by a Bernoulli Single-Server Queuing Process with a frame size of 3 seconds. The average arrival rate equals 4 per minute. The average time to process a single job (once it's in the system) equals 10 seconds. Whenever there is at least one job in the system, but it is not at full capacity, what is the probability that the number of jobs in the system will increase by one from this frame to the next? Whenever there is at least one job in the system, but it is not at full capacity, what is the probability that the number of jobs in the system will decrease by one from this frame to the next? Whenever there is at least one job in the system, but it is not at full capacity, what is the probability that the number of jobs in the system will stays the same from this frame to the next?

step-by-step explanation:

if you want to solve percent error follow these steps.

1. subtract the accepted value from the experimental value.

2. take the absolute value of step 1.

3. divide that answer by the accepted value.

4. multiply that answer by 100 and add the % symbol to express the answer as a percentage.

if you want to solve percent of change just do this.

1. work out the difference (increase) between the two numbers you are comparing.

2. divide the increase by the original number and multiply the answer by 100. if your answer is a negative number then this is a percentage decrease.

hope this : )

41*67/2=1373.5

step-by-step explanation:

i just used a calculator

step-by-step explanation:

given that there are two triangles j'k'l' and jkl>

the former is got by dilating the second one.

when drawn on a grid, (graph paper) coordinates of vertices of first triangle is

(6,12)j, (9,6)k and (12,9)l

side jk has length = distance between j and k

=

side j'k'=distance between j' and k'

=

ie j'k'l' has sides 1/3 of that of corresponding sides of jkl.

hence scale factor of 1/3 is used to get j'k'l'from jkl

(-5,-28)

i graphed to on desmos.