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Nataliya [291]
4 years ago
11

When creating a visual basic desktop application, the last thing you will do is ________?

Computers and Technology
1 answer:
fiasKO [112]4 years ago
7 0
The last thing you will do is correct runtime errors.
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Select the correct answer.
aliina [53]

Answer:

B

Explanation:

engineers design and carry out test cases and evaluate exit criteria (by following the scope set in the planning phase). They create bug reports describing detected defects and report to the stakeholders on the test findings and the completion status. Testing may be repeated to check for errors.

7 0
3 years ago
The picture that graphically represents the items you use in Windows is called a/an
Novosadov [1.4K]
The answer to this is Bulls icon
6 0
4 years ago
Computer Networks - Queues
lyudmila [28]

Answer:

the average arrival rate \lambda in units of packets/second is 15.24 kbps

the average number of packets w waiting to be serviced in the buffer is 762 bits

Explanation:

Given that:

A single channel with a capacity of 64 kbps.

Average packet waiting time T_w in the buffer = 0.05 second

Average number of packets in residence = 1 packet

Average packet length r = 1000 bits

What are the average arrival rate \lambda in units of packets/second and the average number of packets w waiting to be serviced in the buffer?

The Conservation of Time and Messages ;

E(R) = E(W) + ρ

r = w + ρ

Using Little law ;

r = λ × T_r

w =  λ × T_w

r /  λ = w / λ  +  ρ / λ

T_r =T_w + 1 / μ

T_r = T_w +T_s

where ;

ρ = utilisation fraction of time facility

r = mean number of item in the system waiting to be served

w = mean number of packet waiting to be served

λ = mean number of arrival per second

T_r =mean time an item spent in the system

T_w = mean waiting time

μ = traffic intensity

T_s = mean service time for each arrival

the average arrival rate \lambda in units of packets/second; we have the following.

First let's determine the serving time T_s

the serving time T_s  = \dfrac{1000}{64*1000}

= 0.015625

now; the mean time an item spent in the system T_r = T_w +T_s

where;

T_w = 0.05    (i.e the average packet waiting time)

T_s = 0.015625

T_r =  0.05 + 0.015625

T_r =  0.065625

However; the  mean number of arrival per second λ is;

r = λ × T_r

λ = r /  T_r

λ = 1000 / 0.065625

λ = 15238.09524 bps

λ ≅ 15.24 kbps

Thus;  the average arrival rate \lambda in units of packets/second is 15.24 kbps

b) Determine the average number of packets w waiting to be serviced in the buffer.

mean number of packets  w waiting to be served is calculated using the formula

w =  λ × T_w

where;

T_w = 0.05

w = 15238.09524 × 0.05

w = 761.904762

w ≅ 762 bits

Thus; the average number of packets w waiting to be serviced in the buffer is 762 bits

4 0
3 years ago
Why were video games invented?
Alenkinab [10]
The answer is in the following website: https://www.reference.com/history/were-video-games-invented-e9413d3dc1378766


4 0
3 years ago
The advantages of the DecimalFormat class compared with the NumberFormat class include Group of answer choices
Anon25 [30]

Answer:5. only A and B(precise control over the number of digits to be displayed and control over the presence of a leading zero)

Explanation: DECIMAL FORMAT CLASS is a subclass of Number format class specifically made to format parsing decimal Numbers. It is capable of formating both ARABIC, WESTERN AND INDIC DIGITS. It has a unique Attribute of control over the number of digits to be displayed and control over the presence of a leading zero.

NUMBER FORMAT CLASS is the class for formatting parsing numbers it helps to format According to Specific locale.

3 0
3 years ago
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