Arcade games such as Donkey Kong, Mr. Dol, and Venture were adapted from arcade titles to home consoles by which company?

Write a machine code program for all the following problems. Test to make sure that they work in LC3. It's best to complete each

Usimov [2.4K]

Answer:

Following is given the solution to each part.

All the comments are given above each step so that they can make the logic clear.

I hope it will help you!

Explanation:

8 0

2 years ago

________ is the extent to which differences in observed scale scores reflect true differences among objects on the characteristi

AysviL [449]

Answer: Validity

Explanation:Validity is the type of claim that is made towards the measuring of reliability of any element. The differences appearing in the measurement of the elements displays the validly factor is not up to the mark which indirectly implies that reliability is not high. Validity is the situation in which all the parameters appearing in the element is perfect. If validity is perfect then only perfect reliability is achieved.

7 0

3 years ago

(5 pt.) The name of a variable in the C programming language is a string that can contain uppercase letters, lowercase letters,

lilavasa [31]

Answer:

The different variable in C is 21213316700.

Explanation:

Given value:

Total value = letters + underscore value

Total value = 52 + 1

Total value =53

choice for first character = 53 letters +10 digits

first character = 63

choice for remaining characters

So,

Variable number With one 1 character = 53

Variable number With 2 character = 53 × 63

Variable number With 3 character = 53 × 63²

Variable number With 4 character = 53 × 63³

.

Variable number With 7 character = $53 \times 63^{7}$

Total difference variable = 53 + 53 × 63+ 53 × 63²+ 53 × 63³+....+ $63^{7}$

Total difference variable = 53(1 + 63 + 63²+ 63³+ .... + $63^{7}$ )

Formula:

$1+x+x^2+x^3+x^4+........x^n\\\\\Rightarrow \frac{x^{(n+1)} -1}{x-1}$

Total difference variable

$53 \times \frac{63^{7+1} -1 }{63-1}\\\\53 \times \frac{63^{8} -1 }{63-1} \\\\ \therefore 63^8 = 2.48 \times 10^{14}\\\\53 \times \frac{2.48 \times 10^{14} -1 }{62}\\\\53 \times \frac{1.48 \times 10^{14}}{62}\\\\21213316700$

6 0

3 years ago

In the following four questions, we are sending a 30 Mbit MP3 file from a source host to a destination host. All links in the pa

pantera1 [17]

Answer:

1)we are sending a 30 Mbit MP3 file from a source host to a destination host. All links in the path between source and destination have a transmission rate of 10 Mbps. Assume that the propagation speed is 2 * 108 meters/sec, and the distance between source and destination is 10,000 km. Initially suppose there is only one link between source and destination. Also suppose that the entire MP3 file is sent as one packet. The TRANSMISSION DELAY is:

3 Seconds

2)

we are sending a 30 Mbit MP3 file from a source host to a destination host. All links in the path between source and destination have a transmission rate of 10 Mbps. Assume that the propagation speed is 2 * 108 meters/sec, and the distance between source and destination is 10,000 km. The END TO END DELAY(transmission delay plus propagation delay) is

3.05 seconds

3)

we are sending a 30 Mbit MP3 file from a source host to a destination host. All links in the path between source and destination have a transmission rate of 10 Mbps. Assume that the propagation speed is 2 * 108 meters/sec, and the distance between source and destination is 10,000 km. how many bits will the source have transmitted when the first bit arrives at the destination.

500,000 bits

4)

we are sending a 30 Mbit MP3 file from a source host to a destination host. All links in the path between source and destination have a transmission rate of 10 Mbps. Assume that the propagation speed is 2 * 108 meters/sec, and the distance between source and destination is 10,000 km. Now suppose there are two links between source and destination, with one router connecting the two links. Each link is 5,000 km long. Again suppose the MP3 file is sent as one packet. Suppose there is no congestion, so that the packet is transmitted onto the second link as soon as the router receives the entire packet. The end-to-end delay is

6.1 seconds

5)

we are sending a 30 Mbit MP3 file from a source host to a destination host. All links in the path between source and destination have a transmission rate of 10 Mbps. Assume that the propagation speed is 2 * 108 meters/sec, and the distance between source and destination is 10,000 km. Now suppose that the MP3 file is broken into 3 packets, each of 10 Mbits. Ignore headers that may be added to these packets. Also ignore router processing delays. Assuming store and forward packet switching at the router, the total delay is

4.05 seconds

6)

we are sending a 30 Mbit MP3 file from a source host to a destination host. All links in the path between source and destination have a transmission rate of 10 Mbps. Assume that the propagation speed is 2 * 108 meters/sec, and the distance between source and destination is 10,000 km. Now suppose there is only one link between source and destination, and there are 10 TDM channels in the link. The MP3 file is sent over one of the channels. The end-to-end delay is

30.05 seconds

7)

we are sending a 30 Mbit MP3 file from a source host to a destination host. All links in the path between source and destination have a transmission rate of 10 Mbps. Assume that the propagation speed is 2 * 108 meters/sec, and the distance between source and destination is 10,000 km. Now suppose there is only one link between source and destination, and there are 10 FDM channels in the link. The MP3 file is sent over one of the channels. The end-to-end delay is

30.05 seconds

6 0

3 years ago

Universal Containers has a custom picklist filed with three values on their products. The Admin would like to create a Dynamic B

Agata [3.3K]

Answer:

The Admin should Create a rule with three conditions, each that compares the configuration attribute filed with a static value.

8 0

2 years ago