...Case Study 2 Ricky Smith Liberty University Over the years there has always been a question of what is the most effective way to teach and learn reading and literature skills. According to Vacca et al. (2015) basal reading programs are the most popular materials used for reading instruction in the country (p. 372). In literature based programs the teacher uses model language for the students. “Literature based instruction approaches accommodate individual student differences in reading abilities and at the same time focus on meaning, interest, and enjoyment” (Vacca et al., 2015, p. 46). I believe that there are different methods of research that I can do to evaluate the effectiveness of a basal reading program and a literature-based program....
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...1. The machine that wants to send out the packet will know the IP address of the destination. It checks the routing table it has to find out which router within its network can send the packet out on its behalf. 2. Based on the IP address of the router, the machine chooses the right network interface card (NIC) that is able to communicate with that router. It then sends the packet to the that NIC, after adding the Layer 2 protocol data to the packet (which should have the MAC address of the router). 3. The NIC should be connected to a switch and sends out the packet to the switch using the Layer 1 protocol to convert the packet into electrical signals that can be sent over the wire. 4. The switch reads the MAC address of the router...
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...Assume that the charts are named as below for the continence to explain. A- First chart which micronaire is 4.28 B- Second chart which micronaire is 4.28 C- First chart which micronaire is 3.38 D- Second chart which micronaire is 3.38 Theta or the degree of thickening of cellulose in the secondary cell wall (thickness of the secondary cell wall) is a measurement of fiber maturity. The perimeter determine the size of the fiber. A fiber with higher perimeter is a coarse fiber while a fiber with less perimeter is a finer fiber. The curve of chart A has skewed to the right side. When we consider the percent of cotton fibers versus theta, it indicates that more percent of cotton fibers have 0.6-0.8 as their theta. Hence more percent of cotton fibers have better thickening of secondary cell walls which means there are more mature fibers in this sample. Considering the percentage of cotton fibers versus the perimeter, more percent of cotton fibers have lower perimeter. So the bivariate distribution of this sample of cotton indicates that the sample contains more mature and finer fibers. In spinning, this sample perform well and there is a better dye uptake. The curve of chart B has skewed to left which indicates that more percent of fibers have less theta value. So a large percent of fibers may have less degree of thickening in their secondary cell wall/ less quantity of cellulose. Therefore the percent of immature fiber is higher than the percent of mature fibers. But more...
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...Action we take in order to reduce the Collision domain In this case, we are segmenting the LAN to share the network resources. Basically, collide occurs when more than one device tries to send a packets on network at the same time. Company want to reduce the collision. There are fallowing certain steps as given below: SWITCH: switch reduces or even eliminates, significantly the number of collisions of LAN. Switches do not create a single shared bus. Switches interpret the bits in receiving frame and send that frame out the one required port, rather than all ports. Switches with one device cabled to each port of the switch allow using of full duplex operation. It means that double the bandwidth of 100 to 200mps. Bridges: A bridges is used to connect multiple LAN segments together, forming multiple collision domains. A collision happens on Ethernet network when multiple devices by separating two or more segments into multiple domain collision domains. Bridges reduces the number of collision that could occur on a LAN segments....
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...Description:- Task 1:- a. The detailed design of a simple addressing solution (with diagram(s) and step by step workout of the calculations) that leaves room for growth is easy to administer and has the minimum wastage of available IP address space. Answer:- NAT (Network address translation) is a process by which a router can provide IP address to the computers. NAT is a process by which it allocates the IP address to a system, after remapping it with the network in a private area, so that one IP address could be allocated to one system, and the same IP address range will be used to connect all the computer system in one location. In this case study one building will have one router; the router is a networking device which provides IP address to its connected device in order to connect the...
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...1. Switching infrastructure congestion, corruption, or collisions. a. Congestion Congestion cause network slow down where the ports that are oversubscribed. Oversubscribed port is where the port on the switches is attempting to deliver data at a higher rate than the port bandwidth can support or allow. When oversubscribed port occurs, some of the packets must end up being dropped. Those client packets are being dropped will experience a slow-down in their file transfers or in the responsiveness of the network-based applications they are running. Overall network performance congestion could cause even though by subscribed high bandwidth ISP to internet access but hardware cannot utilizes the bandwidth ISP provide. b. Corruption If packets is corrupted by faulty cabling, electrical interference, or switch hardware faults then the corrupted packets will be dropped by the receiving switch. If corrupt packets at high rate it will cause a slow-down in network performance. Because servers/hosts require to resend the dropped packets again to the clients. If this issues did not fix it will cause the network traffic busy because the same data is resend again. It will occupied the network bandwidth cause another client on the network experience slow network. c. Collisions...
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...value. One of the problems that will occur is the collision. There are two or more text that produce the same hash value. In this case the MD5 inputs along regardless, will produce along the 128-bit hash value. That means very many input possibilities, infinite, but probably only a hash value of 2^ 128. There will be two or more input text that has the same hash value. It turns have MD5 weaknessesthat allow searched two files that have the same hash value with a short time. While the workings of the SHA-1 message given extra to make the length in multiples of 512 bits (512 lx). The number of bits is the origin of k bits. Add just enough bits to 64-bits lack of a multiple of 512 (512-64=448), which is also called congruent with 448 (mod 512). Then add 64 bit long message stating. Initiation 5 MD variable length of 32 bits, namely a, b, c, d, e. Message is divided into blocks of 512 bits, and each block od processed. Then the outputs of each block are combined with the output of the next block, thus obtained output(diggest). 7....
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...Tichaona Zvidzayi; 613z0022 Question 1 a) How SSL(Secure Socket Layer) works The reasons why SSL exists are encryption and identification (Vaughan-Nichols, 2013). Encryption hides the information moving between two computers and identification makes sure that you are talking to the right person. If the server cannot verify itself, the client (browser/server) displays a security warning because someone might be pretending to be a legit website for personal gains. Vaughan-Nichols, 2013 SSL stages: The client initiates a connection with the web-server and prompts the web-server to verify if it is a legit website. The website will then attempt to verify itself by sending a copy of SSL certificate. The client will then confirm if it can trust the website or not,...
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...difference in the proportion between the similar and dissimilar pair groups. Both dark rearing and binocular deprivation totally prevented the increase in the proportion of synchronous pairs (Figure 5C). The CCG peak area in synchronous pairs was not significantly different between the similar and dissimilar pair groups at P13-15 (Figure 5D). During normal development, the CCG peak area in synchronous pairs that showed similar, but not dissimilar, stimulus-feature preference increased significantly (Dunn’s test, P < 0.05 for the similar pair group). As a result, the CCG peak area in the similar pair group was far larger than that in the dissimilar pair group at P24-28. Thus, feature-selective synchronization seems to develop remarkably during 1-2 weeks after eye opening. The developmental increase in the strength of synchronization, which took place selectively in the similar pair group, was not observed after dark rearing or binocular deprivation (Figure 5D). The CCG peak area in the similar pair group showed no increase after dark rearing and an increase only at a trend level after binocular deprivation (Dunn’s test, P > 0.05), compared with that at P13-15. No significant difference was found in the CCG peak area between the similar and dissimilar pair groups after either type of deprivation. These observations indicate that, in the upper layer, strong synchronization is established selectively in neuron pairs sharing a similar preference for visual stimulus-features only when raised...
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...The plan would include steps necessary to mitigate such attacks and restore full service as soon as possible. Steps for disaster recovery should be included along with business continuity. These would include such items as a complete set of backup files stored elsewhere, in case of physical damage to the site; from a tornado, for example. Acceptable risk and loss would also be discussed. In the tornado example, it is possible that a certain amount of hardware (if not all of it) could be destroyed in the storm. Will backup equipment be available? Will new equipment be purchased instead to cover losses? If so, there should be an awareness that the new equipment will not be immediately available, and also will not be immediately usable once it arrives (it needs to be configured and attached to the network) if backup files are...
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...1) Inter-process communication is the activity of data sharing across multiple and very often specialized processes using communication protocols. In this process, usually clients and servers are being involved, where the client requests the data and the server responds to the requests. 2) The maximum size of a UDP datagram is theoretically speaking 65,535 bytes (8 byte header and 65,527 byte data). Using packet-based protocol such as UDP is advantageous when error checking and correction is not necessary or is performed in the application layer. Time-sensitive applications such as VoIP or online games usually use UDP because dropping packets is prefered over waiting for delayed packets. Using stream protocol has the advantage of being reliable and provides ordered and error-checked delivery of stream of octets. 3) TCP makes sure that data reaches its destination, that it reaches it without any duplication and that it reaches it in time. In TCP, all the work is done by the operating system, even the debugging. It automatically breaks up data into packets. Good failure recovery TCP doesn’t interrupt existing services when it adds networks High error-rate handling Platform independence Low data overhead 4) htons() host to...
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...Mq, which receives the query, transmits a query according to the following steps: 1. If Mq receives the query for the first time then 2. . Store Query path and hop count as its Parent Query path 3. Store the nodeID at the end of Query path as its parent 4. Set RD for replying data items 5. Add Mq’s nodeID to the end of Query path and send the query to neighbor nodes 6. Else 7. Store Query path and hop counts as its Neighbor Query path 8. Store the nodeID at the end of Query path as its neighbor 9. end if Here, hop count denotes the number of hops to the query-issuing node. Then, Mq sets a waiting time for reply (RD). As hop count increases, RD decreases. When Mq receives the query later again, it stores the ID of the query sender node as its neighbor node, as well as, the Query path and the number of hops. When its RD has passed, each node sends back a reply message, which includes its own node identifier (Sender nodeID), the identifier of the next node along the reply route (Dest nodeID), an encrypted id (Enc ID), a list of the data items (including their scores) and the node identifiers of the nodes possessing them (Data list), and a list summarizing the reply message routes, i.e., a list of the pairs of sender and next node identifiers (Forwarding Route). Following are the steps to send the reply to a message. The reply will be send after the RD has elapsed. 1. Select the Neighbor with minimum hopCount as the DestNode 2. If more than one node...
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...Written assignment Unit 7 Course: Epidemiology Instructor: Donielle Hyde Date of submission: 20-03-2024. 1. What is the difference between a. and a. Why is it important to find people that have high cholesterol in the community? The identification of high-cholesterol persons within the population is critical to early diagnosis and avoidance of cardiovascular illnesses. The chance of having major illnesses like a heart attack or stroke is greatly increased by high cholesterol. Through screening for elevated cholesterol levels, medical professionals can promptly initiate measures, such as drug therapy and lifestyle adjustments, to reduce this risk and enhance cardiovascular well-being. When a patient knows that with time their cholesterol...
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... Figure 2 Each node pair performs a WSA handshake to select a TxSlot of a SCH . Nodes only switch to the selected SCH during the selected TxSlot in order to broadcast their own EMG packets and avoid missing the EMG packets on the CCH in another time. Nodes maintain the Neighbor Information List (NIL) and Channel Usage List (CUL). The NIL stores the SCH and TxSlot used by the neighbors while the CUL shows the available TxSlots of each SCH. Based on the NIL, a node knows when its neighbor is on the CCH in order to perform a WSA handshake. The CUL is used to select the common TxSlot during the WSA negotiation. 7.OPERATION OF MAC If an EMG packet arrives at the MAC layer, nodes try to broadcast it on the CCH in the current CCHI (or SCHI) and then rebroadcast it in the next SCHI (or Figure 3 NIL CCHI). When a node has service packets to exchange, it sends the WSA including its CUL. Upon receiving the WSA, the receiver selects the common TxSlot and SCH based on the CULs of both sender and receiver. Then, the receiver sends the ACK indicating the selected [TxSlot ,SCH] to the sender. The sender sends the RES (Reservation) to confirm the [TxSlot ,SCH] selected by the receiver. Both sender and receiver switch to the selected SCH in the selected TxSlot to exchange their service packets. Neighbor nodes, which overhear the ACK or RES messages, update their NILs and CULs as shown in Figure 3, Figure 4. ...
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...the equals sign should be placed inside the missing number’s box. For this student’s remediation I would teach them the equality principle and show the student that they have to have the same amount of lines on each side of the equals sign. For student B, they are having difficulty with simple addition. For this student, I would recommend that the teacher work with the student on simple one-digit addition, and then trying to rework the problems. 7. • add two plus three in the ones place, write five below the line in the ones place • add six plus eight in the tens place, write the ones (four) below the line in the tens place and regroup the one in the tens place to above the hundreds place in the problem • add the regrouping number (1) to the one and two in the hundreds place, and write the number four in the hundreds place below the line • The solution to this problem is 445. 8. Based on the three lists provided in the text book, sets a and c are unacceptable lists. The reason that set a is an unacceptable list, is that it includes subtraction problems, and addition problems that are adding three-digit numbers to two-digit numbers. Also on this list are problem that have more regrouping than what the student know. To fix this set, the teacher needs to get rid of the subtraction problems, and the addition problems that do not just have regrouping in the ones to the tens place. The reason the set c is unacceptable, is that it also has subtraction problems, but it also...
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