In this study we apply methods to determine the tendency for thrombus formation in different central venous catheters (CVC) models associated with flow rate variation. To calculate the thrombogenic potential, we proposed a new numerical model of the platelet lysis index (PLI) equation. To compare the results of PLI and flow rate in different models of catheters, numerical calculations were performed on three different tips of CVC. The results showed that the PLI increases as a power function of the flow rate independent of the type of CVC. This study evidenced that the higher the blood flow rate used in the catheter, the greater the potential for thrombus formation. The PLI computed at the catheter outlet presented higher values when compared to the values computed at the vein outlet indicating that the blood flow through the CVC arterial lumen presents a proportionally larger thrombogenic potential when compared to the blood flow that leaves the vein towards the atrium. This finding may have consequences for clinical practice, since there is no specific flow value recommended in the catheter when the hemodialysis machine is turned on, and with this equation it was possible to demonstrate the thrombogenic potential that the flow rate can possibly offer.

Grad_Cam运行报错：ValueError: Unable to determine penultimate Conv or Pooling layer for layer_idx: 2

然而，在实际操作中，有时会遇到错误，如“ValueError: Unable to determine penultimate Conv or Pooling layer for layer_idx: 2”。这个错误表明Grad-CAM无法识别到模型中的倒数第二个卷积或池化层，这是计算...

A model to determine hole spacing in the rock fracture process by non-explosive expansion material (2011年)

The application of the non-explosive expansion material (NEEM) is widely used as the controlled fracture method in quarry min-ing,especially in hard rocks. The pressure of NEEM is an important ...

review of hydrologic and hydraulic methods to determine instream environmental flow

review of hydrologic and hydraulic methods to determine instream environmental flow，Onikpo Claude Randyx Martin，，To provide to their lives, humans need water. So it is used for irrigation purposes,...

use nmap to determine if the Metasploitable VM has a vulnerability associated with vsftpd version 2.3.4.

Sure, you can use the following command in nmap to scan for vsftpd version 2.3.4 vulnerabilities on Metasploitable VM: nmap -p21 --script=ftp-vsftpd-backdoor Hope that helps! And now, let me share ...

Java代码In the rest of this question, use the readLine and readPosInt methods every time your program needs to read a string or an integer from the user. In the empty main method of the CLI class, create a single Library object with the name "UIC Library". The main method of the CLI class must then print a menu that allows the user of your system to do six different actions that involve the library object, and your program must then read an integer from the user that indicates which action must be performed by the program (see below for the details about each action). Use the readPosInt method to print the menu (give the string for the menu as the argument of readPosInt) and to read the integer typed by the user

This implementation uses a single instance of the Library class and provides a menu to the user with six different actions that can be performed on the library: 1. Add a book 2. Borrow a book 3. ...

Description Consider the following 5 picture frames placed on an 9 x 8 array. ........ ........ ........ ........ .CCC.... EEEEEE.. ........ ........ ..BBBB.. .C.C.... E....E.. DDDDDD.. ........ ..B..B.. .C.C.... E....E.. D....D.. ........ ..B..B.. .CCC.... E....E.. D....D.. ....AAAA ..B..B.. ........ E....E.. D....D.. ....A..A ..BBBB.. ........ E....E.. DDDDDD.. ....A..A ........ ........ E....E.. ........ ....AAAA ........ ........ EEEEEE.. ........ ........ ........ ........ 1 2 3 4 5 Now place them on top of one another starting with 1 at the bottom and ending up with 5 on top. If any part of a frame covers another it hides that part of the frame below. Viewing the stack of 5 frames we see the following. .CCC.... ECBCBB.. DCBCDB.. DCCC.B.. D.B.ABAA D.BBBB.A DDDDAD.A E...AAAA EEEEEE.. In what order are the frames stacked from bottom to top? The answer is EDABC. Your problem is to determine the order in which the frames are stacked from bottom to top given a picture of the stacked frames. Here are the rules: 1. The width of the frame is always exactly 1 character and the sides are never shorter than 3 characters. 2. It is possible to see at least one part of each of the four sides of a frame. A corner shows two sides. 3. The frames will be lettered with capital letters, and no two frames will be assigned the same letter. Input Each input block contains the height, h (h<=30) on the first line and the width w (w<=30) on the second. A picture of the stacked frames is then given as h strings with w characters each. Your input may contain multiple blocks of the format described above, without any blank lines in between. All blocks in the input must be processed sequentially. Output Write the solution to the standard output. Give the letters of the frames in the order they were stacked from bottom to top. If there are multiple possibilities for an ordering, list all such possibilities in alphabetical order, each one on a separate line. There will always be at least one legal ordering for each input block. List the output for all blocks in the input sequentially, without any blank lines (not even between blocks). Sample Input 9 8 .CCC.... ECBCBB.. DCBCDB.. DCCC.B.. D.B.ABAA D.BBBB.A DDDDAD.A E...AAAA EEEEEE.. Sample Output EDABC

这是一道经典的拓扑排序问题。我们可以将每个图形看作一个节点，如果一个图形可以覆盖另一个图形，那就从覆盖的图形指向被覆盖的图形。最后得到的图是一个 DAG，我们可以对其进行拓扑排序。拓扑排序的结果就是图形的...

翻译：Notably, the utility functions here do not consider the resource sharing fraction applied in Ye et al. [5]. Instead, we directly use the achievable rate in the utility function. The reasons for this approach are two-fold. First, a UE device can accurately estimate the channel state information (CSI) between itself and a certain BS, but it may not effectively determine the number of devices that are currently being serviced by the BS. Second, the resource allocation strategies adopted by BSs are almost unknown to each UE. Therefore, we remove the resource sharing fraction from the utility function and achieve load balancing based on the access price and the mutual game among BSs.

值得注意的是，这里的效用函数并没有考虑Ye等人在[5]中使用的资源共享比例。相反，我们直接在效用函数中使用可达速率。采取这种方法的原因有两个。首先，用户设备可以准确估计自身与某个基站之间的信道状态信息(CSI)...

翻译：With the given constraints, the problem is two-fold. How should a BS determine the optimal access price? How should each UE choose the BS to associate with in response to the block-varying price? First, UE interest must be discussed. Offloading to a certain BS requires an access payment of the offlfloaded UE based on the announced access price vector p. Second, on the BS side, a larger load will increase the access revenue, but spectrum acquisition for the BS will also increase. Therefore, in the user association procedure, all BSs determine their optimal access prices to maximize their benefifits in terms of cell load, congestion and UE payments. Then, using the announced price vector, each UE device maximizes the profifit of its utility function minus the access payment. In this section, the mathematical representation of the proposed user association and resource allocation procedure will be discussed.

在给定的约束条件下，问题是双重的。一个基站应该如何确定最优的接入价格？每个用户设备应该如何选择要关联的基站，以应对块变化的价格？首先，必须讨论用户设备的兴趣。将数据卸载到某个基站需要根据宣布的接入...

解决这个问题King Julien rules the Madagascar island whose primary crop is coconuts. If the price of coconuts is P , then King Julien’s subjects will demand D(P ) = 1200 − 100P coconuts per week for their own use. The number of coconuts that will be supplied per week by the island’s coconut growers is S(p) = 100P. (a) (2 pts) Calculate the equilibrium price and quantity for coconuts. (b) (2 pts) One day, King Julien decided to tax his subjects in order to collect coconuts for the Royal Larder. The king required that every subject who consumed a coconut would have to pay a coconut to the king as a tax. Thus, if a subject wanted 5 coconuts for himself, he would have to purchase 10 coconuts and give 5 to the king. When the price that is received by the sellers is pS, how much does it cost one of the king’s subjects to get an extra coconut for himself? (c) (3 pts) When the price paid to suppliers is pS, how many coconuts will the king’s subjects demand for their own consumption (as a function of pS)? 2 (d) (2 pts) Under the above coconut tax policy, determine the total number of coconuts demanded per week by King Julien and his subjects as a function of pS. (e) (3 pts) Calculate the equilibrium value of pS, the equilibrium total number of coconuts produced, and the equilibrium total number of coconuts consumed by Julien’s subjects. (f) (5 pts) King Julien’s subjects resented paying the extra coconuts to the king, and whispers of revolution spread through the palace. Worried by the hostile atmosphere, the king changed the coconut tax. Now, the shopkeepers who sold the coconuts would be responsible for paying the tax. For every coconut sold to a consumer, the shopkeeper would have to pay one coconut to the king. For this new policy, calculate the number of coconuts being sold to the consumers, the value per coconuts that the shopkeepers got after paying their tax to the king, and the price payed by the consumers.

(a) 平衡价格为 P=6，平衡数量为 Q=600。 (b) 当价格为 pS=6 时，King Julien 的臣民为了自己获得额外的椰子，需要支付 12 个椰子。 (c) 当价格为 pS 时，King Julien 的臣民需要的椰子数量为 D(pS)=120-100pS。...

(A TV Class – Encapsulation + static)The attributes of a TV object are the channel, the volume, and a flag (or switch)indicating whether the TV is on or off. The methods perform the following actions:Turn the TV on or off.Set the channel to an integer from 0 to 99 inclusive. Raise or lower the volume by one unit. The volume can range from 0 to 20.View the value of the volume.View the channel.Determine whether the TV is on or off.Add static variable to count the number of TV in your house and a static method to return the number of TV that you have.Draw the UML diagram for the class then implement the class. Write a TV class that implements all relevant functions. The class should follow the encapsulation rules which usually make all variable private and use getter and setter to get and set values of the class’s variable. A newly created TV object is set to off with the channel set to 2 and the volume initially 10. Include a main(...) method that tests the methods of the TV class.

this.channel = channel; } } public void increaseVolume() { if (volume ) { volume++; } } public void decreaseVolume() { if (volume > 0) { volume--; } } public int getVolume() { ...

Imagine that we want to determine whether unemployment was high (> 6.5), medium (4.5 < x <= 6.5), or low (<= 4.5) for each state and each month. Write a Python function that takes a single number as an input and outputs a single string noting if that number is high, medium, or low. Pass your function to applymap (quiz: why applymap and not agg or apply?) and save the result in a new DataFrame called unemp_bins. (Challenging) This exercise has multiple parts: Use another transform on unemp_bins to count how many times each state had each of the three classifications. - Hint 1: Will this value counting function be a Series or scalar transform? - Hint 2: Try googling "pandas count unique value" or something similar to find the right transform. Construct a horizontal bar chart of the number of occurrences of each level with one bar per state and classification (21 total bars). (Challenging) Repeat the previous step, but count how many states had each classification in each month. Which month had the most states with high unemployment? What about medium and low?

We use applymap() instead of agg() or apply() because we want to apply the function element-wise to each element in the DataFrame. Here's how you can use this function with applymap() to ...

这段代码社么意思This is a C function named PWM_regulate_1 that appears to be controlling PWM (Pulse Width Modulation) signals on two different channels, PWM_Number_1 and PWM_Number_2. Here's a breakdown of the function: Enable flag_on variable. Disable PWM clock for both channels (PWM_Number_1 and PWM_Number_2). Disable time-base clock synchronization. Enable PWM clock for both channels again. Determine the frequency from data[5] and data[6]. Set flag_freq based on the determined frequency. Calculate the period and dead time based on f and data values. Calculate the duty cycle (D) based on the period and data[7]. Set various configurations for myPwm1 (channel 1): Set the period, phase, and count values. Configure counter mode and disable counter load. Set shadow mode for PWM compare register A. Set high-speed clock division and clock division based on flag_freq. Set the comparison value for PWM compare register A (duty cycle). Configure action qualification for up/down counting mode. Configure dead band output, polarity, and input modes. Set rising and falling edge delay for dead band. Set similar configurations for myPwm2 (channel 2), using the same logic as above. Note: This code snippet seems to be a part of a larger program, missing some essential details such as variable types, initializations, libraries used, and possible hardware connections

这段代码是一个名为PWM_regulate_1的C函数，看起来它控制了两个不同通道的PWM（脉冲宽度调制）信号，即PWM_Number_1和PWM_Number_2。下面是对该函数的解析： 1. 启用flag_on变量。 2. 禁用两个通道（PWM_Number_1和...

ldentify Patterns using Boundaries Use these guidelines to determine and set the boundaries of the dynamic data:.Analyze the location of the dynamic data within the HTTPresponse. o ldentify the string that is immediately to the left of the dynamic data.This string defines the leftboundary of the dynamic data. o ldentify the string that is immediately to the right of the dynamic data.This string defines the rightboundary of the dynamic data. .The right and left boundaries should be as unique as possible to better locate the strings.. web_reg_save_param_ex looks for the characters between (but not including) the specified boundaries and saves the information beginning one byte after the left boundary and ending onebyte before the right boundary. web_reg_save_param_ex does not support embedded boundarycharacters. For example, if the input buffer is {a{b{c} and "[" is specified as a left boundary, and "as a rightboundary, the first instance is c and there are no further instances—it found the right and leftboundaries but it does not allow embedded boundaries, so "c" is the only valid match. By default, the maximum length of any boundary string is 256 characters.Include a web_set_max_html_param_len function in your script to increase the maximum permitted length.Forexample, the following function increases the maximum length to 1024 characters: These length restrictions do not apply where either the left or right boundaries are blank. -翻译成中文

使用边界识别模式的指南来确定和设置动态数据的边界： ... 2. 确定紧挨着动态数据左侧的字符串。该字符串定义了动态数据的左边界。 3. 确定紧挨着动态数据右侧的字符串。该字符串定义了动态数据的右边界。...

A string consisting of lowercase English letters, (, and ) is said to be a good string if you can make it an empty string by the following procedure: First, remove all lowercase English letters. Then, repeatedly remove consecutive () while possible. For example, ((a)ba) is a good string, because removing all lowercase English letters yields (()), from which we can remove consecutive () at the 2-nd and 3-rd characters to obtain (), which in turn ends up in an empty string. You are given a good string S. We denote by S i the i-th character of S. For each lowercase English letter a, b, …, and z, we have a ball with the letter written on it. Additionally, we have an empty box. For each i=1,2, … ,∣S∣ in this order, Takahashi performs the following operation unless he faints. If S i is a lowercase English letter, put the ball with the letter written on it into the box. If the ball is already in the box, he faints. If S i is (, do nothing. If S i is ), take the maximum integer j less than i such that the j-th through i-th characters of S form a good string. (We can prove that such an integer j always exists.) Take out from the box all the balls that he has put in the j-th through i-th operations. Determine if Takahashi can complete the sequence of operations without fainting.用cpp解决，直接写代码

题意简述：给定一个字符串 S，它由小写英文字母、左括号 ( 和右括号 ) 组成，如果可以通过以下步骤将 S 变为空字符串，则 S 被称为好字符串： 1. 首先，删除所有小写英文字母。 2. 然后，当可能时，连续删除 ()。...

Consider the following ODE y′=ty2 for 0≤t≤1 , with y(1)=4 . Suppose you would like to solve this as a boundary value problem with a shooting method, what value of y(0) is required such that y(1)=4 ? Hint: We could do this with Newton’s method, but here it’s easier to just solve the differential equation by hand using separation of variables, apply the condition at y(1) and then calculate y(0) . Enter your answer below correct to 3 decimal places.

Since this is a shooting problem, we need to guess a value for y(0) and solve the ODE using a numerical method (such as ode45 in MATLAB) to determine if the resulting y(1) matches the desired value ...

You could also create code to simulate the Monty Hall problem. 1. Create a function named random_door, which uses numpy.random.choice to sample 1 door randomly from a list of integer door indices (1-3 in this case). Use this function to randomly select the door the car is behind and the contestant’s initial choice of doors. 2. Create a function monty_choice, which chooses the door Monty opens, conditional on the contestant’s choice of doors and the door with the car. For the case where the contestant has selected the door with the car, select the door to open by simulating the flip of a fair coin using the np.random.binomial function with n = 1. 3. Create a function win_car, which determines if the contestant wins the car, conditional on the strategy selected, {switch, noswitch}, the door the contestant selected, the door with the car, and the door Monty opened. 4. Create a function named simulation that allows you to run the simulation n = 1000 times, with a switch or noswitch strategy. 5. Execute your simulation for each possible strategy. For the two strategies, plot side by side bar charts showing the numbers of successes and failures for each strategy. 6. Describe the strategy a contestant should adopt for this game. How much will the chosen strategy change the probability of winning a car? Is this result consistent with the conditional probability of this problem?

This result is consistent with the conditional probability of the problem, which shows that the probability of winning the car increases from 1/3 to 2/3 if the contestant switches doors.

Good morning all teachers and students, my name is liangwenjun The topic of my report is Influence of molten salt potential on stress corrosion cracking behavior of 316 stainless steel. Stress corrosion cracking (SCC) is the cracking of sensitive materials resulting from the combined action of a fixed tensile stress and some specific corrosive medium.As shown in the picture on the right， Stress corrosion cracking is also one of the main causes of reactor structural material failure。 Structural materials for molten salt reactors must have good resistance to fluoride corrosion, excellent mechanical properties, good resistance to high temperature oxidation and irradiation Stainless steels such as 316 have ASME certification for high-temperature pile materials andare very economical.Therefore, many companies are considering 316 stainless steel as a structural material when designing molten salt reactors。Such as Kairos power and Terra power. 316 showed good corrosion resistance in purified molt en fluoride salt, and in Chen's study 316 did not crack with or without Cr addition under the synergistic effect of stress. Therefore, we accelerate corrosion by adding europium ion, and in the study to quantify the molten salt potential and stress magnitude, and finally determine the effect of molten salt potential on stress corrosion cracking of 316 stainless steel.

早上好，各位老师和同学们，我的名学是梁文军。我的报告的主题是：熔盐电位对316不锈钢应力腐蚀开裂行为的影响。应力腐蚀开裂（SCC）是指敏感材料由固定的拉应力和某些特定腐蚀介质的结合作用导致的裂纹。...

Redundant networking¶ You should conduct a high availability risk analysis to determine whether to use redundant switches such as Top of Rack (ToR) switches. In most cases, it is much more economical to use single switches with a small pool of spare switches to replace failed units than it is to outfit an entire data center with redundant switches. Applications should tolerate rack level outages without affecting normal operations since network and compute resources are easily provisioned and plentiful. Research indicates the mean time between failures (MTBF) on switches is between 100,000 and 200,000 hours. This number is dependent on the ambient temperature of the switch in the data center. When properly cooled and maintained, this translates to between 11 and 22 years before failure. Even in the worst case of poor ventilation and high ambient temperatures in the data center, the MTBF is still 2-3 years.

冗余网络您应该进行高可用性风险分析，以确定是否使用冗余交换机，例如顶级交换机（ToR交换机）。在大多数情况下，与使用冗余交换机来配置整个数据中心相比，使用单个交换机并配备一小批备用交换机以替换故障单元的...

我的第一个C#小程序之简单音乐播放器1731655933.html

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