manual configuration hp lj 1200 printer stuck on user password request

I tried to configure an HP lj 1200 printer on parallel port using the Gnome Printer configuration tool on ubuntu 19.04. After having manually chosen the appropriate model, driver and having clicked on ‘Apply’ I am asked for user and password. My user and password (sudo priviledges) do not work.

Any help is appreciated, thank you in advance.

500 Directory submissions manual for $3

500 Directory submissions manually500 Directory submissions Hi, Welcome to my services, I will submit your business or website into niche directories manually. Directory submission is very essential for business or website. After work done I will send you an excel file with all links or screenshot links. BENEFITS OF MY SERVICEI am very faster100% Manually submission100% Satisfaction GuaranteeIt will improve your website RankingsYou will Get more traffic

by: Libinvarambathu
Created: —
Category: Directory Submission
Viewed: 87


manual SEO & Social Campaigns- 3000 From profile Backlink- 5000 Wiki Backlink -Unlimited Social web for $2

Quality Backlinks! All campaigns on this page designed as a link wheel campaigns. In this kind of campaign, we submit the top quality links directly to your website and some other quality backlinks for those submitted orders (also called Link Pyramids). The tier1 links (direct links) on all campaigns will work on our Premium Sites List, which consists of High DA (Domain Authority) sites & Top world wide web 2.0s. CHECK THE PREMIUM SITES LIST So order now and sent me your link with keyword..Thanks

by: asifsorkar05
Created: —
Category: Web 2.0
Viewed: 35


Test de Wald manual

Intento entender el test de Wald en regresión logística, pero al usar la fórmula me salen una cosa distinta. La fórmula que uso es la siguiente:

S = (XtVX)-1

Donde X es la matriz r x k+1 de las variables independientes (con la constante en primer lugar) y V es la matriz diagonal cuyos elementos son vii = pi(1-pi), siendo p la probabilidad calculada para cada elemento. Las raíces cuadradas de los elementos diagonales serían los desvíos estándar de los parámetros.

Lo saqué de http://www.real-statistics.com/logistic-regression/significance-testing-logistic-regression-coefficients/ pero es básicamente la misma en todos los lugares que busqué. Ahora, la intento sacar mediante esta fórmula y me da una cosa distinta a la que me sacaría un modelo. Use R para crear un dataset con datos aleatorios:

set.seed(5674)  prueba <- cbind.data.frame(a = sample(seq(0,3,length.out = 5000), size = 1000,                                       replace = TRUE),                            b = sample(seq(0,3,length.out = 5000), size = 1000,                                       replace = TRUE),                            c = sample(seq(0,3,length.out = 5000), size = 1000,                                       replace = TRUE),                            d = sample(seq(0,3,length.out = 5000), size = 1000,                                       replace = TRUE),                            e = sample(seq(0,3,length.out = 5000), size = 1000,                                       replace = TRUE))   prueba$  y <- sapply(1:1000, function(x){   ex <- exp(0.3*prueba$  a[x] + 0.25*prueba$  b[x] - 0.4 * prueba$  c[x] -                0.15 * prueba$  d[x] - 0.05*prueba$  e[x])   x1 <- ex/(ex+1)   x2 <- 1/(ex+1)   t <- sample(0:1, size = 1, prob = c(x2,x1))   return(t[1]) }) 

Ahora creo el modelo, y saco la probabilidad predicha:

modelo <- glm(y ~ a+b+c+d+e, data = prueba) w <- modelo$  fitted.values w1 <- w*(1-w) 

Luego utilizo la formula que puse antes para sacar la matriz de varianzas – covarianzas:

matriz <- cbind(cons = rep(1,1000), as.matrix(prueba[,1:5]))  matriz1 <- apply(matriz, 2, function(x){x * w1})  varianza <- t(matriz1) %*% matriz varianza1 <- solve(varianza) sqrt(varianza1[1,1]) 

La última línea es para ver el desvío estándar de la variable a. Me da esto:

[1] 0.07855672 

Lo cual es distinto de lo que me informa el modelo:

summary(modelo)[[12]] 
               Estimate Std. Error    t value     Pr(>|t|) (Intercept)  0.46925485 0.06153214  7.6261742 5.641569e-14 a            0.08539726 0.01776354  4.8074451 1.764783e-06 b            0.06246717 0.01793220  3.4835200 5.164320e-04 c           -0.09870667 0.01752367 -5.6327620 2.307966e-08 d           -0.04879819 0.01752240 -2.7849038 5.456082e-03 e           -0.01540819 0.01715283 -0.8982882 3.692495e-01 

¿Hay algún error en la fórmula?

100 Unique Domains Manual Blog Comments Backlinks With Da 20 plus for $20

What would you advantage if you took this provider? You can take this provider to enhance your website or to get a variety of site visitors. I am imparting 100 Unique Domain DoFollow Blog Comments Backlinks with DA 20plus. When Google bots are crawling a website they search for outside hyperlinks factor to different websites. If this link is “dofolow” then search engines like google will follow the link and so hyperlink juice gets handed. Links that count number as points, pushing search engine optimization link juice and boosting the web page rank of the related-to sites, assisting them to pass higher inside the SERPs as an end result. I will give you 100%Dofollow blog comments with High WHY CHOICE ME? 1.According to the latest GOOGLE Update 2019 2.100% DoFollow, 100%Approval Rates. 3.100% Manual &Live Links 4.Unlimited URL and keywords 5.100% Google panda Safe 6.Backlinks on DA 20+Pages 7.100% creative works. 8.manual and white hat SEO 9. Provide a quality blog comment post. 10. We will deliver our service as promised. 11.Detailed ExcelReport on Order Completion 12.24/7 CustomerSupport These web sites have a big traveller. Therefore after I createa remark. With your hyperlink then the blog visitor redirected your web site and you may have a large visitor. So It’s the best way to increase traffic to your website. Thanks for regards, Shariful

by: SEOShariful
Created: —
Category: Blog Comments
Viewed: 131


I will high authority backlinks only manual SEO link building for $30

★Get the Best service at a reasonable price best ranking effects on SEOCLERK.COM★ Note: Work is 100% save we create manual backlinks from high authority sites mostly DA 90 to 100. What services we are providing? ✔️All sites indexed Mostly from 90 DA above sites. ✔️All Domains are Unique. ✔️100% White Hat and Manual work. ✔️More authority and rankings for your website. ✔️The latest Google updates Service. ✔️ Mixed Backlinks(Dofollow as well). ✔️Unlimited URLs + Keywords acceptable. Why You choose my Backlinks ✔️Only Gain -No Loss. ✔️Quality work. ✔️Highly Expert Team. ✔️Prefer Quality over quantity. ✔️Best Customer Support. bvb ✔️Life Time Backlinks. ✔️Full manually work report at the end of work report in excel file. Thank you! ★ ★ We are friendly feel free to Contact Us.★ ★ ✔️✔️✔️ORDER NOW✔️✔️✔️ With Extras get 100% ranking effects in Google

by: promotionsite
Created: —
Category: Link Building
Viewed: 143


Manual high quality 10 bookmark for your website for $3

About me: Hi, 1. You get 100% Quality Service 2. Unlimited Revision 3. 24/7 VIP Support 4. Any Questions you can ask me before or after order. 5. Fast delivery 6. You get feedback by MS Excel Sheet You get Top ranking Social Media Bookmarking This List is below: 1. Facebook 2. Twitter 3. LinkedIn 4. Instagram 5. Tumblr 6. Mix. For you 7. Reddit 8. VK 9. OK 10. Scoop.it 11. GrowthHackers 12. And more

by: braveseo
Created: —
Category: Forum Posts
Viewed: 136


Manual Regression Tree using Python

I wrote a code to create a regression tree for a synthetic train data of size Np. The idea is, first I have the source node (which consists of all set of points) represented as a dictionary {'points':..., 'avg':..., 'left_node':..., 'right_node', 'split_point': }. The left and right nodes are the leafs after the splitting process of the whole data (source). split_point is for information about the best split. Then I loop to get deeper tree with maximum number of nodes specified before, also I set that a node must have more than 5 points in order it can be split.

This way, If I want to predict a point (x',y'), I can just start from source node source and check which region the point lies (left_node or right_node), ..and then continuing down the tree. Because all left_nodes and right_nodes values have the same structure as source….

Also, the form function is used to find the best split, the best split is the one with the smallest form(reg_1, avg1, reg_2, avg2). This is a greedy algorithm to find the best split.


I would like to know better ways to perform it..without external modules. But this is intended to be taught to high school students.


Full code:

import math import random import matplotlib.pyplot as plt   def form(region_1, av1, region_2, av2):     return sum([(i[1]-av1)**2 for i in region_1]) \            + sum([(i[1]-av2)**2 for i in region_2])  Np = 400 x_data = [abs(random.gauss(5, 0.2) + random.gauss(8, 0.5)) for i in range(Np)] y_data = [abs(random.gauss(10, 0.2) + random.uniform(0, 10)) for i in range(Np)] value = [abs(random.gauss(4, 0.5)) for i in range(Np)]  data = [((i,j), k) for i,j,k in zip(x_data, y_data, value)]  fig, ax = plt.subplots()  ax.plot(x_data, y_data, 'o')  fig.show()   ###### Splitting from the source node (all data)  source = {'points': data, 'avg': sum([i[1] for i in data])/Np, \           'split_point': None, 'left_node': None, 'right_node': None} forms = []  for x in x_data:     var = x     region_1 = [j for j in data if j[0][0] <= var]     region_2 = [j for j in data if j not in region_1]      if len(region_1) > 0 and len(region_2) > 0:          av1 = sum([i[1] for i in region_1])/len(region_1)         av2 = sum([i[1] for i in region_2])/len(region_2)          f = form(region_1, av1, region_2, av2)         leaf_1 = {'points': region_1, 'avg': av1}         leaf_2 = {'points': region_2, 'avg': av2}         forms.append( (leaf_1, leaf_2, ('x', var), f) )  for y in y_data:     var = y     region_1 = [j for j in data if j[0][1] <= var]     region_2 = [j for j in data if j not in region_1]      if len(region_1) > 0 and len(region_2) > 0:          av1 = sum([i[1] for i in region_1])/len(region_1)         av2 = sum([i[1] for i in region_2])/len(region_2)          f = form(region_1, av1, region_2, av2)         leaf_1 = {'points': region_1, 'avg': av1}         leaf_2 = {'points': region_2, 'avg': av2}         forms.append( (leaf_1, leaf_2, ('y', var), f) )   sorted_f = sorted(forms, key = lambda x: x[3]) best_split = sorted_f[0] source['split_point'] = best_split[2] source['left_node'] = best_split[0] source['right_node'] = best_split[1]   ##### Splitting from the 2 leafs and so on..  leafs = [source['left_node'], source['right_node']]  all_nodes = [leafs[0], leafs[1]]  max_nodes = 1000  while len(all_nodes) <= max_nodes:     next_leafs = []     for leaf in leafs:         if (len(leaf['points']) > 5):              xx = [i[0][0] for i in leaf['points']]             yy = [i[0][1] for i in leaf['points']]             rr = [i[1] for i in leaf['points']]             vv = [((i,j), k) for i,j,k in zip(xx, yy, rr)]             forms = []              for x in xx:                 var = x                 region_1 = [j for j in vv if j[0][0] <= var]                 region_2 = [j for j in vv if j not in region_1]                  if len(region_1) > 0 and len(region_2) > 0:                      av1 = sum([i[1] for i in region_1])/len(region_1)                     av2 = sum([i[1] for i in region_2])/len(region_2)                      f = form(region_1, av1, region_2, av2)                     leaf_1 = {'points': region_1, 'avg': av1}                     leaf_2 = {'points': region_2, 'avg': av2}                     forms.append( (leaf_1, leaf_2, ('x', var), f) )              for y in yy:                 var = y                 region_1 = [j for j in vv if j[0][1] <= var]                 region_2 = [j for j in vv if j not in region_1]                  if len(region_1) > 0 and len(region_2) > 0:                      av1 = sum([i[1] for i in region_1])/len(region_1)                     av2 = sum([i[1] for i in region_2])/len(region_2)                      f = form(region_1, av1, region_2, av2)                     leaf_1 = {'points': region_1, 'avg': av1}                     leaf_2 = {'points': region_2, 'avg': av2}                     forms.append( (leaf_1, leaf_2, ('y', var), f) )              sorted_f = sorted(forms, key = lambda x: x[3])             best_split = sorted_f[0]             leaf['split_point'] = best_split[2]             leaf['left_node'] = best_split[0]             leaf['right_node'] = best_split[1]              print(leaf['split_point'])              next_leafs.append(leaf['left_node'])             next_leafs.append(leaf['right_node'])              print("\n")      leafs = next_leafs     all_nodes.extend(leafs)     if len(leafs) == 0:         break 

Related Post Widget manual Post selection

I am trying to create a sidebar widget that shows Posts based on the selection from an ACF field.

The selected Posts can be from any Post category or CPT, using the Post Object field in ACF.

Depending on the Post selected, the front end should just show the Post Title and Excerpt, currently only the Post_ID numbers are showing.

How do I take the Post_ID numbers to show the related Post Title and Excerpt?

Current Code –

class CF_Custom_Post_widget extends WP_Widget {  /** * Sets up the widgets name etc */ public function __construct() { $  widget_ops = array(      'classname' => 'cf_related_posts', //'my_widget     'description' => 'Show Related Posts', ); parent::__construct( 'cf_related_posts', 'CF Related Custom Posts', $  widget_ops ); }  /** * Outputs the content of the widget * * @param array $  args * @param array $  instance */ public function widget( $  args, $  instance ) {  $  title = apply_filters( 'widget_title', $  instance['title'] );  echo $  args['before_widget']; if ( ! empty( $  title ) ) echo $  args['before_title'] . $  title . $  args['after_title'];   // widget ID with prefix for use in ACF API functions $  widget_id = 'widget_' . $  args['widget_id'];  $  title = get_field( 'title', $  widget_id ) ? get_field( 'title', $  widget_id ) : '';  the_field( 'sidebar-recent_posts', $  widget_id);  echo $  args['after_widget'];  }  /** * Outputs the options form on admin * * @param array $  instance The widget options */ public function form( $  instance ) {  // outputs the options form on admin if ( isset( $  instance[ 'title' ] ) ) { $  title = $  instance[ 'title' ]; } else { $  title = __( 'New title', 'cf_related_posts' ); }  ?> <p> <label for="<?php echo $  this->get_field_id( 'title' ); ?>"> <?php _e( 'Title:' ); ?></label> <input class="widefat" id="<?php echo $  this->get_field_id( 'title' );  ?>" name="<?php echo $  this->get_field_name( 'title' ); ?>" type="text"  value="<?php echo esc_attr( $  title ); ?>" /> </p> <?php        } /** * Processing widget options on save * * @param array $  new_instance The new options * @param array $  old_instance The previous options * * @return array */ public function update( $  new_instance, $  old_instance ) { // processes widget options to be saved $  instance = array(); $  instance['title'] = ( ! empty( $  new_instance['title'] ) ) ?  strip_tags( $  new_instance['title'] ) : ''; return $  instance;    } }   /** * Register Widget */ function register_cf_widget() { register_widget( 'CF_Custom_Post_widget' ); } add_action( 'widgets_init', 'register_cf_widget' );