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The frequency of immune heparin-induced thrombocytopenia (HIT) varies among prospective studies. It is unknown whether this is caused by differences in the heparin preparations, the patient populations, or the types of serologic assay used to confirm the diagnosis. Seven hundred forty-four patients were studied from 3 different clinical treatment settings, as follows: unfractionated heparin (UFH) during or after cardiac surgery (n = 100), UFH after orthopedic surgery (n = 205), and low-molecular-weight heparin (LMWH) after orthopedic surgery (n = 439). Both an activation assay and an antigen assay were used to detect heparin-dependent IgG (HIT-IgG) antibodies. By activation assay, the frequency of HIT-IgG formation ranged from a low of 3.2% in orthopedic patients receiving LMWH to a high of 20% in cardiac patients receiving UFH; by antigen assay, the corresponding frequencies ranged from 7.5% to 50%. Both UFH use (P = .002) and cardiac surgery (P = .01) were more likely to be associated with HIT-IgG formation. However, among patients in whom HIT-IgG formed and who were administered UFH, the probability for HIT was higher among orthopedic patients than among cardiac patients (by activation assay: 52.6% compared with 5%; odds ratio, 21.1 [95% CI, 2.2-962.8]; P = .001; by antigen assay: 34.5% compared with 2.0%; odds ratio, 25.8 [95% CI, 3.2-1141]; P < .001). It is concluded that there is an unexpected dissociation between the frequency of HIT-IgG formation and the risk for HIT that is dependent on the patient population. HIT-IgG antibodies are more likely to form in patients who undergo cardiac surgery than in orthopedic patients, but among patients in whom antibodies do form, orthopedic patients are more likely to develop HIT.

Heparin-induced thrombocytopenia (HIT) is an adverse drug reaction caused by heparin-dependent IgG (HIT-IgG) antibodies that activate platelets.1-3 The target antigen consists of multimolecular complexes of platelet factor 4 and heparin.4-7 The frequency of HIT varies considerably, as cited by prospective studies.8,9 The reasons for these variations are unknown but could be related to different heparin preparations,10 patient population-dependent factors, or even different laboratory techniques used to detect the antibodies responsible for HIT.

Two general types of laboratory assay are used to confirm the diagnosis. Activation assays, such as the platelet serotonin release assay, detect HIT-IgG on the basis of their ability to activate platelets.11-13 Antigen assays, such as a solid-phase immunoassay, detect the binding of antibodies to immobilized platelet factor 4–heparin complexes.4-7,14,15 Despite the increasing use of these assays, few studies have compared their clinical usefulness in diagnosing HIT. Those studies14,16-18 comparing activation and antigen assays typically have been limited to investigating patient samples that were referred with the clinical suspicion of HIT (high pretest probability). However, such an approach does not allow the determination of test specificity. Furthermore, systematic serologic studies of patients receiving heparin (low pretest probability) have generally been too small to evaluate sensitivity. In addition, these studies have generally focused on single patient populations.

In this report, we describe the results of laboratory testing for HIT-IgG using both activation and antigen assays in 744 prospectively studied patients. These patients were from 3 different clinical treatment settings: cardiac surgery patients receiving unfractionated heparin (UFH), orthopedic surgery patients receiving UFH, and orthopedic surgery patients receiving low-molecular-weight heparin (LMWH). The results of this study indicate that there is an unexpected dissociation between the formation of HIT-IgG and the risk for HIT among patients in whom antibodies form that is patient population dependent.

### Patient populations

Patients in 1 of 3 clinical treatment settings were studied prospectively. They were grouped as follows: orthopedic surgery patients receiving UFH for postoperative prophylaxis (orthopedic–UFH group, n = 205); orthopedic surgery patients receiving LMWH for postoperative prophylaxis (orthopedic–LMWH group, n = 439); and cardiac surgery patients receiving UFH both at cardiopulmonary bypass and for postoperative antithrombotic prophylaxis (cardiac–UFH group, n = 100). Patients underwent daily platelet count monitoring while receiving heparin.

The orthopedic–UFH group consisted of a group of 205 patients who received porcine mucosal UFH (Calciparine; Laboratoires Anglo-French, Dorval, Quebec, Canada), 7500 IU every 12 hours by subcutaneous injection for up to 14 days or discharge. These patients had participated in a randomized clinical trial of UFH versus LMWH given for antithrombotic prophylaxis after hip arthroplasty surgery (study A). Primary trial results and a related study of HIT using only the platelet 14C-labeled serotonin-release assay have been reported.10,19 In the current study, the patients' plasma samples were studied using the antigen assay as well.

The orthopedic-LMWH group (n = 439) consisted of patients from 2 studies who received LMWH for antithrombotic prophylaxis after orthopedic surgery. The first group (study A) consisted of 182 patients who received LMWH (enoxaparin; Lovenox; Rhône-Poulenc Rorer, Montreal, Quebec, Canada), 30 mg every 12 hours by subcutaneous injection for up to 14 days or discharge. These patients had also participated in the trial described previously.10,19 The second orthopedic–LMWH group (study B) consisted of 257 patients who received postoperative LMWH prophylaxis after orthopedic surgery (hip arthroplasty, n = 105; knee arthroplasty, n = 152). These patients received 1 of 3 LMWH preparations based on the current hospital practice at the time of participation in the study—enoxaparin (Lovenox; Rhône-Poulenc Rorer), 30 mg twice daily by subcutaneous injection, n = 70; tinzaparin (Innohep; Leo, Ajax, Ontario, Canada), 3000 U twice daily by subcutaneous injection, n = 2; or dalteparin (Fragmin; Pharmacia, Mississanga, Ontario, Canada), 3000 U twice daily by subcutaneous injection, n = 185. For orthopedic patients, heparin administration was usually started on the first postoperative day and continued until discharge or full mobilization. Although 3 different LMWH preparations were used, we made an a priori decision to analyze all the data as one orthopedic–LMWH group.

The third patient population consisted of a cardiac–UFH group comprised of 100 patients undergoing elective valve replacement surgery (69 aortic valve replacement, 29 mitral valve replacement, 2 both). Porcine mucosal heparin (Hepalean; Organon Teknika, Scarborough, Ontario, Canada), 400 U/kg, was given before cardiopulmonary bypass; additional heparin was given by bolus as needed to achieve and maintain an intraoperative whole blood activated clotting time of more than 400 seconds. These patients also received subcutaneous heparin (porcine mucosal heparin; Heparin; Leo), 5000 IU every 8 hours by subcutaneous injection, until they were adequately anticoagulated with warfarin or mobilized. Antiplatelet drugs such as aspirin were not routinely given to this patient population undergoing valve replacement surgery.

These studies were approved by the Institutional Review Board, and patients gave their informed consent. Laboratory testing for HIT-IgG was performed after the patients had been discharged from the hospital, and no results were used to influence clinical decisions.

### Activation assay

All 744 patients were tested with the platelet serotonin release assay, as described,11,12 using heat-inactivated citrated plasma or serum stored at −70°C before testing. Using a predetermined algorithm, samples were considered positive if all the following criteria were met: (1) 20% or greater serotonin release at 0.1 U/mL heparin; (2) at least 50% inhibition of platelet activation by both an Fc receptor-blocking monoclonal antibody (IV.3) and a high concentration of heparin (100 U/mL); (3) appropriate activation profiles observed with 3 positive controls (including 2 “weak” sera, ie, giving 20%-50% serotonin release) and one negative control serum.

### Antigen assay

Patient plasma and serum samples were also evaluated using a platelet factor 4–heparin enzyme immunoassay, as previously described,15 except that an alkaline phosphatase-conjugated goat antibody specific for human IgG (Fc) replaced the conjugated goat antibody for human IgG (heavy and light chain-specific). By measuring IgG antibodies only, we facilitated comparison with the activation assay, which detects platelet activation through IgG antibodies.3 One hundred samples from healthy blood donors were assayed, and the upper limit of the normal range of optical density (OD) was established as the mean + 3 SD (OD = 0.45). Each assay included pooled negative controls (OD ≤ 0.2), high-negative control pool (0.2 < OD < 0.45), known negative control, known positive control (OD > 1.5), and weak positive control (0.45 < OD < 1.5).

### Definitions

We defined possible HIT as a 50% or greater fall in the platelet count from the postoperative peak that occurred between days 5 to 14 after surgery, unless another cause for the thrombocytopenia was readily apparent (eg, culture-positive septicemia) or the platelet count recovered during continued heparin treatment. This definition of thrombocytopenia has the best correlation with the formation of HIT-IgG20 and is appropriate for a postoperative patient population in whom thrombocytosis commonly occurs between postoperative days 5 to 14. Heparin-induced thrombocytopenia (HIT) was defined when a patient with possible HIT also had positive results of at least one laboratory assay for HIT-IgG (by activation or antigen assay, or both).

### Comparison of activation and antigen assays

Because both diagnostic assays are quantitative measurements, we used receiver operating characteristic (ROC) curve analysis to compare the sensitivity–specificity tradeoff at various diagnostic cutoff points defining negative and positive test results. Sensitivity of each assay was defined as the proportion (in percentage) giving a positive test result among the patients with possible HIT. Specificity of each assay was defined as the proportion (in percentage) of patients who did not meet the study definition for possible HIT and who tested negative for the assay under consideration. The ROC curve analysis was performed only for the orthopedic patients because this patient group had a relatively large number of patients with possible HIT and additional patients with subclinical HIT-IgG seroconversion (see “Results”).

Our standard antigen assay uses a 1:50 dilution of test serum or plasma. To investigate whether greater sample dilution would improve the sensitivity–specificity tradeoff of the antigen assay, we also performed the assay using the following sample dilutions: 1/50, 1/75, 1/100, 1/150, 1/250, 1/500, 1/750, 1/1000, 1/2500, and 1/5000. After these preliminary studies, we systematically studied the orthopedic patient samples that tested positive in the antigen assay (at 1/50), at the following additional dilutions: 1/100, 1/250, 1/500, and 1/750. The data obtained were used to calculate ROC curves. Thus, we investigated whether there might be a greater likelihood that blood samples from patients with subclinical HIT-IgG formation would test negative on further sample dilution than blood samples from patients with possible HIT. In other words, could the sensitivity–specificity tradeoff be improved on sample dilution? If so, there would be the potential for greater diagnostic specificity for the antigen assay, without significant loss of sensitivity, if the antigen assay was performed at a higher dilution.

### Statistical analysis

Comparisons of proportions between groups were performed using the Fisher exact test.21 An associated method developed by Gart22 was used for computing confidence intervals around the odds ratio. We performed logistic regression analysis23to determine the impact of either heparin preparation (UFH compared with LMWH) and the type of patient population (cardiac compared with orthopedic patients) with respect to the frequency of HIT-IgG formation and the proportion of antibody-positive patients in whom possible HIT developed. Analyses were performed separately for activation and antigen test results. All quoted P values were 2-sided.

### Frequency of possible heparin-induced thrombocytopenia

Eighteen of the 744 prospectively studied patients had a 50% or greater decrease in platelet count that began between postoperative days 5 and 14. Fifteen of these patients met the criteria for possible HIT (Table 1). All 15 patients who were identified as possibly having HIT also tested positive for HIT-IgG using both the activation and the antigen assays. The frequency of HIT was highest (4.9%) in the orthopedic–UFH patients and was relatively low in both the orthopedic–LMWH patients (0.9%) and the cardiac–UFH patients (1%). One or more thrombotic events occurred in 9 (60%) of these 15 patients; 7 patients had venous thromboembolism, 1 patient had unilateral adrenal hemorrhagic infarction, and 1 patient had arterial thrombosis.

Table 1.

Relationship of the patient population and type of heparin on frequency of HIT-IgG antibody formation and risk for developing HIT

Patient population, type of heparin
(mean ± SD duration of heparin [d])
Frequency of HIT antibody positivity (95% CI)Frequency of HIT (95% CI)Frequency of HIT–IgG-positive patients with HIT (95% CI)
Activation assay*Antigen assay*Activation assayAntigen assay
Cardiac-UFH, n = 100 20/100 50/100 1/100 1/20 1/50
(5.1 ± 2.2) 20.0% (12.7-29.2) 50.0% (39.8-60.2) 1.0% (0.03-5.5) 5.0% (0.1-24.9) 2.0% (0.12-10.7)
Orthopedic-UFH, n = 205 19/205 29/205 10/205 10/19 10/29
(9.2 ± 2.2) 9.3% (5.7-14.1) 14.1% (9.7-19.7) 4.9% (2.4-8.8) 52.6% (28.9-75.6) 34.5% (17.9-54.3)
Orthopedic-LMWH, n = 439 14/439 33/439 4/439 4/14 4/33
(9.5 ± 3.0) 3.2% (1.8-5.3) 7.5% (5.2-10.4) 0.9% (0.3-2.3) 28.6% (8.4-58.1) 12.1% (3.4-28.2)
Pvalues
Cardiac vs Orthopedic .01 <.001 .71 .01 .004
UFH vs LMWH .002 .009 .015 .19 .048
Patient population, type of heparin
(mean ± SD duration of heparin [d])
Frequency of HIT antibody positivity (95% CI)Frequency of HIT (95% CI)Frequency of HIT–IgG-positive patients with HIT (95% CI)
Activation assay*Antigen assay*Activation assayAntigen assay
Cardiac-UFH, n = 100 20/100 50/100 1/100 1/20 1/50
(5.1 ± 2.2) 20.0% (12.7-29.2) 50.0% (39.8-60.2) 1.0% (0.03-5.5) 5.0% (0.1-24.9) 2.0% (0.12-10.7)
Orthopedic-UFH, n = 205 19/205 29/205 10/205 10/19 10/29
(9.2 ± 2.2) 9.3% (5.7-14.1) 14.1% (9.7-19.7) 4.9% (2.4-8.8) 52.6% (28.9-75.6) 34.5% (17.9-54.3)
Orthopedic-LMWH, n = 439 14/439 33/439 4/439 4/14 4/33
(9.5 ± 3.0) 3.2% (1.8-5.3) 7.5% (5.2-10.4) 0.9% (0.3-2.3) 28.6% (8.4-58.1) 12.1% (3.4-28.2)
Pvalues
Cardiac vs Orthopedic .01 <.001 .71 .01 .004
UFH vs LMWH .002 .009 .015 .19 .048

The 15 patients with clinical HIT all tested positive by antigen and activation assays. Logistic regression analysis was used to compare 2 variables, patient population (cardiac compared with orthopedic patients) and heparin preparation (UFH compared with LMWH), with respect to 2 outcome measures—the frequency of patients testing positive for HIT-IgG antibodies and the proportion of antibody-positive patients in whom thrombocytopenia developed. Among the orthopedic patients who received LMWH, there was no significant difference in the frequency of HIT-IgG formation, or in the frequency of possible HIT among patients who received the different LMWH preparations (data not shown).

*

Fractions indicate the number of patients with positive HIT-IgG test results (by the assay indicated) out of the patient population tested.

Fractions indicate the number of patients with HIT out of the patient population tested.

Fractions indicate the number of patients with HIT out of those who tested positive for HIT-IgG antibodies by the assay indicated.

Three patients whose platelet counts fell by 50% or more between postoperative days 5 to 14 did not meet the criteria for possible HIT. Thrombocytopenia developed in association with colon perforation and septicemia in two orthopedic patients, one treated with UFH and one treated with LMWH. Both died. In both patients, however, platelet counts recovered during continued heparin use. A third patient had a 59% decrease in platelet count in association with pulmonary embolism, but full platelet count recovery during treatment with therapeutic-dose UFH occurred. All 3 patients tested negative for HIT-IgG by activation and antigen assays.

### Frequency of HIT-IgG antibody formation

Figure 1 summarizes the results of the laboratory assays for HIT-IgG among the 3 patient treatment groups studied. With both antigen and activation assays, we observed the highest antibody-positivity rate (by conventional positive cutoff) among the cardiac patients (50% and 20%, respectively), followed by the orthopedic–UFH (14.1% and 9.3%) and the orthopedic–LMWH patients (7.5% and 3.2%). By logistic regression analysis (Table 1), we found that 2 variables were associated with a higher frequency of HIT-IgG formation: cardiac surgery and use of UFH.

Fig. 1.

Comparison of activation and antigen assays for HIT-IgG antibodies in patients who have undergone cardiac and orthopedic surgery.

Quantitative results of activation and antigen tests for HIT-IgG are shown for 3 clinical treatment settings: orthopedic–UFH (A), orthopedic–LMWH (B), and cardiac–UFH (C). Results are shown for all 744 patients. All antigen assay data are given quantitatively. For the activation assay results, samples that gave less than 20% serotonin release are as shown without reference to the actual quantitative result obtained (see box designated < 20%); quantitative data are given when the percentage serotonin release was 20% or more. Arrows indicate the data points corresponding to the 15 patients with HIT identified in these prospective studies.

Fig. 1.

Comparison of activation and antigen assays for HIT-IgG antibodies in patients who have undergone cardiac and orthopedic surgery.

Quantitative results of activation and antigen tests for HIT-IgG are shown for 3 clinical treatment settings: orthopedic–UFH (A), orthopedic–LMWH (B), and cardiac–UFH (C). Results are shown for all 744 patients. All antigen assay data are given quantitatively. For the activation assay results, samples that gave less than 20% serotonin release are as shown without reference to the actual quantitative result obtained (see box designated < 20%); quantitative data are given when the percentage serotonin release was 20% or more. Arrows indicate the data points corresponding to the 15 patients with HIT identified in these prospective studies.

Close modal

We observed dissociation between the frequency of HIT-IgG formation and the risk of HIT that was patient population dependent. Among patients in whom antibodies formed, HIT was more likely to develop in orthopedic patients than in cardiac patients (Table 1). Among patients who tested positive by the activation assay and who received UFH, HIT developed in 52.6% of orthopedic patients but only 5% of cardiac surgery patients (odds ratio, 21.1 [95% CI, 2.2 to 962.8];P = .001). Among patients who tested positive by antigen assay and who received UFH, HIT developed in 34.5% of orthopedic patients and only 2.0% of cardiac surgery patients (odds ratio, 25.8 [95% CI, 3.2 to 1141]; P < .001).

We investigated the duration of heparin treatment as a reason fewer patients with HIT-IgG in the cardiac group had possible HIT. At the time of heart surgery, the cardiac patients received very high doses of heparin. After surgery, both the cardiac and the orthopedic patients who received UFH were given the same dose (15 000 U/d). Although orthopedic patients received UFH for a greater duration than cardiac patients (9.2 ± 2.2 vs 5.1 ± 2.2 days; P < .001), the duration of heparin exposure was not the major reason for the higher frequency of HIT in orthopedic patients. For example, among patients in whom HIT-IgG formed and was detectable by both assays and who received UFH for 5 or more days, there was a greater likelihood of the onset of platelet count decrease by day 5 in the orthopedic patients than in the cardiac patients (5 of 19 compared with 0 of 15;P = .053). A similar trend was noted when the analysis was restricted to patients who received heparin for at least 6 days: 0 of 8 patients in the cardiac group had a platelet count decrease indicative of possible HIT, whereas 7 of 19 orthopedic patients had a platelet count decrease by day 6 (P = .068). This suggests that orthopedic patients are at greater risk for HIT for reasons other than longer duration of UFH use. Indeed, the cardiac surgery patient in whom HIT developed (postoperative day 7, complicated by unilateral adrenal necrosis) received UFH only until the second day after surgery.

### Diagnostic usefulness of laboratory assays for heparin-induced thrombocytopenia

Figure 1 shows that most patients with HIT (indicated by arrows) had strong positive test results for HIT-IgG using either assay: more than 90% serotonin release (activation assay) and greater than 1.0 OD (antigen assay). Figure 2 compares these 2 assays by sensitivity–specificity tradeoff analysis (ROC curve analysis) at various cutoff points between positive and negative results for the orthopedic patients. Because activation and antigen assays were positive at their conventional cutoff points (more than 20% serotonin release and greater than 0.45 OD, respectively) for all 15 patients with possible HIT identified in the prospective studies, the sensitivity of each assay for HIT was high (100%). However, because the antigen assay was more likely to be positive in patients in whom HIT did not develop, the activation assay had superior operating characteristics (ie, greater diagnostic specificity) over all values of the diagnostic cutoff between positive and negative tests. The activation assay remained superior even when the antigen assay was performed at greater dilutions. For orthopedic patients, test specificity (at the conventional cutoff points) for the activation and antigen assays was 97% and 92%, respectively. For cardiac patients, the corresponding specificities were only 81% and 51% because possible HIT developed in fewer cardiac surgery patients, even though HIT-IgG antibodies formed in more.

Fig. 2.

Comparison of activation and antigen assays for HIT by receiver operating characteristics (ROC) curve analysis.

For the orthopedic patients, the activation and antigen assays are compared at various cutoff points between negative and positive results. Activation assay (SRA, serotonin release assay, thick solid line). Although the sensitivity-specificity trade-off analysis was performed at 5% increments of serotonin release, the data points for cutoff 20% or more, 50% or more, and 90% or more serotonin release are shown. Antigen assay (EIA, enzyme immunoassay, thin solid line indicating standard dilution of blood sample $150$; thin broken lines indicating higher dilutions [1/100, 1/250, 1/500, 1/750], as indicated). Although the sensitivity–specificity trade-off analysis was performed at 0.1-OD increments, the data points for cutoff of 0.5 or more, 1.0 or more, and 2.0 or more are shown. The ROC curve analysis suggests that both the activation and the antigen assays are highly informative for diagnosing HIT in this patient population. For comparison, the inset shows the full-scale ROC curve analysis for both activation and (at conventional 1/50 dilution) antigen assays together with a (theoretical) noninformative assay (shown as dashed line). The data indicate that the operating characteristics of the antigen assay are not improved by performing the assay using a greater dilution of patient serum/plasma.

Fig. 2.

Comparison of activation and antigen assays for HIT by receiver operating characteristics (ROC) curve analysis.

For the orthopedic patients, the activation and antigen assays are compared at various cutoff points between negative and positive results. Activation assay (SRA, serotonin release assay, thick solid line). Although the sensitivity-specificity trade-off analysis was performed at 5% increments of serotonin release, the data points for cutoff 20% or more, 50% or more, and 90% or more serotonin release are shown. Antigen assay (EIA, enzyme immunoassay, thin solid line indicating standard dilution of blood sample $150$; thin broken lines indicating higher dilutions [1/100, 1/250, 1/500, 1/750], as indicated). Although the sensitivity–specificity trade-off analysis was performed at 0.1-OD increments, the data points for cutoff of 0.5 or more, 1.0 or more, and 2.0 or more are shown. The ROC curve analysis suggests that both the activation and the antigen assays are highly informative for diagnosing HIT in this patient population. For comparison, the inset shows the full-scale ROC curve analysis for both activation and (at conventional 1/50 dilution) antigen assays together with a (theoretical) noninformative assay (shown as dashed line). The data indicate that the operating characteristics of the antigen assay are not improved by performing the assay using a greater dilution of patient serum/plasma.

Close modal

Immune HIT continues to be one of the most important IgG-mediated drug reactions that physicians must manage. The association between anticoagulant-induced platelet count decrease and unexpected arterial and venous thrombosis represents a paradox of considerable clinical interest. Another confounding factor is the unusual composition of the target antigen of the disorder, which is a stoichiometrically defined complex between a native platelet protein, platelet factor 4, and heparin.4-7 Antibodies against this antigenic complex can be detected using either of 2 classes of assay. One type measures the HIT-IgG antibody-induced activation of platelets in a heparin-dependent fashion,11-13 and the other detects HIT-IgG that recognize immobilized platelet factor 4–heparin antigen.4-7,14,15

Despite these recent insights into the pathogenesis of HIT, there remain important unresolved issues. For example, it is unknown why the frequency of HIT differs among patients in prospective studies.8,9 Explanations include differences in risk for forming, and clinical effects arising from, HIT-IgG antibodies that are patient population dependent. Differences in the diagnostic significance between the 2 different tests for HIT-IgG could also explain differences between studies. For these reasons, we performed both activation and antigen assays on blood samples from 744 patients studied prospectively for HIT who were treated either with UFH or LMWH, after either cardiac or orthopedic surgery.

The results of this study indicate a patient population-dependent dissociation between the risk for HIT-IgG formation and the risk for HIT among patients in whom antibodies formed. The highest frequency of HIT-IgG formation occurred after cardiac surgery: 50% of these patients had antibodies detected by antigen assay, and 20% had antibodies detected by activation assay. However, the clinical syndrome of HIT was uncommon in this group. In contrast, only 14.1% and 9.3%, respectively, of patients who received UFH after orthopedic surgery had HIT-IgG detected by antigen and activation assay. Yet this was the patient population in whom HIT was most likely to develop. Fewer orthopedic patients developed antibodies (7.5% and 3.2%, respectively) when treated with LMWH. Logistic regression analysis showed that patient population and type of heparin preparation determined antibody formation (Table 1).

Although cardiac surgery patients had the highest frequency of antibody formation, among those patients who developed antibodies, the orthopedic patients were about 20 times more likely (by odds ratio) to have HIT. Other investigators have also shown high frequencies of antibody formation in cardiac surgery patients without associated thrombocytopenia.24-26 In 2 of these studies,24,25 postoperative heparin prophylaxis was not given, so the risk for HIT might have been reduced. In the third study,26 no thrombocytopenia occurred among 16 patients receiving postoperative UFH who had positive antigen assay. Because these studies did not include large comparative patient populations, it was not possible to determine whether there were population-dependent differences in HIT risk.

Our study was not designed to determine the explanation for patient population-dependent differences in frequency and clinical significance of HIT-IgG formation. Reasons for the high frequency of antibody formation among cardiac patients could include high doses of intraoperative heparin or the release of platelet factor 4 from platelets during contact between blood and the cardiopulmonary bypass apparatus27,28; together these could increase the risk for antibody formation. Alternatively, various proinflammatory factors associated with orthopedic surgery could influence the pathogenicity of antibodies in this patient population.29 Aspirin use could not have accounted for the differences because the cardiac surgery population underwent valve replacement surgery and was routinely given anticoagulant rather than antiplatelet drugs.

We observed that activation and antigen assays are very sensitive for HIT. However, the antigen assay was more likely to detect antibodies that did not cause thrombocytopenia, ie, the diagnostic specificity for clinical HIT was lower. In addition, the operating characteristics of the antigen assay did not improve when performed using samples tested at a greater dilution (Figure 2). This finding, plus the observation that most patients with possible HIT had antibodies that caused strong platelet activation (serotonin release greater than 90%; see Figure1), suggests that the activation assay may be better able than the antigen assay to detect pathogenic antibodies with potent platelet-activating properties. Indeed, preliminary HIT antibody purification studies by Pouplard et al30 suggest that only a relatively small subset of anti-PF4–heparin antibodies has a potent platelet-activating profile.30 If activation assays are better able to detect clinically significant HIT antibodies, the situation would appear to parallel somewhat that of laboratory testing for the antiphospholipid antibody syndrome: a biologic assay (the nonspecific lupus inhibitor) has greater specificity for thrombosis than the antigen assay (anticardiolipin antibodies).31 Our data also suggest that the patient population influences the diagnostic significance of HIT-IgG test results. For either activation or antigen assay, a positive result in an orthopedic patient may be more specific for clinical HIT.

The current study indicates that HIT is a more complicated syndrome than might be expected for a drug-induced, immune-mediated thrombocytopenic disorder. There exist patient population-dependent differences in risk for HIT-IgG formation and in the thrombocytopenic potential of antibodies. There also exist differences in the diagnostic usefulness of the 2 different classes of assays to detect these antibodies. The complex nature of this syndrome may relate to the compound nature of the responsible antigen, in which both drug and autologous protein concentrations could vary in different clinical settings.

We thank Jean I. Russett and Judith Johnson for organizing the collection of plasma samples in these patients, Carol A. Smith for technical assistance with some of the assays, and Prof Robin Roberts for assistance in the statistical analysis.

Supported by operating grants A2449 and B3763 from the Heart and Stroke Foundation of Ontario (T.E.W.) and by a Research Scholarship from the Heart and Stroke Foundation of Canada (T.E.W.).

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## Author notes

Theodore E. Warkentin, Hamilton Regional Laboratory Medicine Program, Hamilton Health Sciences Centre, Hamilton General Site, 237 Barton St East, Hamilton, Ontario L8L 2X2, Canada; e-mail: twarken@fhs.mcmaster.ca.


text:
#### Cut Tree Construction from Massive Graphs

##### Takuya Akiba, Yoichi Iwata, Yosuke Sameshima, Naoto Mizuno, Yosuke Yano

The construction of cut trees (also known as Gomory-Hu trees) for a given graph enables the minimum-cut size of the original graph to be obtained for any pair of vertices. Cut trees are a powerful back-end for graph management and mining, as they support various procedures related to the minimum cut, maximum flow, and connectivity. However, the crucial drawback with cut trees is the computational cost of their construction. In theory, a cut tree is built by applying a maximum flow algorithm for $n$ times, where $n$ is the number of vertices. Therefore, naive implementations of this approach result in cubic time complexity, which is obviously too slow for today's large-scale graphs. To address this issue, in the present study, we propose a new cut-tree construction algorithm tailored to real-world networks. Using a series of experiments, we demonstrate that the proposed algorithm is several orders of magnitude faster than previous algorithms and it can construct cut trees for billion-scale graphs.

arrow_drop_up

format: plain


text:
# Finding the second focus of an ellipse and its directrix

My principle question is how do I find the other focus of the ellipse if one is (0,0)??

For context I have done the following:

A) The ellipse $E$ has eccentricity $\frac12$, focus $(0,0)$ with the line $x=-1$ as the corresponding directrix. Find an equation for $E$.

I have attempted this using $\frac{\lvert PF\rvert}{\lvert Pl\rvert}=\frac12$.

Where $P$ is an arbitrary point, $F$ is the focus and $l$ is the directrix and have ended up with the equation $$\tfrac34x^2 + \tfrac34y^2 + \tfrac12x -\tfrac14=0$$ It is the next part I am having difficulty with.

B) Find the other focus and directrix of E.

I know that in the general equation of an ellipse, the foci are given by $(c,0)$ and $(-c,0)$ with $c^2=a^2-b^2$ but I am not sure how to use this information when my equation for $E$ is not in the general form and the focus is $(0,0)$. How can it be $(0,0)$ if they're symmetric about the origin??

I tried to put my equation for $E$ in the general form but got stuck so perhaps my attempt is wrong?

• math.stackexchange.com/questions/49517/… – lab bhattacharjee Nov 8 '16 at 12:42
• Thank you, this is what I have mentioned in part B) but this does not help for a focus of (0,0) as it seems to imply the other focus is (0,0) with a directrix of x=(a^2)/0 which is wrong. Thanks anyway! – harry55 Nov 8 '16 at 12:59
• Your equation for the ellipse is clearly incorrect: both squared terms have the same coefficient, so the eccentricity of the conic described by this equation can’t be $\frac12$. Moreover, the center that one computes from this equation lies between the given focus and directrix, so the equation can’t be correct. What you’ve found is the equation of a circle that’s tangent to the directrix with center on the $x$-axis. – amd Jul 18 '17 at 20:15

In the different formulas you have different coordinate systems. The coordinate system used by your problem statement and presumably the equation you found has one focus in the origin. The symmetric focus coordinates you quote assume that the origin is in the center of symmetry, and furthermore that the first coordinate axis agrees with the major axis of the ellipse.

There are different ways to go about this. If you have computed $c$, you can just use the quoted formula to conclude that the distance between the center and the focus is $c$, so the distance between foci is $2c$, so if you know the direction where the other focus has to lie, you can easily compute its position.

If you don't have that $c$ yet, you could also start by finding the center of the ellipse. As your ellipse is axis-aligned, this is easy to do by completing the square. Once you have the center, then the second focus and directrix are what you get by reflecting the first focus and directrix in that center.

Or you could take the equational approach. You know that things have to be symmetric, so as the first directrix is one unit left of the first focus, the second directrix is one unit right of the second focus. Assume the second focus at $(f,0)$ and the second directrix at $x=f+1$. Find a second formula of the ellipse, depending on $f$, and then make the choose $f$ in such a way that the two formulas agree.

• Thanks, I have tried to complete the square to find the center of the ellipse but have ended up with (x+1/3)^2 + y^2=4/9 which is just a circle center (-1/3,0), radius 2/3 which I know is wrong. Can you see where I've gone wrong? – harry55 Nov 9 '16 at 14:20
• @harry55: I take it that my answer to your separate question on this should resolve this issue here. – MvG Nov 9 '16 at 15:25

The foci and directrices are symmetic with respect to the minor axis, so once you have the center, you can find the other focus and directrix by reflection in the line through this point parallel to the known directrix. The following solution doesn’t require first finding the equation of the ellipse.

Let $D$ be the intersection of the directrix and its perpendicular through the focus $F$ (i.e., with the major axis of the ellipse). One vertex $A$ of the ellipse is between these two points. For this vertex, $|A-F|=e|D-A|$, so $A={1\over1+e}F+{e\over1+e}D$. For the other vertex $A'$, we have $|A'-F|=e|A'-D|$, from which $A'={1\over1-e}F-{e\over1-e}D$. The center of the ellipse is their midpoint $$C=\frac12(A+A')={1\over1-e^2}F-{e^2\over1-e^2}D.$$ In this problem, $F=(0,0)$ and $D=(-1,0)$, so the center of the ellipse is at $${1\over1-\left(\frac12\right)^2}(0,0)-{\left(\frac12\right)^2\over1-\left(\frac12\right)^2}(-1,0)=-\frac14\cdot\frac43\cdot(-1,0)=\left(\frac13,0\right).$$ From this, we easily find that the other focus is $\left(\frac23,0\right)$ with corresponding directrix $x=\frac53$.

If you have the equation of the conic in hand, you can find its center by differentiating and setting the partial derivatives to zero, i.e., $$\frac32x+\frac12=0\\ \frac32y=0$$ which has solution $x=-\frac13$, $y=0$. This point is between the given focus and directrix, so as I mention in my comment to your question, this means that the equation that you’ve come up with is incorrect. That aside, once you’ve found the correct center point $(c,0)$, the other focus is the reflection in this point, namely, $(2c,0)$, with corresponding directrix $x=2c+1$.

In order to find both x- coordinates, set $y=0$ in the equation you got and solve for two $x$ values.

$$3x^2 + 2x -1 = (3 x -1) (x+1) = 0$$

$$(x= \frac13, \, x=-1)$$

Other focus is found.

Now the foci and directrix centre points have same midpoint. Find other directrix.


text:
# Comparing Automated Methods to Detect Explicit Content in Song Lyrics

The Parental Advisory Label (PAL) is a warning label that is placed on audio recordings in recognition of profanity or inappropriate references, with the intention of alerting parents of material potentially unsuitable for children. Since 2015, digital providers {--} such as iTunes, Spotify, Amazon Music and Deezer {--} also follow PAL guidelines and tag such tracks as {}explicit{''}... Nowadays, such labelling is carried out mainly manually on voluntary basis, with the drawbacks of being time consuming and therefore costly, error prone and partly a subjective task. In this paper, we compare automated methods ranging from dictionary-based lookup to state-of-the-art deep neural networks to automatically detect explicit contents in English lyrics. We show that more complex models perform only slightly better on this task, and relying on a qualitative analysis of the data, we discuss the inherent hardness and subjectivity of the task. read more

PDF Abstract

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text:
## Intermediate Algebra for College Students (7th Edition)

Divide 3 on both sides of the equation to obtain: $y=\frac{18}{3} \\y=6$ RECALL: The graph of the equation $y=k$ is a horizontal line. Every point on the line has the same y-value. Thus, to graph $y=6$, you can plot the points (0, 6), (-3, 6), and (3, 6). Then, connect them using a line.


text:
# Revision history [back]

I'm not sure what you're asking. Let us know if this is not it.

sage: arctan(tan(2*pi/3))
arctan(-sqrt(3))

Notice Sage doesn't automatically simplify this, although

sage: arctan(0)
0
sage: arctan(1)
1/4*pi

You have to do this.

sage: arctan(tan(2*pi/3)).simplify()
-1/3*pi

To get other branches, you'll want to think about arctan2.

sage: arctan2(tan(2*pi/3),1).simplify()
-1/3*pi
sage: arctan2(-tan(2*pi/3),-1).simplify()
2/3*pi

You can do

sage: arctan2?

for more details.


text:
# Field of Values and Eigenvalues

Let $A$ be an $n \times n$ matrix. Define the field of values of $A$, denoted $W(A)$, as

$W(A) := \{c \in \mathbb{C} : \exists x \in \mathbb{C}^n, \|x\|_2 = 1, x^H Ax = c \}$

The question is, suppose one knows the spectra and singular values of $A$, are there any nontrivial bounds for the distance between an eigenvalue $\lambda$ and the edge of the field of values $\partial W(A)$?

-
It might be. We know that the numerical radius $r(A)$, which is the radius of the smallest disk containing $W(A)$, satisfies $r(A)\ge\frac12\|A\|$. Besides, $\|A\|$ is the largest singular value. Therefore, if the eigenvalues are all zero, we have that the distance of $\lambda=0$ to $\partial W(A)$ is larger than or equal to half the largest singular value. – Denis Serre May 9 '11 at 20:25


text:
# Going Blind, Group or Ring?

I have several problems approaching the following problem: Let's assume that we have Alice, Bob and Cathy. Alice wants to present a message to Cathy, showing her that someone else has signed it (Bob), but without disclosing Bob's identity. She needs to show that the signature does not belong to her.

If we use blind signatures. then Alice could be the one who signed it, so we cannot be sure that it was signed from someone else.

By using group signatures, more or less we have the same problem, sure there's the group manager who could certify that, but we break the anonymity...

With ring signatures, we have the needed anonymity, Bob presents a message having himself along with others, however, if you want to make it efficient, the ring has to be quite small.

Any other ideas?

-
Basically, Alice would have to prove that she does NOT know the secret key used for signing. And unless I'm missing something, that should be impossible, because the instance of Alice that knows the key can always simulate an instance of Alice that has less knowledge. –  Maeher Feb 2 at 18:18
simple: just have Alice not be part of the (signature) group. $\;$ –  Ricky Demer Feb 2 at 21:50
To Maeher: If you're thinking with blinding, then yes that's the loophole that I see. To Ricky Not having Alice in the signature group would mean that if you have n use users, then you need to generate n groups, which for large number of users is not practical, as you might have users joining and leaving. –  absinthe Feb 3 at 0:06
How about a zero-knowledge argument of knowledge of a signing key for one verification key and a signature for a different verification key? $\;$ –  Ricky Demer Feb 3 at 0:18
Your problem could be addressed by including an identity in the keys and replacing "different verification key" with "verification key for a different identity". $\:$ Additionally, I now realize that the protocol would need to be concurrent zero-knowledge. $\:$ (Although I would normally recommend replacing stage 2 of that paper's protocol with something that has a ppt knowledge extractor, such a modification is not necessary in this case.) $\;\;\;$ –  Ricky Demer Feb 3 at 1:00


text:
# .60x=3

## Simple and best practice solution for .60x=3 equation. Check how easy it is, and learn it for the future. Our solution is simple, and easy to understand, so dont hesitate to use it as a solution of your homework.

If it's not what You are looking for type in the equation solver your own equation and let us solve it.

## Solution for .60x=3 equation:

.60x=3We simplify the equation to the form, which is simple to understand .60x=3We move all terms containing x to the left and all other terms to the right.  + 0.6x=+3We simplify left and right side of the equation.  + 0.6x=+3We divide both sides of the equation by 0.6 to get x. x=5`

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text:
24th of September, 2008

1. Let $\rho$ denote the charge density, and $\omega$ the angular velocity.

Let us divide the cylinder into thin cylindrical shells. A shell of radius $r$ will have a surface current density $K=\rho \omega rdr$. To find the magnetic field of such a shell, imagine that we bend it into a torus with perimeter $p$. In the limit $p\to \infty$ the torus is an infinitely long cylindrical shell, and we do not need to distinguish between its inner and outer perimeters. Using Ampère’s law and integrating along the “centre ring” of the torus we find that $B={\mu }_{0}K$ inside the torus. This also holds for a straight cylindrical shell. Outside $B=0$.

Summing up the contributions of the shells we find that the magnetic field at distance $r$ from the axis of the cylinder is

$B\left(r\right)=\rho \omega {\mu }_{0}{\int }_{r}^{R}{r}^{\prime }d{r}^{\prime }=\rho \omega {\mu }_{0}\phantom{\rule{3.26212pt}{0ex}}\left({R}^{2}-{r}^{2}\right)∕2.$

2. The potential created by a bounded charge distribution at point $\stackrel{\to }{R}$ is
$V\left(\stackrel{\to }{R}\right)=\frac{1}{4\pi {\epsilon }_{0}}\int \frac{\rho \left(\stackrel{\to }{r}\right)}{\left|\stackrel{\to }{R}-\stackrel{\to }{r}\right|}\phantom{\rule{3.26212pt}{0ex}}{d}^{3}\stackrel{\to }{r}.$

For simplicity let us consider $\frac{1}{4\pi {\epsilon }_{0}}\equiv 1$ below.

Let the components of vectors $\stackrel{\to }{R}$ and $\stackrel{\to }{r}$ be denoted

$\stackrel{\to }{R}=\left(X,Y,Z\right)=\left({X}_{1},{X}_{2},{X}_{3}\right)$

$\stackrel{\to }{r}=\left(x,y,z\right)=\left({x}_{1},{x}_{2},{x}_{3}\right),$

and let $F\left(\stackrel{\to }{r}\right)=1∕\left|\stackrel{\to }{R}-\stackrel{\to }{r}\right|$. Now let us expand $F\left(\stackrel{\to }{r}\right)$ in Taylor series, and write the potential as

 $\begin{array}{cc}\begin{array}{rl}V\left(\stackrel{\to }{R}\right)& =\underset{{V}^{\left(0\right)}}{\underbrace{\int F\left(0\right)\rho \left(\stackrel{\to }{r}\right)\phantom{\rule{3.26212pt}{0ex}}{d}^{3}\stackrel{\to }{r}}}\phantom{\rule{3.26212pt}{0ex}}+\\ & +\underset{{V}^{\left(1\right)}}{\underbrace{\int \left(\sum _{i}{\frac{\partial F}{\partial {x}_{i}}|}_{\stackrel{\to }{r}=0}\phantom{\rule{0em}{0ex}}{x}_{i}\right)\rho \left(\stackrel{\to }{r}\right)\phantom{\rule{3.26212pt}{0ex}}{d}^{3}\stackrel{\to }{r}}}\phantom{\rule{3.26212pt}{0ex}}+\\ & +\underset{{V}^{\left(2\right)}}{\underbrace{\int \left(\frac{1}{2}\sum _{i,j}{\frac{{\partial }^{2}F}{\partial {x}_{i}\partial {x}_{j}}|}_{\stackrel{\to }{r}=0}\phantom{\rule{0em}{0ex}}{x}_{i}{x}_{j}\right)\rho \left(\stackrel{\to }{r}\right)\phantom{\rule{3.26212pt}{0ex}}{d}^{3}\stackrel{\to }{r}}}\phantom{\rule{3.26212pt}{0ex}}+\cdots \end{array}& \end{array}$ (1)

Here ${V}^{\left(0\right)}$ is the monopole term in the expansion of the potential, ${V}^{\left(1\right)}$ is the dipole term, ${V}^{\left(2\right)}$ the quadrupole, etc.

Let us define $A$, ${A}_{i}$, ${A}_{ij}$, … in the following way:

$\begin{array}{llll}\hfill \frac{A}{R}& =F\left(0\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \frac{{A}_{i}}{{R}^{3}}& ={\frac{\partial F}{\partial {x}_{i}}|}_{\stackrel{\to }{r}=0}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \frac{{A}_{ij}}{{R}^{5}}& ={\frac{{\partial }^{2}F}{\partial {x}_{i}\partial {x}_{j}}|}_{\stackrel{\to }{r}=0}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \dots & \phantom{\rule{2em}{0ex}}& \hfill \end{array}$

Now the terms of the potential’s expansion can be written as

$\begin{array}{llll}\hfill {V}^{\left(0\right)}& =\frac{1}{R}A\int \rho \left(\stackrel{\to }{r}\right)\phantom{\rule{3.26212pt}{0ex}}{d}^{3}\stackrel{\to }{r}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {V}^{\left(1\right)}& =\frac{1}{{R}^{3}}\sum _{i}{A}_{i}\int {x}_{i}\rho \left(\stackrel{\to }{r}\right)\phantom{\rule{3.26212pt}{0ex}}{d}^{3}\stackrel{\to }{r}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {V}^{\left(2\right)}& =\frac{1}{{R}^{5}}\sum _{i,j}\frac{{A}_{ij}}{2}\int {x}_{i}{x}_{j}\rho \left(\stackrel{\to }{r}\right)\phantom{\rule{3.26212pt}{0ex}}{d}^{3}\stackrel{\to }{r}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \dots & \phantom{\rule{2em}{0ex}}& \hfill \end{array}$

Note that the coefficients $A$, ${A}_{i}$, … only depend on $\stackrel{\to }{R}$, but not $\stackrel{\to }{r}$. If we carried out the tedious calculations, for the first few $A$-constants we would get $A=1$, ${A}_{i}={X}_{i}$, ${A}_{ij}=3{X}_{i}{X}_{j}-{\delta }_{ij}{R}^{2}$, where ${\delta }_{ij}$ is the Kronecker delta.

The monopole term in the potential is just proportional to the total charge $\int \rho \left(\stackrel{\to }{r}\right)\phantom{\rule{3.26212pt}{0ex}}{d}^{3}\stackrel{\to }{r}=Q$, and does not depend on $\stackrel{\to }{r}$, hence it does not depend on the choice of origin of our coordinate system. What happens to the dipole term if the origin of the coordinate system is changed, or, equivalently, the charges are shifted by a vector $\stackrel{\to }{a}=\left({a}_{1},{a}_{2},{a}_{3}\right)$?

 $\begin{array}{cc}\begin{array}{rl}{V}^{\left(1\right)}& =\frac{1}{{R}^{3}}\sum _{i}{A}_{i}\int \left({x}_{i}-{a}_{i}\right)\rho \left(\stackrel{\to }{r}\right)\phantom{\rule{3.26212pt}{0ex}}{d}^{3}\stackrel{\to }{r}\\ & =\frac{1}{{R}^{3}}\left(\sum _{i}{A}_{i}\int {x}_{i}\rho \left(\stackrel{\to }{r}\right)\phantom{\rule{3.26212pt}{0ex}}{d}^{3}\stackrel{\to }{r}-\sum _{i}{A}_{i}{a}_{i}\int \rho \left(\stackrel{\to }{r}\right)\phantom{\rule{3.26212pt}{0ex}}{d}^{3}\stackrel{\to }{r}\right)\\ & =\frac{1}{{R}^{3}}\left(\sum _{i}{A}_{i}\int {x}_{i}\rho \left(\stackrel{\to }{r}\right)\phantom{\rule{3.26212pt}{0ex}}{d}^{3}\stackrel{\to }{r}-Q\sum _{i}{A}_{i}{a}_{i}\right)\end{array}& \end{array}$

Note that if the total charge (and the monopole term in the potential) is 0, then ${V}^{\left(1\right)}$ does not change when the charges are shifted.

Now suppose that ${V}^{\left(1\right)}=0$ for any $\stackrel{\to }{R}$, and thus $\int {x}_{i}\rho \left(\stackrel{\to }{r}\right)\phantom{\rule{3.26212pt}{0ex}}{d}^{3}\stackrel{\to }{r}=0$ for each $i$. When the charges are shifted, ${V}^{\left(2\right)}$ will be

 $\begin{array}{cc}\begin{array}{rl}{V}^{\left(2\right)}& =\frac{1}{{R}^{5}}\sum _{i,j}\frac{{A}_{ij}}{2}\int \left({x}_{i}-{a}_{i}\right)\left({x}_{j}-{a}_{j}\right)\rho \left(\stackrel{\to }{r}\right)\phantom{\rule{3.26212pt}{0ex}}{d}^{3}\stackrel{\to }{r}\\ & =\frac{1}{{R}^{5}}\sum _{i,j}\frac{{A}_{ij}}{2}\int {x}_{i}{x}_{j}\rho \left(\stackrel{\to }{r}\right)\phantom{\rule{3.26212pt}{0ex}}{d}^{3}\stackrel{\to }{r}-\frac{1}{{R}^{5}}\sum _{i}\left(\sum _{j}{A}_{ij}{a}_{j}\right)\int {x}_{i}\rho \left(\stackrel{\to }{r}\right)\phantom{\rule{3.26212pt}{0ex}}{d}^{3}\stackrel{\to }{r}+\\ & \phantom{\rule{1em}{0ex}}+\frac{1}{{R}^{5}}\sum _{i,j}\frac{{A}_{ij}{a}_{i}{a}_{j}}{2}\int \rho \left(\stackrel{\to }{r}\right)\phantom{\rule{3.26212pt}{0ex}}{d}^{3}\stackrel{\to }{r}\end{array}& \end{array}$

Here we used the fact that ${A}_{ij}={A}_{ji}$ to simplify the result. Note again that if $\int \rho \left(\stackrel{\to }{r}\right)\phantom{\rule{3.26212pt}{0ex}}{d}^{3}\stackrel{\to }{r}=0$ and $\int {x}_{i}\rho \left(\stackrel{\to }{r}\right)\phantom{\rule{3.26212pt}{0ex}}{d}^{3}\stackrel{\to }{r}=0$, then ${V}^{\left(2\right)}$ does not change when shifting the charges.

It can be shown in a similar way that any ${V}^{\left(n\right)}$ will be independent of the choice of origin of the coordinate system if ${V}^{\left(i\right)}=0$ for all $i and all $\stackrel{\to }{R}$.

3. Adding one more “step” to the infinite ladder will not change it. Let ${R}_{L}$ denote the resistance of the ladder.

The configuration of figure 1 must have resistance ${R}_{L}$ too:

${R}_{L}=R+{\left(\frac{1}{R}+\frac{1}{{R}_{L}}\right)}^{-1}$

The positive solution of this equation is

${R}_{L}=R\frac{1+\sqrt{5}}{2}.$

4. If we let a current $I$ flow into point $A$, the current though the resistor between $A$ and $B$ will be $I∕4$ (because of the symmetrical arrangement of the resistors, see fig. 2). Therefore the voltage drop across resistor $AB$ is ${U}_{1}=RI∕4$.

Similarly, if a current $I$ is flowing out from point $B$, the voltage drop across resistor $AB$ is ${U}_{2}=RI∕4$.

According to the principle of superposition, when we have both a current $I$ flowing into point $A$ and a current $I$ flowing out from point $B$, the voltage drop across $AB$ will be ${U}_{total}={U}_{1}+{U}_{2}=2×RI∕4$. The resistance between points $A$ and $B$ is ${U}_{total}∕I=R∕2$.

5. Let us calculate the force between the wires in both cases:
1. Using Gauß’s law we find the electric field at distance $r$ from a wire $E=1∕2\pi {\epsilon }_{0}\phantom{\rule{3.26212pt}{0ex}}\lambda ∕r$ ($\lambda$ is the linear charge density). The force on unit length $ds$ of the other wire is
$\frac{dF}{ds}=\frac{1}{2\pi {\epsilon }_{0}}\frac{{\lambda }^{2}}{r}.$

2. The moving wires have a current $I=v\lambda$. It is easy to find the magnetic field using Ampère’s law (integrate along circular loops around the wire): $B={\mu }_{0}∕2\pi \phantom{\rule{3.26212pt}{0ex}}v\lambda ∕r$. The force due to the magnetic interaction is attractive, so the total force is
$\frac{dF}{ds}=\frac{1}{2\pi {\epsilon }_{0}}\frac{{\lambda }^{2}}{r}-\frac{{\mu }_{0}}{2\pi }\frac{{v}^{2}{\lambda }^{2}}{r}.$

The results in points (a) and (b) are not identical. However, the two presented situations are exactly the same setup, viewed from different reference frames.

This is a good illustration of Maxwell’s equations not being invariant to the classical Galilei–transformation (${x}^{\prime }=x-vt$; ${t}^{\prime }=t$). If we take into account the change in charge density due to the relativistic Lorentz–contraction, ${\lambda }^{\prime }=\lambda ∕\sqrt{1-{v}^{2}∕{c}^{2}}$, it is easy to show that the force is the same in both reference frames. (Use the relation ${\epsilon }_{0}{\mu }_{0}=1∕{c}^{2}$.)

Back to the main page


text:
# Christopher Spencer Physics 180D Professor Gary Williams 14 May 2014

Resonant frequencies measured at (0.69,1.93,2.92,3.69,4.31,4.82) Hz. Q factors varied from the sweeps and from measurements of alpha largely but this can be due to the effect of high amplitudes on the higher frequencies and the natural error in trying to measure the time decaying exponential of a wave that has been turned off from the amplifier.

# Introduction

Dispersive water waves can be seen of surfaces of water due to the inclusion of gravity and surface tension terms in the dispersion relation. These can be created in the lab using a container filled with water to a height,h, and mixed with some salt and fotoflow. Using a function generator and and amplifier sweeps from 0.5-10 Hz and made and observed resonant frequencies are found. Further investigation of the data, one can find the phase velocities of these dispersive waves and q factors. Then by changing the apparatus around, solitions can be made and observed by driving the container at resonant frequencies and adding pertubations to the surface can induce a self induced wave packet.

# Theory

At low frequencies water is an incompressible fluid and the density is constant. At the surface the pressure oscillation is $$\delta p=0$$ due to a plane wave not being able to propogate because of boundary conditions. When factoring gravity in then there is a pressure oscillation that culminates in a bulge in the water. Now $$\delta p=\rho g \delta h$$, where $$\rho$$ is the liquid density and $$\delta h$$ is the height of the wave. Mass conversation is described by $$\nabla ^2 \phi=0$$ where $$\phi$$ is the velocity potential. Using eulers equation it is found the the pressure oscillation becomes $$\delta p=-\frac{\partial \phi}{\partial t} \rho$$. We can look at the boundary conditions for this lab are at z=0,h the velocity of the fluid is $$v_z=0$$ at z=0 and $$\delta p=\rho g \delta h$$ at z=h.The velocity potential takes on the form $\phi (x,z,t)=(A\cosh kz+ b\sinh kz)e^{i(kz-\omega t)}$ After applying the boundary conditions listed above, the dispersion relation for these waves are shown to be $\omega ^2=gk\tanh(kh)$ where k is the wavenumber=$$\frac{2\pi}{\lambda}$$, h is the height of the wave and $$\omega$$ is the angular frequency. For a shallow water wave make the assumption $$kh<<1$$, then recall that the phase velocity is $$v_{phase}=\frac{\omega}{k}$$ and in the shallow water limit $$v_{phase}=\sqrt{gh}$$. For the general dispersion relation, the phase velocity is $v_{phase}=\sqrt{\frac{g\tanh kh}{k}}$ At higher frequencies the wavelength becomes comparable the length of the water so include the surface tension in the dispersion relation as $\omega^2=k(g+\frac{\delta}{\rho} k^2)\tanh(kh)$

# Apparatus and Procedure

Water waves were observed in a container that had water up to a height of $$0.03$$ meters and the length of the container was $$.38$$ meters long. A function generator that is connected to an amplifier that drives a speaker that is connected to the glass container at one end. The container is on a bearing stage that allows the container to move with the frequency that is driving the speaker. The movement will cause the surface to oscillate with the oscillator and if a resonant mode is reached, a maximum on the surface oscillation will be seen and this is detected by two wires on the other side of the container that conduct due to the salt in the water. The resistance between the two wires will oscillate with the oscillation in the height of the water. This is then connected to a wheatstone bridge which is a circuit used to find an unknown resistance by balancing two legs of a bridge circuit. Of the four resistances in the wheatstone bridge one is unknown and one of the known resistances is variable. If the ratio of the resistances in the known leg is equal to the ratio of the resistances in the unknown leg, then no current will flow through the galvanometer. If theres an inbalance than you can vary the variable resistance until no current flows. A lock in amplifier is used to measure the voltage from the equal ratios of resistance. This voltage is the reference voltage and the two voltages from the waves themselves are used at A and -B as shown in the diagram below. after filtering the output and converting AC to DC,a DC rms amplitude that is equivalent to the waves amplitude is fed into LabView to be recorded. Data was taken this way for sweeps from 0.5-3.5 Hz,3.5-7 Hz, and 7-10 Hz. The function generator was set to a sweep time of 1999s. From these sweeps, resonances were found and recorded. Note that when searching for resonances between 7-10 Hz nonlinear and odd behavior was observed. After 3 times trying to get that data it was decided that these frequencies were too difficult to get and used 0.5-5 Hz for data analysis.


text:
emacs-orgmode
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## Re: [O] Collapse LaTeX source before start of main document?

 From: Chris Malone Subject: Re: [O] Collapse LaTeX source before start of main document? Date: Thu, 5 May 2011 10:23:50 -0400

Hi Matt and Suvayu,

@Matt: I can get your example to work fine for html export, but not
LaTeX export where the entire contents of the headline are removed as
well.  =org-version=:  7.5 (release_7.5.105.g8d0c) if that makes any
difference.

@Suvayu: Thanks for the suggestion.  That is essentially what I am
doing now with the #+LaTeX_HEADER lines, but the problem is that I
have a bunch of things which must follow the \begin{document} in
LaTeX.  I know I could put this in a separate file and just \input{}
it; I wanted to see if there was a way to keep everything together in
a single file, but also have the ability to fold this preliminary
stuff.

Chris

On Wed, May 4, 2011 at 6:25 PM, Suvayu Ali <address@hidden> wrote:
> Hi Chris,
>
> I just finished writing my master's thesis in org-mode.
>
> On Wed, 4 May 2011 11:36:58 -0400
>
>> I'm working on using org mode for my PhD thesis.  I'd like to do this
>> in one large file where each headline is a single chapter.  Naturally
>> in a thesis there needs to be a lot of front matter: title page,
>> abstract, signature page, etc.  Right now, I'm doing this with
>> something like:
>
> For all formatting stuff, I just used
>
>
> For my abstract I used
>
>
> where abstract.tex is a simple tex file with something like:
>
> \abstract{ My abstract }
>
> If you wish to include this inside the \begin{document}..\end{document}
> block then you can use the same latex snippet but without
>
> I hope this helps and good luck with the thesis. :)
>
> --
> Suvayu
>
> Open source is the future. It sets us free.
>
>


text:
# If c is the velocity of light, h is Planck’s constant and G is Gravitational constant are taken as fundamental quantities…

Q: If c is the velocity of light, h is Planck’s constant and G is Gravitational constant are taken as fundamental quantities. Them the dimensional formula of mass is

Sol. c = [LT-1 ] …(i)

h = [ML2 T-1 ] …(ii)

G = [M-1 L3 T-2 ] …(iii)

Solving (ii) and (iii)

h/G = [(ML2 T-1)/(M-1 L3 T-2 )]

=[M2 L-1 T1 ]

Substituting (i) in above

h/G = M2/c

⇒ [M] = [h1/2 G-1/2 c1/2 ]


text:
# stagnation temperature - Swedish translation – Linguee

Av Bo Almqvist - Kungl. Gustav Adolfs Akademien

Specifikationer  stagnation från engelska till svenska. Redfox Free är ett gratis lexikon som innehåller 41 språk. The rate of methyl mercury (MeHg) production during summer stagnation are more important than possible temperature changes in small polyhumic lakes with  The staggered, hydraulic control avoids a possible, short-term melt stagnation and moulding and can be combined in one housing with temperature control. Recommended hot water temp.: 60°C cold position to prevent stagnation in the connecting lines. Let the water run until the water temperature is constant.

Effective thermal capacity, incl. fluid (per gross area, AG). C/m2. kJ/(Km2). Transversal. Infrared Thermal Imaging Technique for Temperature Measurement in Various Energy ANALYSIS OF STAGNATION REGION HEAT TRANSFER OF TWO  Stagnation temperature (°C). 78.9.

## Climate induced thermocline change has an effect on the

Consider the Here T0 is called the stagnation (or total) temperature. The pressure a fluid attains when brought to rest isentropically is called the stagnation pressure P0. For ideal gases with constant specific heats, P0 is related to the static pressure of the fluid by $\frac{p_0}{p}=(\frac{T_0}{T})^{k⁄k-1}$ In aviation, stagnation temperature is known as total air temperature and is measured by a temperature probe mounted on the surface of the aircraft. Total air temperature - Wikipedia Since there is no downstream turbine, a ramjet combustor can safely operate at stoichiometric fuel:air ratios, which implies a combustor exit stagnation temperature of the order of 2400 K for kerosene. 2017-12-07 Solar collectors to reach very high temperatures, particularly during power failures or periods when there is minimal energy demand.

### TO GROW ON - svensk översättning - bab.la engelskt-svenskt

Grâce à la fonction Safety, il est possible de prérégler la température premier demi-litre pour des boissons, le matin ou après une période de stagnation. Europa de senaste decennierna och än ses ingen stagnation på efterfrågan. Introducing a central receiver system for industrial high-temperature process  There is no single thermometer measuring the global temperature. alltså in för att förneka alla tendenser till ekonomisk stagnation eller kris. „The Elusive Absolute Surface Air Temperature (SAT)“ in deutsch:“ Die i fraktskeppens maskinrum, medan satelliterna visade stagnation.

gravity force). Several properties that can be represented by this theoretical process which include temperature, pressure, and density et cetera and denoted by the subscript "$$0$$.'' stagnation temperature: translation. stabdymo temperatūra statusas T sritis Energetika apibrėžtistemperatūra statusas T sritis Energetika apibrėžtis In thermodynamics and fluid mechanics, stagnation temperature is the temperature at a stagnation point in a fluid flow. At a stagnation point the speed of the fluid is zero and all of the kinetic energy has been converted to internal energy and is added to the local static enthalpy . 2014-06-12 · For the perfect gas, the stagnation temperature is derived from the isentropic total temperature equation: Tt = T * [1 + M^2 * (gamma-1)/2] where Tt is the total temperature, T is the static temperature at a given altitude, M is the Mach number, and gamma is the ratio of specific heats for a calorically perfect gas and has a constant value of 1.4.
Hur tar jag tag i mitt liv

Stagnation is a real problem and Varmesystem Villa Aston Villa are by turning the temperature (room temperature indicator I think) down! och befolkningskris med en påtaglig ekonomisk och kulturell stagnation som följd. A 2´500-year long temperature-sensitive tree-ring record in far north-. Dr. Andrew Huberman Exposure to light at a temperature minimum to and oils alleviate stagnation and congestion with the energy Hidden  Temperature is also important for orchid care. While orchids tolerate cooler or warmer temperatures throughout their normal growing season, they need cooler  därmed motverkar stagnation och syrebrist.

2014-06-12 Thus, for a stagnation temperature of 1050 K, the static temperature is very low, only 50 K. This explains the term “cold” used for this type of wind tunnel. Usually the heated air is not kept in the tunnel but ejected to the atmosphere via a 3-way valve. This “ stagnation temperature ” occurs at approximately 80°C for a simple flat-plate solar collector. If water, for example, is passed through passages in the plate collector, then it will stabilize at a lower temperature and the water will extract some of the energy in being usefully heated up. Stagnation temperature.
Klippan dhl

For use with dangerous fluids temperature rating is –20°C +60°C and  collectors Open to ambient collectors Results and reporting Standard stagnation temperature Objective Testing under stagnation conditions Measurement and  impinging jets with denser or sparse locations, the temperature of the movement on surface temperature and heat flux was captured in stagnation point of jet in. Absorber capacity [litres]: 0,85 - Maximum operating pressure [bar]: 10 - Net weight [Kg]: 36 - Stagnation temperature [°C]: 201  such an event and how much the sea ice is lost due to growth stagnation during the period of the enhanced forcing and temperature increase  Recommended hot water temp.: 60°C cold position to prevent stagnation in the connecting lines. Let the water run until the water temperature is constant. Evaluation of the impact of stagnation temperatures in different prototypes of low concentration PVT solar panels2015Inngår i: ISES Solar World Congress 2015,  Absorbing coating: SS-ALN-Cu.

It is the temperature at the point at which a streamline is brought to rest isentropically. It means that if the continuity of a streamline is obstructed by some object and while doing so the process follwed is isentropic then the point will at wh Stagnation Temperature - YouTube. Stagnation Temperature. Watch later. Share. Copy link. Info.
Söka jobb vad ska man säga

malvinerna
nils lundwall


text:
# Math Help - Hello everyone, I need your help...

1. ## Hello everyone, I need your help...

Hello, studying for a test and I have a review question that I just can't seem to get.. if anyone can help me it would be greatly appreciated...

A container of oil is supposed to hold 1000 ml of oil. We need to be sure that the standard deviation of the oil containers is less than 20 ml. We randomly select 10 cans of oil with a mean of 997 ml and a standard deviation of 32 ml. Use these samples to construct a 95% confidence interval for the true value of the standard deviation.

2. Originally Posted by iza0iza
Hello, studying for a test and I have a review question that I just can't seem to get.. if anyone can help me it would be greatly appreciated...

A container of oil is supposed to hold 1000 ml of oil. We need to be sure that the standard deviation of the oil containers is less than 20 ml. We randomly select 10 cans of oil with a mean of 997 ml and a standard deviation of 32 ml. Use these samples to construct a 95% confidence interval for the true value of the standard deviation.
Read this: Estimating a population standard devation or variance

3. Originally Posted by iza0iza
Hello, studying for a test and I have a review question that I just can't seem to get.. if anyone can help me it would be greatly appreciated...

A container of oil is supposed to hold 1000 ml of oil. We need to be sure that the standard deviation of the oil containers is less than 20 ml. We randomly select 10 cans of oil with a mean of 997 ml and a standard deviation of 32 ml. Use these samples to construct a 95% confidence interval for the true value of the standard deviation.
n is small (n < 30), so use the following confidence interval:

$\bar{x} \pm t_{\alpha/2}(n-1)\frac{s}{\sqrt{n}}$

Where $\bar{x} = 997, s = 32, \alpha = 0.05, n = 10$

4. Originally Posted by colby2152
n is small (n < 30), so use the following confidence interval:

$\bar{x} \pm t_{\alpha/2}(n-1)\frac{s}{\sqrt{n}}$

Where $\bar{x} = 997, s = 32, \alpha = 0.05, n = 10$
Hi colby. Unfortunately, the question asks for a confidence interval for the population standard deviation, not for the population mean ...... (I nearly made the same mistake).

Since n is small (n < 30), the required confidence interval is of the form:

$\frac{(n-1)s^2}{\chi^2_{\frac{\alpha}{2}, ~ df=n-1}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1 - \frac{\alpha}{2}, ~ df=n-1}}$.

Note that there are two critical values that must be read from the appropriate tables (due to the asymmetry of the chi-square distribution .....)


text:
# Why doesn't linear algebra approach work for Iso

## Recommended Posts

Well, I'm still working on the top half of my map, to figure out which tile my mouse clicked on. I think this approach would work, but it yields ridiculous numbers. My approach is as following: Since my tiles are 64 * 32, and I set (0,0) at (512,0) screen coord. I create an equation for the left most line: x = -2y + 512; Then I create an equation for my mouse, x = 1/2y + N; and then I figure out the N and find point of intersection -2y + 512 = 1/2y + N; so I find y, and use that y to find the x at Point of intersection. To find the isometric x, I look at how many 64s is the x at P.O.I away from 512 and to find y, I look at how many 64s is the mouse x away from the P.O.I x I think this method makes perfect sense, but it's simply not working right now. Anyone see anything wrong with it?

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Quote:
 Then I create an equation for my mouse, x = 1/2y + N;

The mouse cursor is not constrained to a certain line. What is N?
Moreover, what are the tilemap's conventions? Are you perhaps mixing up screen pixel coordinates and tilemap coordinates?

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I don't understand what you mean, but I can assure you that linear algebra works perfectly fine for isometric projections.

Assuming your tiles are 64 pixels wide and 32 pixels high, your world x-axis is pointing down and right on the screen, world y-axis is pointing down and left:
 ---------------------------------> screen x-axis|                /\  |               /  \  |              /    \  |             /      \  |     world |/_      _\| world|    y-axis              x-axis||Vscreen y-axis

You'll get a projection matrix (projecting world coordinates to screen coordinates) as follows:
[32 -32]
[16 16]

Your world (x,y) projected into screen coordinates (x',y') and adjusted 512 pixels will be
x' = 32*x - 32*y + 512
y' = 16*x + 16*y

When you want to find out where in world-coordinates your mouse cursor is, you use the inverse of the projection matrix:
[ 1/64 1/32]
[-1/64 1/32]

To calculate the mouse position in world coordinates (x',y') from it's screen coordinates (x,y):
x' = (x-512)/64 + y/32
y' = -(x-512)/64 + y/32

// Edit: a bit of a miscalculation fixed

[Edited by - Luctus on August 11, 2008 11:30:51 AM]

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Quote:
Original post by LorenzoGatti
Quote:
 Then I create an equation for my mouse, x = 1/2y + N;

The mouse cursor is not constrained to a certain line. What is N?
Moreover, what are the tilemap's conventions? Are you perhaps mixing up screen pixel coordinates and tilemap coordinates?

oh, I'm setting up an equation of the line perpendicular to the x = -2y + 512 line that goes through the point where the mouse is at

It doesn't really concern the tilemap tho, cuz I was just taking the mouse coordinate and use the equation to figure out which tile it should be

I put on a tilemap, but that's only for visual aid

[quote]Original post by Luctus
I don't understand what you mean, but I can assure you that linear algebra works perfectly fine for isometric projections.

Assuming your tiles are 64 pixels wide and 32 pixels high, your world x-axis is pointing down and right on the screen, world y-axis is pointing down and left:
You'll get a projection matrix (projecting world coordinates to screen coordinates) as follows:
[32 -32]
[16 16]

Your world (x,y) projected into screen coordinates (x',y') and adjusted 512 pixels will be
x' = 32*x - 32*y + 512
y' = 16*x + 16*y

When you want to find out where in world-coordinates your mouse cursor is, you use the inverse of the projection matrix:
[ 1/64 1/32]
[-1/64 1/32]

Lol, I need to catch up on linear algebra :(
summer made me forget everything, I was actually doing somethign simpler

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# Given cyclohexane in a chair conformation, substitute two of the H labels with Br to construct the mosi stable isomer

###### Question:

Given cyclohexane in a chair conformation, substitute two of the H labels with Br to construct the mosi stable isomer and conformation of 1,3-dibromocyclohexane Br H H H H

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What are the vertices of AA'B'C' produced by Tz3. 6> (AABC) = AA'BC' ? 0 A A' (6,6) B' (6,4), â‚¬' (3.3) 0 A' (0_6)B' (0,-8),C' (-3.9) A' (0.6) B' (0,4) C' (-3 3) 6), B' (6.-8),C' (3,...
##### Are the zeros/roots of a parabola the inverse of the x-intercepts?
Are the zeros/roots of a parabola the inverse of the x-intercepts?...
##### Questions: Determine which type of mathematical object each of the following is (You are encouraged to review the definitions from the slides of the last lecture:) (Choose one for each column:)Rejection Region: Type Error:Level:Type 2 Error:Power:A numberA numberA number.A number.A number.A set;A set:A set:A set:A set:A function_A function_A functionA functionfunction:
Questions: Determine which type of mathematical object each of the following is (You are encouraged to review the definitions from the slides of the last lecture:) (Choose one for each column:) Rejection Region: Type Error: Level: Type 2 Error: Power: A number A number A number. A number. A number. ...
##### Use a convergence test of your choice to determine whether the following series converges or diverges....
Use a convergence test of your choice to determine whether the following series converges or diverges. 0 Σ ke 5k k= 1 Select the correct choice below and fill in the answer box to complete your choice. (Type an exact answer.) A. The limit of the terms of the series is This is not 0, so the ser...
##### Sales for Ross Pro’s Sports Equipment are expected to be 51,000 units for the coming month....
Sales for Ross Pro’s Sports Equipment are expected to be 51,000 units for the coming month. The company likes to maintain 10 percent of unit sales for each month in ending inventory. Beginning inventory is 14,000 units. How many units should the firm produce for the coming month?   ...
##### The blood bank director should:(Select the best answer this is tricky think about it)Use the screening test because it has high sensitivity (95%) so it picks Up most people who have the disease.Use the screening test because it has high PVN (99.9%) and you want to know what blood is really safe (negative) to transfuse:Not use the screening test because it has very low PVP (1.86%) and you want to know what blood is really not safe to transfuse:Not use the screening test because the sensitivity is
The blood bank director should: (Select the best answer this is tricky think about it) Use the screening test because it has high sensitivity (95%) so it picks Up most people who have the disease. Use the screening test because it has high PVN (99.9%) and you want to know what blood is really safe (...
##### The table below presents the actual monthly sales for company over the first six months of 1986. Using a 3-month moving average; the mean squared error for this forecast isMonthsActual Sales 270 263 275January 1986 February 1986 March 1986April 1986262May 986250 278June 1986192.332.7815.6522.7813.22
The table below presents the actual monthly sales for company over the first six months of 1986. Using a 3-month moving average; the mean squared error for this forecast is Months Actual Sales 270 263 275 January 1986 February 1986 March 1986 April 1986 262 May 986 250 278 June 1986 192.33 2.78 15.6...
##### How many grams of Ca does 32 grams of O2 (oxygen)have to react to give 72 grams of CaO? Answer is given 40 grams but if 32 grams of oxygen has to react then it means that there are two oxygen atoms and then the formula will be CaO2
How many grams of Ca does 32 grams of O2 (oxygen)have to react to give 72 grams of CaO? Answer is given 40 grams but if 32 grams of oxygen has to react then it means that there are two oxygen atoms and then the formula will be CaO2...
##### A researcher was attempting to quantify the amount of dichlorodiphenyltrichloroethane (DDT) in spinach with gas chromatography using chloroform internal standard. To begin, the researcher examined a sample containing 6.71 mg/L DDT standard and 2.90 mg/L chloroform the internal standard, producing peak areas of 3679 and 11885, respectively: Then, the researcher collected 11.25 g of spinach, homogenized the sample, and extracted the DDT using established method (assume 100% extraction), producing
A researcher was attempting to quantify the amount of dichlorodiphenyltrichloroethane (DDT) in spinach with gas chromatography using chloroform internal standard. To begin, the researcher examined a sample containing 6.71 mg/L DDT standard and 2.90 mg/L chloroform the internal standard, producing pe...
##### AedeadnnolFede (queetienWhat Is the cou-ct namuoel Us compound KSCN?Seectone: Potassium sulfur cyanldePotasslum cyanate: Potasolum cyanlde Potoeelum thlooyenete
Aedeadnnol Fede (queetien What Is the cou-ct namuoel Us compound KSCN? Seectone: Potassium sulfur cyanlde Potasslum cyanate: Potasolum cyanlde Potoeelum thlooyenete...
##### Question 2 of 5 (1 point) A random sample of 45 cars in the drive-thru of...
Question 2 of 5 (1 point) A random sample of 45 cars in the drive-thru of a popular fast food restaurant revealed an average bill of $19.59 per car. The population standard deviation is$6.29. Estimate the mean bill for all cars from the drive-thru with 93% confidence. Round your answers to two deci...
##### Lim?(x?3)??(?x-?3)/(x^2-9)?lim?(x?4)??(x-4)/(?(x+5)-3)?
lim?(x?3)??(?x-?3)/(x^2-9)? lim?(x?4)??(x-4)/(?(x+5)-3)?...
##### Blackboard ecu edu bbcswebdav/pid-10850417-dt-content-rid-65051329_V/courses MATH (9) Suppose (G(r) +I/ g(z). Then which of the statements below is true (a) C(r) is an antiderivative of g(r) _ 1 (6) g(1) - 1 is an antiderivative of G(r) G(r) Is the derivative of g(r). (d) g(m has no antiderivative (e) None of the above T blackboardecuedu bbcswebdav/pid-10850417-dt-content-rid-65051333_Vcourses MATH?(10) Suppose G(r)' 31 = g(r). Then which of the statements below is true: (a) G(r) is an anti
blackboard ecu edu bbcswebdav/pid-10850417-dt-content-rid-65051329_V/courses MATH (9) Suppose (G(r) +I/ g(z). Then which of the statements below is true (a) C(r) is an antiderivative of g(r) _ 1 (6) g(1) - 1 is an antiderivative of G(r) G(r) Is the derivative of g(r). (d) g(m has no antiderivative (...


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# Summarize the number and amount of reimbursements between $24 and$25 by employee number. Do the ...

Summarize the number and amount of reimbursements between $24 and$25 by employee number. Do the same for amounts between $4,900 and$5,000. (Hint: Create filters for each of these dollar ranges.) Print the results.

Analyze the results and provide possible explanations for the results. Identify reimbursements for certain employees that may need further investigation.