Model: sucharush/MNLP_M3_document_encoder
Author: sucharush
Pipeline: sentence-similarity
Library: sentence-transformers
Tags: sentence-transformers, safetensors, bert, sentence-similarity, feature-extraction, generated_from_trainer, dataset_size:98112, loss:MultipleNegativesRankingLoss, arxiv:1908.10084, arxiv:1705.00652, base_model:thenlper/gte-small, base_model:finetune:thenlper/gte-small, model-index, text-embeddings-inference, endpoints_compatible, region:us
Base models (finetune): id: thenlper/gte-small
Card metadata: tags: sentence-transformers, sentence-similarity, feature-extraction, generated_from_trainer, dataset_size:98112, loss:MultipleNegativesRankingLoss
base_model: thenlper/gte-small
widget: source_sentence: How does a photocell control outdoor lighting?
sentences: To solve this problem, we can use the binomial probability formula:

P(X = k) = C(n, k) * p^k * (1-p)^(n-k)

where:
- P(X = k) is the probability of exactly k successes (faulty keyboards) in n trials (laptops produced)
- C(n, k) is the number of combinations of n items taken k at a time (n! / (k!(n-k)!))
- p is the probability of success (5% or 0.05)
- n is the number of trials (400 laptops)
- k is the number of successes (20 faulty keyboards)

However, we want to find the probability of at least 20 faulty keyboards, so we need to find the sum of probabilities for k = 20, 21, 22, ..., 400.

P(X >= 20) = 1 - P(X < 20) = 1 - Σ P(X = k) for k = 0 to 19

Now, we can calculate the probabilities for each value of k and sum them up:

P(X >= 20) = 1 - Σ C(400, k) * 0.05^k * 0.95^(400-k) for k = 0 to 19

Using a calculator or software to compute the sum, we get:

P(X >= 20) ≈ 1 - 0.0184 = 0.9816

So, the probability that at least 20 laptops will have a faulty keyboard is approximately 98.16%., A photocell controls outdoor lighting by detecting the level of ambient light. It automatically turns the lights on when it becomes dark and off when it becomes light, functioning as a light-dependent switch for energy efficiency and convenience., Glycosylation with β-N-acetylglucosamine (O-GlcNAcylation) is one of the most complex post-translational modifications. The cycling of O-GlcNAc is controlled by two enzymes: UDP-NAc transferase (OGT) and O-GlcNAcase (OGA). We recently reported that endothelin-1 (ET-1) augments vascular levels of O-GlcNAcylated proteins. Here we tested the hypothesis that O-GlcNAcylation contributes to the vascular effects of ET-1 via activation of the RhoA/Rho-kinase pathway. Incubation of vascular smooth muscle cells (VSMCs) with ET-1 (0.1 μM) produces a time-dependent increase in O-GlcNAc levels. ET-1-induced O-GlcNAcylation is not observed when VSMCs are previously transfected with OGT siRNA, treated with ST045849 (OGT inhibitor) or atrasentan (ET(A) antagonist). ET-1 as well as PugNAc (OGA inhibitor) augmented contractions to phenylephrine in endothelium-denuded rat aortas, an effect that was abolished by the Rho kinase inhibitor Y-27632. Incubation of VSMCs with ET-1 increased expression of the phosphorylated forms of myosin phosphatase target subunit 1 (MYPT-1), protein kinase C-potentiated protein phosphatase 1 inhibitor protein (protein kinase C-potentiated phosphatase inhibitor-17), and myosin light chain (MLC) and RhoA expression and activity, and this effect was abolished by both OGT siRNA transfection or OGT inhibition and atrasentan. ET-1 also augmented expression of PDZ-Rho GEF (guanine nucleotide exchange factor) and p115-Rho GEF in VSMCs and this was prevented by OGT siRNA, ST045849, and atrasentan., source_sentence: A torus has a major radius of 5 cm and a minor radius of 3 cm. Find the volume of the torus.
sentences: To find the Hausdorff dimension of the Koch curve, we can use the formula:

Hausdorff dimension (D) = log(N) / log(1/s)

where N is the number of self-similar pieces and s is the scaling factor.

For the Koch curve, each line segment is divided into four segments, each of which is 1/3 the length of the original segment. Therefore, N = 4 and s = 1/3.

Now, we can plug these values into the formula:

D = log(4) / log(1/3)

D ≈ 1.2619

So, the Hausdorff dimension of the Koch curve is approximately 1.2619., To find the volume of a torus, we can use the formula:

Volume = (π * minor_radius^2) * (2 * π * major_radius)

where minor_radius is the minor radius of the torus and major_radius is the major radius of the torus.

Given that the major radius is 5 cm and the minor radius is 3 cm, we can plug these values into the formula:

Volume = (π * 3^2) * (2 * π * 5)

Volume = (π * 9) * (10 * π)

Volume = 90 * π^2

The volume of the torus is approximately 282.74 cubic centimeters., The purpose of the present study was to elucidate the mechanisms of action mediating enhancement of basal glucose uptake in skeletal muscle cells by seven medicinal plant products recently identified from the pharmacopeia of native Canadian populations (Spoor et al., 2006). Activity of the major signaling pathways that regulate glucose uptake was assessed by western immunoblot in C2C12 muscle cells treated with extracts from these plant species. Effects of extracts on mitochondrial function were assessed by respirometry in isolated rat liver mitochondria. Metabolic stress induced by extracts was assessed by measuring ATP concentration and rate of cell medium acidification in C2C12 myotubes and H4IIE hepatocytes. Extracts were applied at a dose of 15-100 microg/ml. The effect of all seven products was achieved through a common mechanism mediated not by the insulin signaling pathway but rather by the AMP-activated protein kinase (AMPK) pathway in response to the disruption of mitochondrial function and ensuing metabolic stress. Disruption of mitochondrial function occurred in the form of uncoupling of oxidative phosphorylation and/or inhibition of ATPsynthase. Activity of the AMPK pathway, in some instances comparable to that stimulated by 4mM of the AMP-mimetic AICAR, was in several cases sustained for at least 18h post-treatment. Duration of metabolic stress, however, was in most cases in the order of 1h., source_sentence: Consider the elliptic curve given by the equation $y^2=x^3-2x+5$ over the field of rational numbers $\mathbb{Q}$. Let $P=(1,2)$ and $Q=(-1,2)$ be two points on the curve. Find the equation of the line passing through $P$ and $Q$ and show that it intersects the curve at another point $R$. Then, find the coordinates of the point $R$.
sentences: Fifteen novel derivatives of D-DIBOA, including aromatic ring modifications and the addition of side chains in positions C-2 and N-4, had previously been synthesised and their phytotoxicity on standard target species (STS) evaluated. This strategy combined steric, electronic, solubility and lipophilicity requirements to achieve the maximum phytotoxic activity. An evaluation of the bioactivity of these compounds on the systems Oryza sativa-Echinochloa crus-galli and Triticum aestivum-Avena fatua is reported here. All compounds showed inhibition profiles on the two species Echinochloa crus-galli (L.) Beauv. and Avena fatua L. The most marked effects were caused by 6F-4Pr-D-DIBOA, 6F-4Val-D-DIBOA, 6Cl-4Pr-D-DIBOA and 6Cl-4Val-D-DIBOA. The IC(50) values for the systems Echinochloa crus-galli-Oryza sativa and Avena fatua-Triticum aestivum for all compounds were compared. The compound that showed the greatest selectivity for the system Echinochloa crus-galli-Oryza sativa was 8Cl-4Pr-D-DIBOA, which was 15 times more selective than the commercial herbicide propanil (Cotanil-35). With regard to the system Avena fatua-Triticum aestivum, the compounds that showed the highest selectivities were 8Cl-4Val-D-DIBOA and 6F-4Pr-D-DIBOA. The results obtained for 6F-4Pr-D-DIBOA are of great interest because of the high phytotoxicity to Avena fatua (IC(50) = 6 µM, r(2) = 0.9616)., To find the equation of the line passing through points $P=(1,2)$ and $Q=(-1,2)$, we first find the slope of the line. Since the y-coordinates of both points are the same, the slope is 0. Therefore, the line is horizontal and its equation is given by:

$y = 2$

Now, we want to find the point $R$ where this line intersects the elliptic curve $y^2 = x^3 - 2x + 5$. Since we know that $y=2$, we can substitute this value into the equation of the curve:

$(2)^2 = x^3 - 2x + 5$

Simplifying, we get:

$4 = x^3 - 2x + 5$

Rearranging the terms, we have:

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

We know that $x=1$ and $x=-1$ are solutions to this equation since they correspond to the points $P$ and $Q$. To find the third solution, we can use synthetic division or factor the polynomial. Factoring, we get:

$(x-1)(x+1)(x-1) = 0$

So, the third solution is $x=1$. Substituting this value back into the equation of the line, we find the corresponding y-coordinate:

$y = 2$

Thus, the third point of intersection is $R=(1,2)$. However, in the context of elliptic curves, we should take the "sum" of the points $P$ and $Q$ as the negative of the third intersection point. Since $R=(1,2)$, the negative of this point is given by $-R=(1,-2)$. Therefore, the "sum" of the points $P$ and $Q$ on the elliptic curve is:

$P + Q = -R = (1,-2)$., The use of geospatial analysis may be subject to regulatory compliance depending on the specific application and the jurisdiction in which it is used. For example, the use of geospatial data for marketing purposes may be subject to privacy regulations, and the use of geospatial data for land use planning may be subject to environmental regulations. It is important to consult with legal counsel to ensure compliance with all applicable laws and regulations., source_sentence: Does sLEDAI-2K Conceal Worsening in a Particular System When There Is Overall Improvement?
sentences: To determine whether the Systemic Lupus Erythematosus Disease Activity Index 2000 (SLEDAI-2K) is valid in identifying patients who had a clinically important overall improvement with no worsening in other descriptors/systems. Consecutive patients with systemic lupus erythematosus with active disease who attended the Lupus Clinic between 2000 and 2012 were studied. Based on the change in the total SLEDAI-2K scores on last visit, patients were grouped as improved, flared/worsened, and unchanged. Patients showing improvement were evaluated for the presence of new active descriptors at last visit compared with baseline visit. Of the 158 patients studied, 109 patients had improved, 38 remained unchanged, and 11 flared/worsened at last visit. In the improved group, 11 patients had a new laboratory descriptor that was not present at baseline visit. In those 11 patients, this new laboratory descriptor was not clinically significant and did not require a change in disease management., To find the dot product of two vectors using their magnitudes, angle between them, and trigonometry, we can use the formula:

Dot product = |A| * |B| * cos(θ)

where |A| and |B| are the magnitudes of the vectors, and θ is the angle between them.

In this case, |A| = 5 units, |B| = 8 units, and θ = 60 degrees.

First, we need to convert the angle from degrees to radians:

θ = 60 * (π / 180) = π / 3 radians

Now, we can find the dot product:

Dot product = 5 * 8 * cos(π / 3)
Dot product = 40 * (1/2)
Dot product = 20

So, the dot product of the two vectors is 20., To determine if hospitals that routinely discharge patients early after lobectomy have increased readmissions. Hospitals are increasingly motivated to reduce length of stay (LOS) after lung cancer surgery, yet it is unclear if a routine of early discharge is associated with increased readmissions. The relationship between hospital discharge practices and readmission rates is therefore of tremendous clinical and financial importance. The National Cancer Database was queried for patients undergoing lobectomy for lung cancer from 2004 to 2013 at Commission on Cancer-accredited hospitals, which performed at least 25 lobectomies in a 2-year period. Facility discharge practices were characterized by a facility's median LOS relative to the median LOS for all patients in that same time period. In all, 59,734 patients met inclusion criteria; 2687 (4.5%) experienced an unplanned readmission. In a hierarchical logistic regression model, a routine of early discharge (defined as a facility's tendency to discharge patients faster than the population median in the same time period) was not associated with increased risk of readmission (odds ratio 1.12, 95% confidence interval 0.97-1.28, P = 0.12). In a risk-adjusted hospital readmission rate analysis, hospitals that discharged patients early did not experience more readmissions (P = 0.39). The lack of effect of early discharge practices on readmission rates was observed for both minimally invasive and thoracotomy approaches., source_sentence: Does systemic administration of urocortin after intracerebral hemorrhage reduce neurological deficits and neuroinflammation in rats?
sentences: Intracerebral hemorrhage (ICH) remains a serious clinical problem lacking effective treatment. Urocortin (UCN), a novel anti-inflammatory neuropeptide, protects injured cardiomyocytes and dopaminergic neurons. Our preliminary studies indicate UCN alleviates ICH-induced brain injury when administered intracerebroventricularly (ICV). The present study examines the therapeutic effect of UCN on ICH-induced neurological deficits and neuroinflammation when administered by the more convenient intraperitoneal (i.p.) route. ICH was induced in male Sprague-Dawley rats by intrastriatal infusion of bacterial collagenase VII-S or autologous blood. UCN (2.5 or 25 μg/kg) was administered i.p. at 60 minutes post-ICH. Penetration of i.p. administered fluorescently labeled UCN into the striatum was examined by fluorescence microscopy. Neurological deficits were evaluated by modified neurological severity score (mNSS). Brain edema was assessed using the dry/wet method. Blood-brain barrier (BBB) disruption was assessed using the Evans blue assay. Hemorrhagic volume and lesion volume were assessed by Drabkin's method and morphometric assay, respectively. Pro-inflammatory cytokine (TNF-α, IL-1β, and IL-6) expression was evaluated by enzyme-linked immunosorbent assay (ELISA). Microglial activation and neuronal loss were evaluated by immunohistochemistry. Administration of UCN reduced neurological deficits from 1 to 7 days post-ICH. Surprisingly, although a higher dose (25 μg/kg, i.p.) also reduced the functional deficits associated with ICH, it is significantly less effective than the lower dose (2.5 μg/kg, i.p.). Beneficial results with the low dose of UCN included a reduction in neurological deficits from 1 to 7 days post-ICH, as well as a reduction in brain edema, BBB disruption, lesion volume, microglial activation and neuronal loss 3 days post-ICH, and suppression of TNF-α, IL-1β, and IL-6 production 1, 3 and 7 days post-ICH., A perfect number is a positive integer that is equal to the sum of its proper divisors (excluding itself). The first perfect numbers are 6, 28, 496, and 8128. Perfect numbers can be generated using the formula 2^(p-1) * (2^p - 1), where p and 2^p - 1 are both prime numbers.

The first five (p, 2^p - 1) pairs are:
(2, 3) - 6
(3, 7) - 28
(5, 31) - 496
(7, 127) - 8128
(13, 8191) - 33,550,336

To find the 6th perfect number, we need to find the next prime number p such that 2^p - 1 is also prime. The next such pair is (17, 131071). Using the formula:

2^(17-1) * (2^17 - 1) = 2^16 * 131071 = 65,536 * 131071 = 8,589,869,056

So, the 6th perfect number is 8,589,869,056., In type theory, the successor function $S$ is used to represent the next number in the sequence. When you apply the successor function $S$ three times to the number 0, you get:

1. $S(0)$, which represents 1.
2. $S(S(0))$, which represents 2.
3. $S(S(S(0)))$, which represents 3.

So, the result of applying the successor function $S$ three times to the number 0 in type theory is 3.
pipeline_tag: sentence-similarity
library_name: sentence-transformers
metrics: cosine_accuracy@1, cosine_accuracy@3, cosine_accuracy@5, cosine_accuracy@10, cosine_precision@1, cosine_precision@3, cosine_precision@5, cosine_recall@1, cosine_recall@3, cosine_recall@5, cosine_ndcg@10, cosine_mrr@10, cosine_map@100
model-index: name: SentenceTransformer based on thenlper/gte-small
results: task: type: logging
name: Logging
dataset: name: ir eval
type: ir-eval
metrics: type: cosine_accuracy@1
value: 0.9291020819957809
name: Cosine Accuracy@1
verified: no, type: cosine_accuracy@3
value: 0.9819315784646427
name: Cosine Accuracy@3
verified: no, type: cosine_accuracy@5
value: 0.9933963129413923
name: Cosine Accuracy@5
verified: no, type: cosine_accuracy@10
value: 0.9984407961111621
name: Cosine Accuracy@10
verified: no, type: cosine_precision@1
value: 0.9291020819957809
name: Cosine Precision@1
verified: no, type: cosine_precision@3
value: 0.32731052615488093
name: Cosine Precision@3
verified: no, type: cosine_precision@5
value: 0.19867926258827848
name: Cosine Precision@5
verified: no, type: cosine_recall@1
value: 0.9291020819957809
name: Cosine Recall@1
verified: no, type: cosine_recall@3
value: 0.9819315784646427
name: Cosine Recall@3
verified: no, type: cosine_recall@5
value: 0.9933963129413923
name: Cosine Recall@5
verified: no, type: cosine_ndcg@10
value: 0.9670096227619588
name: Cosine Ndcg@10
verified: no, type: cosine_mrr@10
value: 0.9565327512887825
name: Cosine Mrr@10
verified: no, type: cosine_map@100
value: 0.9565967419425125
name: Cosine Map@100
verified: no
Likes: 0 | All-time downloads: 11
Created: 2025-06-05 13:59:10