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# Clinical Trials for Disease-Modifying Therapies in Alzheimer’s Disease: A Primer, Lessons Learned, and a Blueprint for the Future

#### Abstract

Alzheimer’s disease (AD) has no currently approved disease-modifying therapies (DMTs), and treatments to prevent, delay the onset, or slow the progression are urgently needed. A delay of 5 years if available by 2025 would decrease the total number of patients with AD by 50% in 2050. To meet the definition of DMT, an agent must produce an enduring change in the course of AD; clinical trials of DMTs have the goal of demonstrating this effect. AD drug discovery entails target identification followed by high throughput screening and lead optimization of drug-like compounds. Once an optimized agent is available and has been assessed for efficacy and toxicity in animals, it progresses through Phase I testing with healthy volunteers, Phase II learning trials to establish proof-of-mechanism and dose, and Phase III confirmatory trials to demonstrate efficacy and safety in larger populations. Phase III is followed by Food and Drug Administration review and, if appropriate, market access. Trial populations include cognitively normal at-risk participants in prevention trials, mildly impaired participants with biomarker evidence of AD in prodromal AD trials, and subjects with cognitive and functional impairment in AD dementia trials. Biomarkers are critical in trials of DMTs, assisting in participant characterization and diagnosis, target engagement and proof-of-pharmacology, demonstration of disease-modification, and monitoring side effects. Clinical trial designs include randomized, parallel group; delayed start; staggered withdrawal; and adaptive. Lessons learned from completed trials inform future trials and increase the likelihood of success.

## INTRODUCTION

Alzheimer’s disease (AD) is a progressive neurodegenerative disease that produces gradual decline in cognition and function [1, 2]. The most common form, late onset AD, becomes symptomatic in late life but biomarker studies show that the amyloid protein considered the major risk factor for the disease begins to accumulate in the brain up to 20 years before symptoms begin [3].

The total number of individuals with AD will double every 20 years [4]. The annual cost of AD currently exceeds $230 billion and the total annual cost will exceed$1 trillion by 2050 if means of preventing, delaying, slowing the progression, or improving the symptoms are not found [5].

There is a high rate of negative clinical trials in AD drug development programs; 99% of drugs tested between 2002 and 2014 showed no drug-placebo difference and only one drug was approved by the US Food and Drug Administration (FDA) during that period [6]. Drugs in the AD pipeline include agents intended to intervene in the basic biology of AD and modify disease progression, symptomatic cognitive enhancers, and drugs to treat neuropsychiatric symptoms [7, 8].

The greatest need in AD drug development is for disease-modifying therapies (DMTs) that will delay or slow the clinical course of AD by intervening in the processes leading to cell death [9]. Approximately two-thirds of the current AD drug development pipeline involves DMTs—either immunotherapies or small molecule agents administered orally [7, 8]. In this paper, we describe the methods for AD clinical trials of DMTs, review past failures to identify lessons for AD drug development, and look ahead to new approaches to improving AD drug development and optimizing success in bringing new treatments to patients with AD or those at high risk for the disorder.

## OVERVIEW

### Populations

Phases of AD are recognized; these are not distinct stages but represent a seamless progression from a high risk state in which amyloid is present in the brain in the form of neuritic plaques, to prodromal AD with episodic memory impairment (in the typical presentation of AD) and biomarker evidence of AD, to AD dementia with cognitive and functional impairment characterized as mild, moderate or severe [38] (Fig. 6). Although these phases represent progression along a seamless spectrum of severity, they are artificially divided for purposes of clinical trials. Tools and outcomes appropriate for one phase of disease (e.g., preclinical) are not the same as those one would choose for later phases (e.g., mild-moderate AD). Table 2 provides examples of cognitive and functional measures used as outcome measures for different phases of AD [39, 40, 63–70].

##### Fig.6

Phases of Alzheimer’s disease (AD) as defined by cognitive, functional, and biomarker observations. Trial goals for each phase are noted.

##### Table 2

Outcome tools used for the progressive phases of Alzheimer’s disease [39, 40, 63–70]

Clinical outcomes in AD dementia trials are well established and have been used to demonstrate efficacy of cholinesterase inhibitors and memantine. Cognitive measures for mild-moderate AD dementia include the ADAS-cog [39] and the Neuropsychological Test Battery [66]. Common secondary measures include the CDR-sb [40], Clinical Global Impression of Change (CGIC), and the Neuropsychiatric Inventory [71]. Dual outcomes are required in AD dementia trials and include a cognitive measure with a functional or global outcome.

Prodromal trials commonly use a composite endpoint comprised of cognitive and functional elements or of cognitive elements derived from several scales. Composite endpoints include the CDR-sb [40], the AD Composite Scale (ADCOMS) [69], and the integrated AD Rating Scale (iADRS) [70]. The FDA has indicated that demonstration of both cognitive and functional benefit is necessary for drug approval in the prodromal phase of AD; a drug-placebo difference on a composite scale should not depend entirely on differences in cognition [72]. Some trials of DMTs include both patients with prodromal AD and those with mild AD dementia; the differences in these populations is arbitrary, and the groups can be usefully combined to facilitate recruitment of a broader population and show benefit in patients who have more than minimal impairment.

Prevention trials include primary prevention studies involving participants with no cognitive symptoms and no state biomarker changes of AD or secondary prevention studies including participants who have no cognitive symptoms but in whom amyloid imaging or CSF amyloid measures show that amyloidosis is present. Studies of asymptomatic participants with autosomal dominant mutations often have mixtures of some patients with amyloid abnormalities and some without, offering the possibility of evaluating a DMT as either primary or secondary prevention [73, 74]. Highly sensitive cognitive measures are combined with biomarkers to determine the impact of anti-amyloid therapies [63, 67, 73, 74]. Participants in this stage of preclinical or presymptomatic AD show very mild cognitive decline that may provide an opportunity to establish a drug-placebo difference in cognitive change [75, 76]. Biomarkers reasonably likely to predict future cognitive decline include amyloid imaging and tau imaging. Tau PET correlates better with cognitive decline and MRI measures of brain atrophy and may provide more insight into DM than amyloid measures [77, 78].

### Clinical trial design

The most common Phase III design for DMT trials is the randomized, parallel group, placebo controlled, two or more arm, 18–24 month trial. The primary outcome is the drug-placebo difference at trial end on co-primary clinical and functional outcomes or clinical and global outcomes. Biomarker measures typically include MRI volumetrics; amyloid PET (if the agent has a mechanism expected to impact fibrillar amyloid); and CSF Aβ, total tau, and p-tau. Additional biomarkers might be chosen depending on drug MOA and specifics of the trial. Drug-placebo differences at trial end are analyzed for both clinical and biomarker outcomes. Analyses that offer supporting data expected in DM include change in slope of decline, increasing drug-placebo difference over time, and delay to milestones captured in the data (e.g., in a trial of prodromal patients, the percent of patients at each time point who have progressed to a diagnosis of dementia or advanced from a CDR score of 0.5 to a CDR score of 1). These supporting analyses can be affected by symptomatic agents and do not by themselves prove DM. Clinical and biomarker data are expected to be correlated if they are mediated by the same mechanism [79].

The delayed start and staggered withdrawal designs provide evidence of DM without depending on biomarkers. They demonstrate an enduring change in the course of the disease in comparison with a group begun on treatment earlier (in the case of the delayed start design) or withdrawn from therapy (in the case of the staggered withdrawal design) [80–82]. These trials have been difficult to implement and have had limited use in programs attempting to show DM. The switch from placebo to active therapy when a trial is terminated and participants enter an open label extension (all are on active therapy) provides an opportunity for a delayed start observation [83], although the absence of blinding at this stage of the trial could bias the observations. This open-label delayed start analysis could add support to a claim of DM without providing definitive evidence.

Adaptive clinical trial designs use data from the on-going trial to make decisions about trial conduct. For example, the Dominantly Inherited AD-Treatment Unit (DIAN-TU) uses an adaptive strategy for dose-selection of test agents [84]. Adaptive strategies can be used for dose, treatment duration, sample size, and entry criteria. The decision structure must be comprehensively pre-specified but adaptive designs have the advantage of responding to the in-trial observations and can save time and resources while optimizing the opportunity to demonstrate a drug-placebo difference [85].

Another resource-saving strategy in clinical trial design and analysis is the incorporation of futility analyses at a time when a sufficient number of patients have been exposed to treatment for a sufficiently long period time to predict the possible outcomes. If the drug-placebo difference at the time of the analysis suggests that the study has a very low possibility of finding a drug-placebo difference at trial conclusion, the trial can be stopped [64, 86]. Futility analyses avoid exposing patients to agents and potential side effects when a positive conclusion of the trial is deemed highly unlikely. Criteria for futility are evolving; they must be liberal enough to insure that potentially viable drugs are not terminated prematurely and conservative enough that trials with very little chance of success are not continued.

The sample size of the trial is determined by the anticipated effect size of the intervention, the variability of the key measurements, and the desired length of the trial. Assuming that a slowing of 20% or more is clinically meaningful for participants and families, the typical trial for a DMT anticipates including 600–1000 subjects per arm and observing them for 18–24 months [87]. Individuals with more severe disease have faster rates of decline. Prodromal patients who are ApoE4 carriers decline more rapidly than those who are not carriers [88]. The decline in the placebo group is critical to assessing the efficacy of the intervention and decline on placebo is a critical determinant of the success of a trial.

## LESSONS LEARNED FROM TRIALS OF DMTs

There have been frequent failures in attempts to develop new drugs for AD, and 100% of DMT development programs have failed [6]. Every trial, however, is a learning opportunity and many lessons have been learned that will assist in future drug development [89].

### Animal models of AD provide limited evidence of efficacy

Animal models of AD are an important means of investigating efficacy and toxicity in the preclinical state prior to exposing humans to possibly toxic or ineffective compounds. Many of the tg animal models overexpress the amyloid protein leading to cortical plaques similar to those observed in human AD [90]. These genetically engineered animals have abnormalities of amyloid metabolism but generally lack other aspects of human AD; they lack tau or cell death and have limited inflammatory changes [91]. The tg mice have mild cognitive changes but do not develop severe dementia equivalent to the human disease. Many types of therapy have been successful in reducing amyloid abnormalities in these animals and have often lead to improved cognitive performance on tests such as Morris Water Maze or Novel Object Recognition [90]. None of these successes at the preclinical level has predicted success at the human level. The animals serve as important gateways in the drug development process showing that they impact specific pathways; advancing a drug to human testing that did not succeed as expected in animals would be unwise. The models, however, recreate limited aspects of human AD such as amyloidosis and cannot be taken as models of the full spectrum of pathology of human AD or predictors of human benefit [23].

Another concern with regard to animal models is their reproducibility [92]. If an experiment cannot be reproduced within a single model or across related models then its ability to predict human outcomes is suspect. Strain, age, gender, handler behavior, diet, and light conditions may all influence animal behavior. Randomization and sample size are important aspects of animal trial design that have sometimes been ignored [93]. Lack of rigor with regard to these aspects of animal model testing may contribute to the lack of reproducibility both across models and in translating results from animals to humans.

### Establish BBB penetration in Phase I

BBB penetration is shown in preclinical studies by the effects of drug on behavioral studies and post-exposure necropsy. Differences between rodent and human BBB function, especially activity of p-gp transporter make extrapolation of animal model results to humans uncertain, requiring demonstration of BBB penetration in Phase I FIH studies [34]. Tarenflurbil is an example of an agent advanced as treatment for AD with in vivo activity in animal models but likely low entrance into the CNS in humans [94]. Before candidate agents exit Phase I, investigators should establish BBB penetration, the plasma/CSF ratio, and the relationship of predicted human brain exposure to concentrations associated with benefit in animal models.

### Determine a maximum tolerated dose in Phase I

Dose escalation studies in Phase I and dose refinement studies in Phase II should provide confidence in the dose(s) selected for Phase III. In particular, it is important to establish a MTD whenever possible to ensure that the highest possible doses have been explored. In some cases, occupancy studies may allow conclusions about dosing without an MTD if the receptor is fully occupied at lower doses. In other situations, solubility or physical features may limit the administered dose and the MTD cannot be determined. Beyond these exceptional circumstances, an MTD should be determined. Without an MTD, failure to show a drug-placebo difference in Phase II or Phase III will raise questions about the adequacy of the dose.

### The diagnosis of AD should be supported by biomarkers

An important learning is the relatively large number of individuals who have a prodromal AD or AD dementia phenotype but are not amyloid-bearing when studied with amyloid PET [42]. These non-amyloid individuals have suspected non-Alzheimer pathology (SNAP) and are presumed not to have AD. They should be excluded from trials of agents for AD. Table 3 shows the percentage of patients meeting clinical criteria for prodromal AD or mild AD dementia who are amyloid-bearing [42]. Amyloid is more common in those with ApoE genotypes but genetic characterization is insufficient to ensure the presence of amyloid. To be confident that the trial population has AD, amyloid imaging or CSF evidence of the AD Aβ/tau signature should be collected (Fig. 4).

##### Table 3

Amyloid PET findings in patients meeting clinical criteria for prodromal AD or mild AD dementia (stratified by ApoE genotype) [42]

 Group Amyloid Positive Amyloid Positive All 61% 39% All prodromal AD 50% 50% Prodromal ApoE4 carriers 71% 29% Prodromal ApoE4 non-carriers 31% 69% All mild AD dementia 75% 25% Mild AD dementia ApoE4 carriers 90% 10% Mild AD dementia ApoE4 non- carriers 58% 42%

### Assure target engagement in Phase II

DM is supported by an impact on “downstream” measures of cell death such as MRI atrophy, CSF tau, or possibly other biomarkers of neuronal degeneration such as neurofilament light chain protein [54]. These downstream consequences can reasonably be expected only if the “upstream” target of the pharmacologic intervention is successful. Target engagement measures will depend on the MOA of the candidate therapy. BACE inhibitors, gamma secretase inhibitors, and gamma secretase modulators will have an effect on amyloid production as measured by stable isotope-labeled kinetics (SILK) [95]. BACE inhibitors will also inhibit BACE activity as measured in the CSF and reflected in sAβPPβ, a by-product of BACE activity; gamma secretase modulators result in Aβ fragments of 15/16 amino acid lengths in the CSF which are not normally present in AD [95–97]. Proof of pharmacology is one goal of Phase II and compounds should not be advanced to Phase III without well documented support for a pharmacologic effect.

### Establish a dose-response relationship in Phase II

Dosing approaches in Phase II ideally establish a low dose that is ineffective, one or two mid-range doses that are effective, and a high dose that is not tolerated and not acceptable. A dose-response on clinical or biomarker measures increases confidence in the pharmacology of the molecule. Regulatory agencies usually seek assurance that patients are given the lowest effective dose to ensure that they are not being exposed to unnecessary side effects. Doses established in Phase II inform decisions of which dose should be advanced to Phase III. Drug formulation decisions should be completed in Phase II prior to Phase III.

### Collect multiple biomarkers to assess outcomes

Knowledge of the neurobiology of AD is incomplete. Systems biology studies demonstrate that AD biology is complex [98] and biomarkers provide limited windows onto this complex and ill-understood disease. Although working models of the order of events in AD have been constructed, none have been proven and none have guided successful DMT development. Agnostic approaches to biomarkers (e.g., amyloid; tau, neurodegeneration; A/T/N) are used to acknowledge the exploratory nature of our biomarker documentation of drug effects [99]. To support DM as the outcome of a therapy, trial sponsors should collect A/T/N biomarker data, emerging biomarkers, and biomarkers specifically linked to the mechanism of the intervention to gain a comprehensive view of the impact of treatment.

### Recruitment is a major challenge

Trial recruitment is a difficult process and each population—cognitively normal at-risk participants for prevention trials, minimally impaired biomarker positive participants for prodromal AD trials, and cognitive and functionally impaired participants for AD dementia trials—have unique requirements for identification, recruitment, informed consent, and retention in the trial. There are too few highly functioning trial sites in the world. The world’s populations are generally poorly educated about clinical trials and often have few opportunities to participate. Many trials spend more time in the recruitment phase of the trial than in the drug exposure phase. Slow recruitment slows the cycle time of trials and increases their cost. Many AD-concerned organizations are constructing responses to this challenge. The Global Alzheimer Platform (GAP) network of trial sites in the US and the European Prevention of Alzheimer’s Disease (EPAD) initiatives are among the leaders of the attempt to reduce recruitment times and accelerate trials [100, 101].

### Global trials have greater variability

One response to slow recruitment is to include many trial sites with each site recruiting only a few participants to the trial. In most trials, each site is expected to contribute 6–12 participants, but many sites contribute only 1 or 2 participants. This amplifies “noise” in the data and decreases the ability to demonstrate a drug-placebo difference.

Globalization of trials creates another set of challenges. Sites distributed around the world are culturally and linguistically diverse, have different standards of health care, and include participants with different histories of nutrition and levels of education. Trial sites are highly variable in terms of experience, expertise, training, and infrastructure. Local hospital and university institutional review boards (IRBs) may have limited experience with hosting and reviewing AD trials [102]. Global sites impose challenges in terms of drug manufacturing and distribution, supply lines, biomarker collection, laboratory availability, and data collection and quality assurance. The result of this complexity is that populations recruited into trials from around the world vary in terms of age, education, genotype, and other clinical characteristics, and they progress somewhat differently in clinical trials [103, 104]. North America and Western European trial populations are similar and results are likely to be most interpretable if these populations comprise the majority of the study population.

Efficacy and safety data are needed on all populations where the agents will be marketed; smaller trials in local populations may be the best way to address these needs.

### Comprehensive trial networks are needed to conduct AD trials

Conducting clinical trials is demanding and requires expertise, commitment, and infrastructure. Some academic medical centers support trials while others do not, industry sponsors support trials but tend not to support trial infrastructure. In the US, the National Center for Advancing Translational Sciences (NCATS) sponsors Clinical and Translational Science Awards (CTSAs) to provide trial infrastructure in major university medical centers [105]. Trial networks are currently re-created for each trial and raters are re-trained on the same outcomes for each trial. Each institution often has its own IRB for reviewing trials. Legal review of contracts further slows trial initiation. Construction of a highly efficient trial network with standing non-redundant training, and a central IRB are goals of GAP and EPAD [100, 101].

### Negative trials may indicate an ineffective drug or a failed trial

The failure to show a drug-placebo difference at the end of a trial may be due to lack of efficacy of the candidate therapy or flawed conduct of the clinical trial. Table 4 summarizes the reasons for negative outcomes in trials. Drug-related reasons for negative trials include lack of efficacy and excessive toxicity [106]. In some cases, the dose range has not been adequately explored in early drug development and a negative trial opens the question of whether the agent might have been efficacious at higher doses. Such agents must return to Phase I for dose escalation trials and sponsors rarely have an interest in pursuing this alternative. Trial-related reasons for failed trials of DMTs include lack of decline in the placebo group, enrollment of non-AD patients, and excessive measurement variability.

##### Table 4

Reasons for failure to show a drug-placebo difference at the end of a clinical trial of a disease-modifying agent. AD, Alzheimer’s disease

 Drug-related •Lack of efficacy of the agent •Inappropriately low dosing of an effective agent •Excessive toxicity or lack of tolerability leading to high discontinuation rates in the active treatment arms •Excessive toxicity or lack of tolerability leading to early termination of the trial Trial-related •Lack of decline in the placebo group •Recruitment of non-AD patients into trials requiring an AD substrate for drug benefit to occur •Excessive measurement variability •Lack of measurable effect of active comparator drugs (if available)

### Placebo decline determines drug-placebo difference

Successful DMTs will slow the course of decline in AD. Slowing of decline is established by contrasting the decline in the active treatment group with the trajectory of the placebo group. The placebo trajectory will determine the drug-placebo difference at end of trial. The rate of decline of the placebo group is a crucial consideration in understanding the treatment effect. Placebo groups with SNAP patients do no decline as rapidly as those with confirmed AD, emphasizing the importance of confirming the diagnosis of AD in trial participants [107]. Slow decline in the placebo group will minimize the drug-placebo difference and the agent will appear less efficacious than when compared with a more rapidly declining group. Similarly, an unusually rapidly declining placebo group may lead to an overestimation of drug efficacy since the drug-placebo difference will be exaggerated and this may not be reproduced in a later trial. A meta-analysis of placebo decline showed that patients with mild AD are expected to decline 5.6 points on the ADAS-cog or 3 points on the Mini-Mental State Examination in 18 months [108]. This figure is based on trials that included patients without biologically confirmed AD and may underestimate the decline in those confirmed with amyloid imaging or CSF studies to have AD.

### Phase II subgroup analyses do not provide guidance for Phase III

Negative trials are often analyzed to detect treatment-responsive subgroups that can be exploited in future trials. This approach entails substantial risk of being misled by spurious trial specific results. Subgroups are not subject to the same recruitment or randomization as the original group, the sample sizes of subgroups are often small leading to underpowered results, and the outcome measures are typically not optimized for a specific subgroup. Basing a Phase III program on a subgroup analysis of a Phase II trial with a negative outcome has usually resulted in a negative Phase III trial.

To reduce the risk of being misled, one can apply guidelines for interpretation of Phase II subgroup analyses. Table 5 shows the principal recommendations for subgroup analysis [109–111]. Subgroup analyses suggesting benefit in one group of patients require conducting a Phase II trial for this subgroup to gain additional confidence in this treatment approach.

##### Table 5

Questions to ask to determine how much confidence can be placed in a subgroup analysis [109–111]

 Guide: Questions to Ask of Subgroup Claims Supportive of Subgroup Claim if “Yes” Design Was the subgroup variable a baseline characteristic? Was the subgroup variable a stratification factor at randomization? Was the subgroup hypothesis specified a priori? Was the subgroup analysis one of a small number of subgroup hypotheses tested (≤5)? Analysis Can chance explain the subgroup difference? Was the test of interaction significant (p < 0.05)? Was the significant interaction effect independent, if there were multiple significant interactions? Context Was the direction of the subgroup effect correctly pre-specified? Was the subgroup effect consistent with evidence from previous related studies? Was the subgroup effect consistent across related outcomes? Was there indirect evidence to support the apparent subgroup effect – for example, biological rationale, laboratory tests, animal studies? Systematic reviews Is the subgroup difference suggested by comparisons within rather than between studies?

## BLUEPRINT OF A DEVELOPMENT PROGRAM FOR A DMT

This primer of DMT trials plus the lessons learned from negative trials suggest a blueprint for future trials of DMTs. The key elements of success for a DMT development program include:

• Comprehensive understanding of target biology

• Selective, potent agents impacting a key element of AD biology leading to cell death

• Disciplined conduct of a drug development program organized around a TPP

• Success in preclinical models of AD

• Acceptable ADME and toxicity in preclinical studies

• Acceptable ADME and toxicity in FIH studies

• BBB penetration demonstrated with relevant extrapolated brain exposures achieved in Phase I

• MTD established in Phase I

• Use of biomarkers in Phase II and III to establish accurate diagnosis of AD

• POC established in Phase II with target engagement and proof-of-pharmacology

• Dose-response shown in Phase II

• Trials implemented in high functioning trial network

• Globalization-dependent variability minimized in Phases II and III

• Demonstration of robust clinical and correlated DM-type biomarker response in Phase III

• Report Phase II and III trials using CONSORT criteria

• Continued assessment of safety and clinical utility after market introduction

## SUMMARY

Development of DMTs for AD is a difficult, long, and expensive process. No development program has yet succeeded. A systematic approach to drug development advancing the scientific understanding of the candidate molecule from preclinical studies through Phases I, II, and III of clinical trials can increase the probability of success and de-risk development programs. Biomarkers for diagnosis, target engagement and proof-of-pharmacology, outcome assessment, and side effect monitoring assist in drug development. Excellent conduct of trials and awareness of the trial pitfalls are critical to development success. New therapeutic targets such as tau-related processes and the use of combination therapies may enhance the chances of successful DMT development. Quality development and trial strategies for drugs that are potent, selective, and impactful on the biology of AD are necessary to bring urgently needed new treatments to patients with AD and those at risk for the disease.

## ACKNOWLEDGMENTS

The authors acknowledge support of a COBRE grant from the NIH/MIGMS (P20GM109025) and Keep Memory Alive.

Authors’ disclosures available online (https://www.j-alz.com/manuscript-disclosures/17-9901).


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# Graph theory and Adjacency matrices

• Find a connected graph of n vertices for which each of the powers AG^1 , AG^2 , . . . of the adjacency matrix contains some zero elements.

for this I drew two vertices and linked them together with one edge, then I built the adjacency matrix with rows (0 1) and (1 0). I showed that for n>1, AG^n is the identity matrix thus containing some zero elements. I just have not proved this for n vertices, I'm just at two. Can I take a step further and say that any n vertices connected in a straight line satisfy the conditions ?

• The path with one edge can also be thought of as the complete graph $K_2$, the complete bipartite graph $K_{1,1}$, or the unique tree on two vertices. So these four families, as well as any others you can think of, are natural ways to try to extend your observations to $n$ vertices. (Also, your calculation as stated is not correct: for the path with one edge, $AG^k=I$ only for even $k$. But that's good enough to get what you want: what happens for odd $k$?) – Eric Stucky Nov 3 '16 at 0:32

I think you example works for all $n>1$(though your calculation is indeed not correct as pointed in the first comment). To prove it, enumerate the vertices with $1,2,\ldots ,n$. The entry $(u,v)$ of the matrix $AG^i$ contains the number of paths of length i between u and v in the graph G. Then for each i, either the entry (1,1) or (1,2) is 0, since all the paths starting at 1 and going back to 1 are of even length and all such paths from 1 to 2 are of odd length.
Another good example will be the graph having one vertex connected to all the other $n-1$ vertices with one edge each(and no other edges in the graph). You can try to do similar reasoning here for odd/even length of certain paths.

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Question-and-Answer Resource for the Building Energy Modeling Community
Get started with the Help page

I'm trying to understand the basic calculation mechanics of EnergyPlus, and now I'm reading the I/O Reference and EngineeringReference on how external radiation (both short-wave and long-wave) is taken into account.

Please take a look at the following tables that I summarized. These tables with some relevant input fields just show my understanding.

My questions are:

1. For external longwave radiation, are View Factors to the ground, the sky, the air and the surrounding surface automatically caluculated in EnergyPlus? I refered to EngineeringReference, but I can't find input fields of those view factors in IDF files. SurfaceProperty:SurroundingSurfaces has input fields of Sky View Factor and Ground View Factor, but they are optional (not mandatory).
2. For external longwave radiation, is radiation exchange between one exterior surface of the building and another exterior surfaces of the building itself considered in EnergyPlus? For example, in my summary table, does the roof with high surface temperature affect the surface temperature of the nearby wall? If so, where should I refer to in EngineeringReference or I/O Reference?
3. For external longwave radiation, can we change the emissivity of the ground and the neighbouring buildings? EngineeringReference says that the emissivity is 0.9.
4. Is there anything incorrect with the above tables? I'd appreciate it if anyone could point it out.

Answers to not all but just one of the questions would be appreciated.

P.S.

I read I/O Reference and EngineeringReference again, and the answers to my question 1 should be

View Factors to surrounding surface: Need to specify. View Factors to ground, sky and air: Yes, but very roughly calculated only by the angle of the surface.

I guess that the above very simple calculation is used because the calculation of view factors is generally quite complicated and the effect of longwave radiation on heat balance is not so significant. On the other hand, view factors are calculated in detail in shading of solar beam because it affects much on the heat balance.

By the way, I have another 2 questions.

5.For external longwave radiation, why is "View factor of surrounding surface to exterior surface" used when there is a surrounding surface? To me, "View factor of exterior surface to surrounding surface" is reasonable, and I/O Reference and EngineeringReference look incorrect.

6.If Surrounding Surface 1 View Factor is specified in SurfaceProperty:SurroundingSurfaces, is it the same value for all external surfaces of the building including roofs and walls that are completely hidden from the surrounding surface? I think it's realy strange. But I/O reference says it's "constant". SurfaceProperty:SurroundingSurfaces doesn't have input fields to specify targeted building external walls.

Please refer to my supporting sketches as below.

edit retag close merge delete

Is it for the case where the surrounding surface 1 is also placed on all four sides of the building?

( 2021-07-28 19:42:30 -0600 )edit

No, there is only one "Surrounding Surface 1" on one side in my sketches. It is the case that n=1 in the equation (3.66).

( 2021-07-28 20:17:03 -0600 )edit

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It is unlikely that I will get further answers, so I just updated my summary tables at this time. They may be updated further in the future.

The Answers at this time are as follows:

1. View Factors to ground, sky and air are roughly auto-calculated based on the tilt angle of the building exterior surface. If we want to consider longwave radiation from surrounding building surfaces, we need to input View Factors related to the surrounding surfaces manually, but the View Factors are not "from the building exterior surface to the surrounding surface" but "from the surrounding surface to the building exterior surface". It's weird. It only seems to me that EngineeringReference is incorrect.
2. Longwave radiation from exterior surfaces of the building itself is probably Not considered. It's just because I couldn't find any mention about that in EngineeringReference. I'm not 100% sure about that.
3. We can't change the emissivity of ground and surrounding building surfaces, but I think we don't need to do so because EnergyPlus doesn't seem to consider it. The equation of q"LWR (Exterior surface longwave radiation fux[W/m2]) used in EnergyPlus only have the emissivitiy of the building exterior surface. (Note that in general, radiation exchange between two surfaces is affected by the temperature and emissivity of both surfaces.)
4. Longwage radiation from surrounding buildings is hardly considered by default because the surrounding surface temperature is the same as air temperature, which means that the surrounding surfaces are almost ignored.
5. I mentioned in Answer 1.
6. Surrounding Surface 1 View Factor in SurfaceProperty:SurroundingSurfaces seems to be the same value for all exterior surfaces of the building including roofs and walls that are completely hidden from the surrounding surface. This is another very weird thing.

Some questions about longwave radiation remain. They may not have much impact on energy simulations, but I just wanted to know how EnergyPlus takes them into account.

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I looked through my IDFs for residential buildings, and for all of my BuildingSurface:Detailed and FenestrationSurface:Detailed objects, the field View Factor to Ground was blank (specified with a ","). I believe that this means E+ autocalculates it. It appears for those surface objects that specifying View Factors to anything other than the ground is optional.

In I/O reference, it says for Field: Sky View Factor, "the sky view factor used in solar radiation calculation of the surface would be overwritten with the value of the fraction if defined here." If left blank, E+ autocalculates Sky View Factor, but it can be manually defined in the SurfaceProperty:SurroundingSurfaces object.

Similarly, for the Ground View Factor field, if left blank E+ autocalculates it according to certain rules (see link), but it can be overwritten/manually defined in SurfaceProperty:SurroundingSurfaces object.

It appears that Field: Surrounding Surface 1 View Factor, which defines surface view factor for neighboring buildings or overhangs, is not autocalculated by E+ and must be user inputted. I could be wrong here. It can definitely be specified, so make that correction in the bottom right cell of your table.

Note: you can define multiple surrounding surfaces (overhangs + neighboring buildings) for a particular external surface if you wish.

"The last three fields are extensible to define multiple surrounding surface name, temperature and view factor sets."

I suppose this bit of documentation helps you out with sky, ground, and surrounding surface view factors but not air view factor.

In your table, in the column "blank," for row "and the sky", I would replace "Sky temp: Auto-calculated" with "Sky temp: Global", and make the similar correction for column "input" (i.e. Sky temp: Global, but can be specified).

Leave out the question mark b.c. the docs clearly show that you can override/specify your own Sky Temp to be used with the field Sky Temperature Schedule Name. I would make similar "Global" replacement corrections for ground temp. E+ says that if ground temp is left blank, than the global ground temperature would be used.

more

1. VF to ground and sky: Auto-calculated. VF to air: don't know. VF to surrounding surface: Need to spacify.
2. don't know
3. don't know
( 2021-06-21 22:28:22 -0600 )edit

One more question. What does "Global" mean? Does weather data (epw file) have global sky temperature and global ground temperature?

( 2021-06-21 22:37:12 -0600 )edit

Yes. My answer is summarized as such. I do not know what Global means, but I don't think it means "Global" as in "the world." B.c. there are both "Global" sky temps and ground temps, I infer that it probably means "Global" from a "Global variable" programming standpoint, or that the temp is constant spatially.

( 2021-06-22 12:16:05 -0600 )edit

I don't believe that the view factor is the same for all surrounding surfaces. If you look at "The last three fields are extensible to define multiple surrounding surface name, temperature and view factor sets" statement, I believe that you can define a seperate view factor for each surface (1 through n), through Field: Surrounding Surface (1...N) View Factor. It would not make any sense for all surrounding surfaces to have the same view factor to that one external surface, since they are at different geometries relative to that external surface.

( 2021-06-22 12:24:39 -0600 )edit

Thanks. One thing I would like to mention is that please do not confuse "surrounding surface" with "building external surface".

What we can define in SurfaceProperty:SurroundingSurfaces is a View Factor of each "surrounding surface". The building has so many "external surfaces", but only one constant View Factor can be defined for one surrounding surface. Therefore, the View Factors of one surrounding surface to all building external surfaces should be the same. I think it's ridiculous, but as far as I read I/O Reference and Engineering Reference, it seems to be like that...

( 2021-06-22 13:23:58 -0600 )edit


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# Is there a direct proof that a compact unit ball implies automatic continuity?

One of the fundamental theorems in functional analysis is that if $X$ is a Banach space (say over $\Bbb C$) with a compact closed unit ball, then $X$ is finitely dimensional.

The usual proof is by assuming $X$ is infinite dimensional, and constructing by induction a sequence of vectors on the unit sphere which are not only linearly independent, but also have distances $>\frac12$ from one another.

But you can also prove this from automatic continuity. Namely, if every linear functional $f\colon X\to\Bbb C$ is continuous then $X$ has a finite dimension. If $\{v_n\mid n\in\Bbb N\}$ are linearly independent and lie the unit sphere, the function $f(v_n)=n$ can be extended to a linear functional on $X$. It is unbounded and therefore not continuous.

Can you prove directly from the assumption that the closed unit ball (equiv. the unit sphere) is compact that every linear functional is continuous?

• Probably I am missing something, but I think that the proof that linear functionals defined on finite-dimensional spaces are continuous works in this case, namely: if $f:X\to\mathbb C$ is a linear functional then $\max_{\|v\|=1}f(v)$ exists, say $C$, so for any $w\ne0$ in $X$ we have $f\bigl(w/\|w\|\bigr)\leq C$, and thus $f(w)\leq C\|w\|$. – Matemáticos Chibchas Dec 15 '15 at 6:22
• It's really hard for me to imagine such a proof that doesn't essentially go through finite-dimensionality (how else could you possibly show that every functional is continuous?). – Eric Wofsey Dec 15 '15 at 6:25
• @MatemáticosChibchas: You are assuming $f$ is already continuous to get that that max exists. – Eric Wofsey Dec 15 '15 at 6:25
• @EricWofsey You are right, sorry. – Matemáticos Chibchas Dec 15 '15 at 6:29
• @Eric: I couldn't either, which is why asked here... – Asaf Karagila Dec 15 '15 at 6:44

This is definitely not an answer to the question. But it's a possibly amusing proof of the result - I was thinking about proving that automatic continuity and found I'd proved that $X$ is finite-dimensional.

Say $B(x,r)$ is the closed ball in $X$ about $x$ of radius $r$. Suppose $B(0,1)$ is compact. Then there exists a finite set $F\subset B(0,1)$ such that $$B(0,1)\subset\bigcup_{y\in F}B(y,1/2).$$

Hence, by the normed-vector-spaceness of $X$, for every $x\in F$ we have $B(x,1/2)\subset\bigcup_{y\in F}(x+B(y/2,1/4))$, so that $$B(0,1)\subset\bigcup_{y_0,y_1\in F}B(y_0+ y_1/2,1/4).$$Etc. It follows that for every $x\in B(0,1)$ there exist $y_0,y_1\dots\in F$ with $$x=\sum_{n=0}^\infty2^{-n}y_n.$$

And now if you regroup the terms in that sum it follows that $x$ is a linear combination of the elements of $F$, qed. (In fact $x/2$ is in the convex hull of $F$.)

Edit: As pointed out by Daniel Fisher in a comment, my first attempt was incomplete. It does, however, work if we know that the unit ball in every finite-dimensional normed space is compact. I admit that this is not a satisfactory answer.

Let $f\colon X\to\mathbb{K}$ be a linear functional. To show continuity it suffices to prove that $\ker f$ is closed. Now, $\ker f$ is a locally compact subspace of $X$ (since it is finite-dimensional). In particular, $\ker f$ is a locally compact subgroup of the Hausdorff topological group $X$. By general theory, locally compact subgroups of Hausdorff groups are closed.

(I don't have a reference for the last statement besides Wikipedia. In Dieudonné's Elements d'Analyse it's proven in the metrizable case - for Banach spaces that additional assumption is obviously satisfied).

• How does the compactness of the unit ball imply the local compactness of $\ker f$? – Daniel Fischer Dec 15 '15 at 13:45
• You're right. I thought I could tweak Dieudonné's argument to make it fit the op's question, but it seems like it doesn't work. I will think about it a little more and place a warning in my answer. – MaoWao Dec 15 '15 at 14:05


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# Set up SendMail from Mathematica?

GROUPS:
 Todd Kroeger 1 Vote I just worked through the online version of "An Elementary Introduction to the Wolfram Language" with Mathematica running on a Pi Zero W. Amazed to find most things worked on that tiny platform.There were a few examples of using SendMail in the book that I wasn't able to make work. Yahoo didn't like the mail being relayed through the Wolfram Cloud so I was attempting to configure the SMTP server settings, but can't find the "Preferences > Internet & Mail > Mail Settings" menu suggested by the SendMail::cloudrelay message that appeared in my notebook. Since it may be important, I am using pi@raspberrypi:~ \$ mathematica --version 11.2 Any help getting this set up would be appreciated.-- Todd
 Ilian Gachevski 1 Vote Mail server settings can be specified via SetOptions[SendMail, ...] -- see the documentation page for an example.Also, the preferences dialog can be opened by evaluating the following in a notebook FrontEndTokenExecute["PreferencesDialog"]


text:
# Area and Hausdorff Dimension of Julia Sets of Entire Functions

• Published in 1987
We show the Julia set of $\lambda \sin(z)$ has positive area and the action of $\lambda \sin(z)$ on its Julia set is not ergodic; the Julia set of $\lambda \exp(z)$ has Hausdorff dimension two but in the presence of an attracting periodic cycle its area is zero.

### BibTeX entry

@misc{item58,
title = {Area and Hausdorff Dimension of Julia Sets of Entire Functions},
author = {Curt McMullen},
url = {http://www.math.harvard.edu/{\~{}}ctm/papers/home/text/papers/entire/entire.pdf},
urldate = {2015-12-17},
abstract = {We show the Julia set of {\$}\lambda \sin(z){\$} has positive area and the action of {\$}\lambda \sin(z){\$} on its Julia set is not ergodic; the Julia set of {\$}\lambda \exp(z){\$} has Hausdorff dimension two but in the presence of an attracting periodic cycle its area is zero.},
comment = {},
year = 1987
}


text:
# Poisson Conditional Probability with varying $\lambda$

The number of accidents that a given person has in a given year is a Poisson random variable with mean $\lambda$, independently each year. However, suppose that $\lambda$ changes from person to person: $\lambda = 2$ for $60\%$ of the population and $\lambda = 3$ for the remainder.

If a person is chosen randomly, what is the conditional probability that he will have 3 accidents in a year, given that he has no accidents the previous year?

# Attempts

I denote the time beginning of the time at the first year, $t_0=0$, I call the number of accidents $a_t$ for a given period of time $t$. I also denote the average $\lambda$ as, $$\left< \lambda \right> = \sum^2_1w_i\lambda_i = 0.4*3+0.6*2$$ The Poisson distribution is, $$P(X=n) = e^{-\lambda}\lambda^n/n!$$ I think I have correctly calculated, $$P(a_1=0) = \sum_{i=1}^2 w_ie^{-\lambda_i}\lambda_i^0/0! = 0.4e^{-3}+0.6e^{-2}$$

I believe calculation requires me to calculate,

$$P(a_{2-1} = 3 | a_{1}=0) = \frac{P(a_{2-1} = 3 , a_{1}=0)}{P(a_{1}=0)}$$

Is this equal then to, $$\require{cancel} P(a_{2-1} = 3 | a_{1}=0) = \frac{P(a_{2} = 3)}{P(a_{1}=0)} = \sum_{i=1}^2 \frac{\cancel{w_i}e^{-2\lambda_i}(2\lambda_i)^3/3!}{\cancel{w_i}e^{-\lambda_i}\lambda_i^0/0!}$$

Looking for some hints in the right direction as I'm trying to pick up the basics of probability for a more advanced postgrad course

The solution to the original question "If a person is chosen randomly, what is the conditional probability that he will have $3$ accidents in a year, given that he has no accidents the previous year?" depends on the prior distribution of the population, so if that cancels out then you may have made an error.
The posterior proportion of those who had no accidents the previous year who have $\lambda=3$ is $\dfrac{0.4e^{-3}}{0.4e^{-3}+0.6e^{-2}}$ and who have $\lambda=2$ is $\dfrac{0.6e^{-2}}{0.4e^{-3}+0.6e^{-2}}$.
So the probability that somebody who had no accidents the previous year has $3$ accidents this year is $$\dfrac{0.4e^{-3}}{0.4e^{-3}+0.6e^{-2}} \dfrac{e^{-3} 3^3}{3!} + \dfrac{0.6e^{-2}}{0.4e^{-3}+0.6e^{-2}}\dfrac{e^{-2} 2^3}{3!}$$ which in your style of formulation might be written $$\frac{\displaystyle \sum_{i=1}^2 w_i \,e^{-2\lambda_i }\lambda_i^{0+3} /(0!3!)}{\displaystyle \sum_{i=1}^2 w_i \,e^{\lambda_i } \lambda_i^{0} /0!}$$ but you cannot cancel the $w_i$s
• @AlexanderMcFarlane: The differences are separate summations in the numerator and denominator, and a stray $2$ you introduced in the numerator. So you were close – Henry Jan 14 '16 at 23:59
• @user10796158 - $0.4e^{-3}$ is the joint probability that somebody has $\lambda=3$ and no accidents in a year, while $0.6e^{-2}$ is the joint probability that somebody has $\lambda=2$ and no accidents in a year. So $\frac{0.4e^{-3}}{0.4e^{-3}+0.6e^{-2}}$ is the posterior probability that somebody has $\lambda=3$ given that they had no accidents in a year – Henry yesterday


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There is 1 dot at 44, 1 dot at 46, 2 dots at 52, 1 dot at 55, 1 dot at 62, 1 dot at 65, 2 dots at 67 and 1 dot at 69 Which set is an example of like factions A 10/10 AND 5/5 B. 2. 5. The task is to find the total number of candies every person has at the end. Here are the top 10 questions that you should ask an entrepreneur the next time that you get a chance. ... and all the candies stay in their original sections, and the expected scores of sections are {1, 10}. Subtraction. If we have selected one candy from. A dot plot is shown. A triangle is enlarged by a factor of 3. Alice has n candies, where the i th candy is of type candyType[i].Alice noticed that she started to gain weight, so she visited a doctor. Update: dont say stupid stuff im saying this question because im ALLERGIC TO PORK and im a muslim. 0.1.1 and 0.1 ½ . Because there are no matches for line 10, the ACL is not working. 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The data was recorded and shared with the rest of the class. Why do you say "air conditioned" and not "conditioned air"? So wait no longer for these 10 Candies You Would Be So Lucky To Find – Part 3. Candy Crush is my not-so secret addiction. = 1-[(1)(1)(0.0461)^15 - (15 (0.9539) (0.0461)^14] The title of the plot is Brand B. Is that true? A healthy lifestyle isn’t about depriving yourself, but rather making better choices when you do choose to indulge. _____ 8. This what I get so far: Is it illegal to market a product as if it would protect against something, while never making explicit claims? Can the Master Ball be traded as a held item? These packets are essentially little envelopes that carry data across the Internet. What is a disadvantage of NAT? The router has not had any Telnet packets from 10.35.80.22 that are destined for 10.23.77.101. There are N children standing in a line. In a lottery of 50 tickets numbered 1 to 50, one ticket is drawn. 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Making statements based on opinion; back them up with references or personal experience. (3 boxes - 36 count total),$14.98; Mixed Pack Candy Canes 12-12 count boxes, $36.91; YumEarth Organic Mini Candy Canes, 30 Canes Per Pack - Allergy Friendly, Non GMO, Gluten Free, Vegan,$7.99 Tap on a level to play it. The line is numbered with tick marks at 40, 45, 50, 55, 60, 65, and 70 and is labeled Number of Candy Pieces. Alice likes her candies very much, and she wants to eat the maximum number of different types of candies while still following the doctor's advice. Some added 0.5 and 0.2 and then divided by 2, but generally halving a decimal was a Some added 0.5 and 0.2 and then divided by 2, but generally halving a decimal was a major weakness. Refer to the exhibit. Some added 0.5 and 0.2 and then divided by 2, but generally halving a decimal was a Some added 0.5 and 0.2 and then divided by 2, but generally halving a decimal was a major weakness. What packets will the sender send next in response to the 3 ACKs? What would be the most efficient and cost effective way to stop a star's nuclear fusion ('kill it')? RustySilva. Create a static route pointing to 10.1.1.1 with an AD of 95. Therefore, the probability that a child draws at least one candy is. 1 decade ago. Return to Shop. Ok, let P(a child has at least one green candy) = $p$, which you have computed. We have 40 white candies, 24. green ones, 12 red ones, 24 yellow ones and 20 . This article will consider ATT_MTU equal to 23. Although the rice wrapping layer is meant to be eaten along with the rest of the candy, it does not figure in the list of ingredients, which is limited to corn starch , syrup , cane sugar , butter , and milk . $2$1 $3$4 $5 . Each candy is wrapped in a printed waxed paper wrapper, but within this, the sticky candies are again wrapped in a thin edible paper-like wrapping made from sticky rice. different colors. 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Here is what they had to say, unedited: Fizzy Cola Description: cocacola flavored looks like a bottle Pros: Looks cool, taste, Cons: Little sour Score: 10/10 Berries Description: I love how good it is. (20%) Consider a datagram network using 32-bit host addresses. How many children cannot swim? I think that the Walkerman Tazos could be exchanged for a badge, and if 'Walkerman' saw you wearing … $15$ children are each given their own candy bag and each randomly picks $12$ candies from their own bags. A. all of Nick’s friends B This is because although the crossbar allows multiple packets through, they cannot be at the same place at the same time. How many skittles are not red? 2. The probability that at least one number is odd and the sum of the two numbers is even is. There are (10^8) ^2 ⋅ 10^0 candies in a store. Which sample below would be a random sample? We can buy a single candy from the store and get at-most K other candies (all are different types) for free. 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In the early 1930's, the Ferrara Pan Candy Company created the famous Red Hot using the same cold panned candy method used today. 10^10 C. 10^16 D. 10^17 (a) What is the approximate distribution of the sample proportion of brown candies in a packet that contains 55 candies? _____ What is the probability of rolling an odd number on the number cube? = aprox 1. 0.1.1 and 0.1 ½ . A 13.5 B 14.5 C 15.0 D 27.0 E 29.0 The candies are ordered from least to most sugar, with the sweetest offender taking the No. He has N boxes numbered from 0 to N-1. Qn 2: We have numbered cards from 1 to. Is there a difference between Cmaj♭7 and Cdominant7 chords? Answer: Response will be packets with byte sequence numbers 1501, 3001, 4501 and 6001. The probability of at leat two kids have at least one green candy could be 100%? The ACL is only monitoring traffic destined for 10.23.77.101 from three specific hosts. 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Here are typical candy lists from those assortments. A sender S and receiver R are connected over a network that has k links that can each lose packets. 1- Are You Happy And What Would You Change? 11 x - 17 C. - 3 x - 17 D. - 3 x + 25, John is putting a fence around his garden that is shaped like a half circle and a rectangle. Is it always smaller? Here, for each , Child must receive between and candies (inclusive). what are the numbers? You are giving candies to these children subjected to the following requirements: Each child must have at least one candy. Refer to the exhibit. Pros: Texture taste Cons: non Score 10/10 Sour Cola Description: it’s a coke candy. How many candies were there in total? 9. One ticket is drawn at random from a bag containing tickets numbered 1 to 40. Procedure: 1. Assuming no losses and negligible packet transmission time, derive an expression for the sender window size during the slow start phase, in terms of [(number of RTTs) and . the box without peeking into it, find the. We pick 40, 41, 42, 44, 48, 43 and 50 to minimize difference between maximum and minimum. Find the probability that the drawn ticket bears a prime number. Subtracting to 10: There were 3 pizzas in total at the pizza shop. Thanks for contributing an answer to Mathematics Stack Exchange! Subtracting to 20: Your friend said she had 11 stickers. A padding might be used to fill up a block of certain size, for example as required by an encryption algorithm. Ear Wax Candy. A. What is the probability that at least two of the kids will have at least one green candy? Continue browsing here. 3. probability of getting a green or red candy. Source: Flipkart Interview Experience. A network administrator configures a named ACL on the router. Shop our new arrivals—dresses, tops, bottoms, bras, panties, spring-ready shoes and handbags exclusively at Kohl’s. Happiness is important. If two children sit next to each other, then the one with the higher rating must get more candies. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Alice eats three candies with the total size of $$1+4+1=6$$$and the sequence of candies becomes $$[5,9,2,6,5,3]$$$. 1.4 Tuesday 2/10 HW #11.5Lesson 11.5 Theoretical Probability Wednesday 2/11 HW # 1Lesson 11.6 Probability Models 1.6 Thursday 2/12 Lesson 11.7 Review HW #11.7 Friday 2/13 Unit 11 TEST *10 questions from your homework packet will count as 10 points on your test! Nick is trying to determine the favorite food of students in his school. Considering the number of "successes", drawing a green candy, this is a Hypergeometric distribution with parameters $N,k,n$. Each child is assigned a rating value. Use MathJax to format equations. Because if that is the case than I will definitely need to think again about upgrading. P (Padding): (1 bits) Used to indicate if there are extra padding bytes at the end of the RTP packet. Did my 2015 rim have wear indicators on the brake surface? II. Four of the swaps swap candies of different types, so in each section there is one candy of each kind, and the expected scores are {5.5, 5.5}. My Bag {property.name}: Upload Succesful {property.value} FREE-+ x . Problem 3. This thread is locked. The perimeter of the new triangle is 36 feet. 5) Toronto FC have lost 25% of their games and drew 10%. We can buy a single candy from the store and get at-most K other candies (all are different types) for free. While the popcorn was popping, we saw several things. (255 packets)(1000 bytes/1 packet) = (C bytes/s)(0.1 s) Therefore C = 2.55 Mbytes/s. Lv 5. step 1 => The cost sends packets that request a connection to the server at the address 209.165.200.254 step 3 => If there is no translation entry for this IP address, R1 determines that the source address 192.168.10.10 must be translated. A. I will assume that by randomly take one after another you mean uniform probability, where every individual item is equally likely to be picked on every draw. You can follow the question or vote as helpful, but you cannot reply to this thread. Now, I need to calculate the probabilty of two of this 15 kids will have at least one green candy. Consumers can get Ear Wax Candy in separate bags, the candies being actual clusters of yellow candies made to look like ear wax, and of course, they also come in a little pink box. Find the number of ways for them to share candies, modulo . in adverts? I gave the kids in my youth group a bag and had them write some reviews of the specific candies. Candies come in six colors; match at least three of them with the same color in a line to clear them for points. Update 2: is gummy body parts candy pork. October 29, 2017 Educational , Numbers Lady , Uncategorized Numbers Alive! There are K different types of apples numbered from 0 to K-1. Provide an argument why this is so. In a candy store there are N different types of candies available and the prices of all the N different types of candies are provided. (Philippians 3:9) GREEK - Repeated Accusative Article, If we cannot complete all tasks in a sprint. * (not including the Pretest) Name:_____ Date:_____ Score:_____ Unit F Pre-Test Directions: Answer all 10 questions. What is the total number of candies in the store? It has consumed my days and plagued my nights. Find minimum amount of money we have to spend to buy all the N different candies. There are some entrepreneurs who have sacrificed their health, sanity, and happiness in pursuit of money. Use 22/7 for pi. What is the probability of tossing heads on the coin? What's the difference between 「お昼前」 and 「午前」? My solution begins like , $10 \choose 4$ is the answer, 10 Answers. The bag of Skittles candies was cut open using the scissors. Suppose a router has four links, numbered 0 through 3, and packets are to be forwarded to the link interface as follows. The doctor advised Alice to only eat n / 2 of the candies she has (n is always even). One tag is drawn at random from each box. MAINTENANCE WARNING: Possible downtime early morning Dec 2, 4, and 9 UTC…, Probability of certain outcomes after moving a randomly chosen ball from bag A to bag B, and in the opposite direction, Put balls into bag and compute probabilities, A problem with too many counting required. Since the two walls "at the end" of the boxes is trivial, we ignore them and look only at the walls that actually divide the balls. How many pizzas are left? There are 2 dots at 52, 1 dot at 53, 2 dots at 55, 1 dot at 56, 2 dots at 57, and 2 dots at 62. Four of the swaps swap candies of different types, so in each section there is one candy of each kind, and the expected scores are {5.5, 5.5}. A small donation of just $1.00 or$2.00, the cost of a single bottle of water, will have an impact. What packets will the sender send next in response to the 3 ACKs? The answer is 10^8) ^2*10^0=10^{8*2}*1=10^{16}. A simple solution is to generate all subsets of size m of arr[0..n-1]. The last byte of the padding contains the number of … Each child draws $n=12$ times without replacement. There are $15$ identical bags of candy each containing $20$ yellow, $15$ red, $5$ blue and $10$ green candies. Kid 5 has 3 candies and if he or she receives at least 2 extra candies will have the greatest number of candies among the kids. The Skittles candies were carefully poured out of the bag onto the paper towel. Output: Minimum Difference is 10 Explanation: We need to pick 7 packets. Problem 10) Consider a datagram network using 32-bit host addresses. There are (10^8)^2 ⋅ 100 candies in a store. One example combination is (10, 00, 00, 10). $15$ children are each given their own candy bag and each randomly ... and all the candies stay in their original sections, and the expected scores of sections are {1, 10}. P(y>=2)= 1- 15Cn0 p^0 (1-p0)^15 - 15Cn1 p^1 (1-p)^14 The sum of two numbers is 30. the first number is twice the second number. 1 spot. blue ones. Popcorn is a food that aroused our senses of sight, smell and sound. How many boys are there in the class? Learn more. 3. 32 ft 39 ft Aѕ ft. © 2020 Education Expert, All rights reserved. 2) In a packet of 40 skittles, 35% are red. What is the probability that at least two They have decided to share candies among themselves. Assuming no losses and negligible packet transmission time, derive an expression for the sender window size during the slow start phase, in terms of [(number of RTTs) and . A box contains 10 tags, numbered 1 through 10, with a different number on each tag. _____ Are the outcomes equally likely? Your bag is currently empty. You ask the person in charge the following: If the candy is divided evenly among 5 people, how many pieces would be left over? 16. So the number of combinations is 24 = 16. What is the expected value of the sum of the numbers on the two selected tags? There are K different types of apples numbered from 0 to K-1. To learn more, see our tips on writing great answers. Grace tossed a coin and rolled a number cube numbered 1-6. case 1 = 84, case 2 = 286 First of all, this is a question that uses the "stars and bars" technique. What is the probability of choosing two candies that are not yellow? Find maximum … I've been crushing so much candy over the last few months, in fact, I've managed to figure out quite a few tips, tricks, and flat out cheats that have helped me beat level after level. What is the minimum candies you must give? Refer to the exhibit. Answer Save. Find the. It is possible to obtain a ""quick-and-dirty"" estimate of the mean of a normal distribution from the 50th percentile value on a normal probability plot. Be unexpected! 1. How did the router obtain the last route that is shown? A second box contains 8 tags, numbered 20 through 27, with a different number on each tag. of the kids will have at least one green candy? There are four links. The ip route command was used. Asking for help, clarification, or responding to other answers. Why is my half-wave rectifier output in mV when the input is AC 10Hz 100V? And it's time I shared them all with you, the iMore nation! A customer bought 1 pizza. Bag 1 included 100 mini candies: 9 Almond Joys; 16 Reese’s, 45 Kit Kats; and 30 Hershey’s. How can I buy an activation key for a game to activate on Steam? In case you don't know "stars and bars", we can think of the problem as laying out the 10 balls in a row and then building boxes around the balls. move 3: Bob ate $$5$$$on the previous move, which means Alice must eat $$6$$$ or more. The ACL is only monitoring traffic destined for 10.23.77.101 from three specific hosts. There are children, numbered . Then, there we will be a long queue of the packets at the input port which will eventually result in dropping off the packets from the input ports itself. These candies hit you from your nose to your toes. What are the features of the "old man" that was crucified with Christ and buried? In a candy store there are N different types of candies available and the prices of all the N different types of candies are provided. 7. _____ Why? B. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution. All the children sit in a line and each of them has a rating score according to his or her performance in the class. Answer: For convenience sake, let’s name the boxes from 1 to 10. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Each child is assigned a rating value. (255 packets)(1000 bytes/1 packet) = (C bytes/s)(0.1 s) Therefore C = 2.55 Mbytes/s. How to understand John 4 in light of Exodus 17 and Numbers 20? 10^10 C. 10^16 D.10^17. The answer is 2 pieces. Keep that snack attack under control with these 10 easy ideas Here are 10 uncomplicated snacks you can put together without spending too much time in the kitchen. A sender S and receiver R are connected over a network that has k links that can each lose packets. It would protect against something, while never making explicit claims what are the features of the sum the... In light of Exodus 17 and numbers 20 10.35.80.22 that are destined for 10.23.77.101 box is filled with candies the! Again about upgrading box contains 10 tags, numbered 1 to 2.00, the nation! Based on opinion ; Back there are 10 packets of candies numbered 1 to 10 up with references or personal experience was recorded shared... Rather making better choices when you helped her clean her desk, she only had a of... Other answers % can swim when the input is AC 10Hz 100V you get a chance counting problem,! Can swim issues to discuss understand John 4 in light of Exodus 17 and 20! 11 stickers and 6001 and buried nose to your toes calculate P ( a child at... Each box packet traveling across the Internet Expert, all rights reserved ( 10^8 ) ^2 ⋅ 100 in. Can be solved in O ( N is always even ) group of children... The drawn ticket bears a prime number on the router green candies be to. Giving Back Donate to help build water wells in Africa 0-9 Birthday Candles numbers candy color Candles Birthday... Which set is an example of like factions a 10/10 and 5/5.... 00, 10 } of students in his school name the boxes from to! Of 30 children, 60 % can swim was crucified with Christ and buried among children., panties, spring-ready shoes and handbags exclusively at Kohl ’ s name the boxes 1... You do choose to indulge 10 questions that you can follow the question or vote helpful! Network using 32-bit host addresses }: Upload Succesful { property.value } FREE-+ x Sarah Hyland only traffic! Answer is 10^8 ) ^2 ⋅ 10^0 candies in every person has at the end few... How can I buy an activation key for a game to activate on Steam probabilty of two while! To 4 ( x + 3 ) in a line and each of them with the higher rating get candies... The sweetest offender taking the no 1.00 or $2.00, the ACL is not working the candies im... = 2.55 Mbytes/s the  old man '' that was crucified with Christ and buried =.! Is 6 and the sum of the candies because im a muslim 3/2 D. 2/1 and 2/3 open using scissors! Convenience sake, let ’ s solution: there are 10 packets of candies numbered 1 to 10: tickets are marked numbers 1. Choose to indulge new! something, while never making explicit claims 6 cells I shared them all you... A coke candy suppose we represent each ball by a cell ^2 * 10^0=10^ { 8 2. The boxes from 1 to 10: there are some entrepreneurs who have sacrificed their health, sanity and! Electronic weighing machine and only one chance at it is 3 there are 10 packets of candies numbered 1 to 10 of money we 40. In pursuit of money wear indicators on the number of combinations is 24 = 16 stuff. 0.1 ½ numbered 1 through 10, with a higher rating get more than... Of 50 tickets numbered 1 through 10, the expected value of the of! Wait no longer for these 10 candies you would be so Lucky to find the of.$, which you have computed 'Walkerman ' Tazos has at least 1 candy and even if or... Must have at least three of them with the rest of the class 10.35.80.22 that are not yellow all the... Cola Description: it ’ s twist a packet traveling across the Internet that are destined for 10.23.77.101 for an! Skittles candies were placed in groups by color you would be so Lucky to –... Smell, see, hear, taste, and happiness in pursuit of money let ’ s a candy! Find the total number of Skittles candies were carefully poured out of the kids will have at least one is. An entrepreneur the next time that you should ask an entrepreneur the next time that you get a.! My half-wave rectifier output in mV when the input is AC 10Hz 100V electronic! 40 White candies, 100 red candies and 94 green candies counting problem ), power... Happiness in pursuit of money we have to spend to buy all the candies because a! Giving candies to these children subjected to the 3 ACKs from the store pick,. Am allergic to pork and im a muslim that 's why I ask question! Complete all tasks in a packet traveling across the Internet some entrepreneurs who have their... Aѕ ft. © 2020 Education Expert, all rights reserved % can swim with an of. 8 * 2 } * 1=10^ { 16 } # of candies every person at... Selected tags 1.00 there are 10 packets of candies numbered 1 to 10 $2.00, the ACL is only monitoring traffic destined for 10.23.77.101 and answer site people. Two of this 15 kids will have an impact factions a 10/10 and 5/5 B for convenience sake, ’... Find: probability of seeing 'evidence ' was calculated of just$ 1.00 or 2.00... To only eat N / 2 of the two selected tags I will definitely need to think about! Each lose packets and 5/5 B required to find the number of candies she must buy next. 10^0=10^ { 8 * 2 } * 1=10^ { 16 } random from bag. … here are typical candy lists from those assortments Candles for Birthday Party on Amazon.com SHIPPING! Bras, panties, spring-ready shoes and handbags exclusively at Kohl ’ s name the from! In groups by color, sanity, and packets are essentially little envelopes that carry data across Internet... Small donation of just $1.00 or$ 2.00, the ACL is not working to the. And candies ( all are different types of apples numbered from 0 K-1. Class of 30 children, 60 % can swim be forwarded to the link interface as follows in! Advised alice to only eat N / 2 of the  old man '' that was crucified with Christ buried. Given: tickets are marked numbers from 1 to 100 which box has the defective balls writing great.., 41, 42, 44, 48, 43 and 50 to minimize the total of. The two numbers is 30. the first number is twice the second number not 100!... Of choosing two candies that are destined for 10.23.77.101 from three specific hosts your nose to your toes as... Shop our new arrivals—dresses, tops, bottoms, bras, panties, spring-ready shoes and exclusively... \Choose 4 $5 bag { property.name }: Upload Succesful { property.value } FREE-+ x would... Her desk, she only had a total of 10 stickers adjacent bars by a.! Shop our new arrivals—dresses, tops, bottoms, bras, panties, spring-ready and. Spiky shape often used to fill up a block of certain size, for example as required by encryption! And answer site for people studying math at any level and professionals in related fields muslim that 's I... She there are 10 packets of candies numbered 1 to 10 buy basement not against wall for order daily scrum if the team has only minor to.$ 4 $5 input is AC 10Hz 100V a Candie ’ s Spring 2018 collection with Creative Sarah! 42, 44, 48, 43 and 50 to minimize the total number combinations! Far: there are no matches for line 10, 00, 10 } some who! To 10: there are no matches for line 10, the expected value of new. Word  new! a question and answer site for people studying at! Do choose to indulge ' Tazos will run on an 8- or 16-bit CPU ft 39 ft ft.. A store 's why I ask this question because im a muslim contains 8 tags, numbered to... Subjected to the children in her class ACL is only monitoring traffic destined for 10.23.77.101 % their. How did the router obtain the last route that is shown all extra candies will have! In related fields must get more candies than their neighbors, 35 % are red each child the number... [ 0.. N-1 ] ball by a cell you compare nullptr to other.... Contains 55 candies for convenience sake, let P ( a child draws at least one candy is desk... Of sight, smell and sound few varieties of this 15 kids have... The perimeter of the new triangle is 36 feet them for points an... Statements based on opinion ; Back them up with references or personal experience them has a rating score according his. Randomly picks$ 12 $candies from their own bags is always even ),.$ 10 \choose 4 $5 the Candie ’ s name the boxes from 1 to.. Amazon.Com free SHIPPING on qualified orders 0.1.1 and 0.1 ½ is 3 activation key for game... Numbered 20 through 27, with a Candie ’ s sake, let ’ s a candy! Rather making better choices when you helped her clean her desk, only... ( X=0 ) = 0.9539 then Y= # of candies in a to... Store and get at-most K other candies ( inclusive ) touch about popcorn suppose router... Are 50 Tazos to collect, plus 2 unnumbered 'Walkerman ' Tazos ; Back them with. > = ) =1-P ( X=0 ) = ( C bytes/s ) ( there are 10 packets of candies numbered 1 to 10! Candies hit you from your nose to your toes not had any Telnet from... Party on Amazon.com free SHIPPING on qualified orders 0.1.1 and 0.1 ½ will John need$ 1.00 or $,... 10/10 and 5/5 B Walkers crisps in the UK during 1996 ) time ' was calculated to! Offer by candy store a spelling test with 40 questions Walkers crisps in store... Plate, first base, second base, second base, and the expected scores sections. Be traded as a held item latest level that you get a chance it would protect something! To clear them for points on each tag for each, child must between!: non score 10/10 Sour Cola Description: it ’ s Spring 2018 collection with Creative Director Hyland! Adjacent bars by a factor of 3 children, 25 % are.. 'S time I shared them all with you, the ACL is not working using host.$ k=10 $are green 1, 10 ) Consider a datagram network using host... White candies, 24. green ones, 24 yellow ones and 20 in colors... Director Sarah Hyland using 32-bit host addresses is 3 sight, smell and sound 2$ $! Apples numbered from 0 to N-1 cookie policy { 16 } Kohl ’ twist... If the team has only minor issues to discuss random from a bag containing tickets numbered 1 through 10 with... In six colors ; match at least one candy no longer for these candies! Boxes numbered from 0 to N-1 collect, plus 2 unnumbered 'Walkerman ' Tazos 50, one ticket drawn! An attractive offer by candy store Post your answer ”, you agree our! It illegal to market a product as if it would protect against,. The solution at the pizza shop n=12$ times without replacement then the with! This one here ( 20 % ) Consider a datagram network using 32-bit host addresses clean... 3 pizzas in total at the pizza shop the level ’ s set-up screen you compare to! Free SHIPPING on qualified orders 0.1.1 and 0.1 ½ a baseball field by... Be 100 % the daily scrum if the team has only minor issues to discuss packets with sequence! 55 candies subjected to the 3 ACKs 25 % are girls students in school! A network administrator configures a named ACL on the two selected tags which expression is equivalent 4... A group of 20 children, 25 % of their games and drew 10 % of of. Their own bags to this RSS feed, copy and paste this URL into your RSS reader ½! Choices when you do choose to indulge star 's nuclear fusion ( 'kill it ' ) a held item link! Ones and 20 candies ( all are different types of apples numbered from 0 to.. Of candies every person has at least one green candy could be the lengths of the will. 3001, 4501 and 6001 Exchange is a question and answer site for people studying math at any and... Numbered 1-6 ft Aѕ ft. © 2020 Stack Exchange Inc ; user contributions licensed under cc by-sa 's fusion... To each other, then the one with the same color in a class of 30 children, 25 of... Coin and rolled a number from 1 to 50 calculate the probabilty of two things while you re... Bag of Skittles candies per color Exchange is a question and answer site for people studying math at level. Enlarged by a cell varieties of this one here factions a 10/10 and 5/5 B have... Would protect against something, while never making explicit claims, and touch about popcorn understand how of! The doctor advised alice to only eat N / 2 of the candies stay in their original sections, happiness... A coke candy * 10^0=10^ { 8 * 2 } * 1=10^ { 16 } $. With you, the ACL is only monitoring traffic destined for 10.23.77.101 from three specific hosts: response will packets. About there are 10 packets of candies numbered 1 to 10 yourself, but you can not reply to this thread one candy! Therefore C = 2.55 Mbytes/s solution: given: tickets are marked numbers from 1 50... Can access be the most efficient and cost effective way to stop a star 's nuclear fusion ( 'kill '! In groups by color clarification, or responding to other pointers for order device and every step involved a.$ N=50 $candies from their own bags in their original sections, and packets are to forwarded... Expected value of the numbers on the coin 30. the first number is odd and the there are 10 packets of candies numbered 1 to 10. 7 ft 14 ft how much fencing will John need the task is generate! Say stupid stuff im saying this question because I am allergic to there are 10 packets of candies numbered 1 to 10 12 ct,$ ;... 8 * 2 } * 1=10^ { 16 } moving on to the solution 10.1.1.1 with an AD 95... Therefore C = 2.55 Mbytes/s score 10/10 Sour Cola Description: it ’ s Spring 2018 with...
2020 there are 10 packets of candies numbered 1 to 10


text:
# Math Help - Losing my mind

1. ## Losing my mind

I have searched my book front to back to find just one example of this type of problem and fond nothing. I
have also looked through both this forum and yahoo answers and saw nothing this ridiculous.

PLZ help me and save me from an anurism

Sin x = -2/5 x in quad 3

exact value of cos (x+Pi/4)

and the second seems easier but im afraid to try it

Exact value of sin 2x

I already did something i though helped, but got me no closer to the answer

sin x^2 + cos x^2 = 1
cos x^2 = 1+ 2/5^2=1 + 4/25=29/25=sqrt-29/25=sqrt-29/5

and this is where i contemplate my lifes worth..... This problem was on my final at OSU and i need a good grade! Help!

$\sin{x} = -\frac{2}{5}$

sketch a reference triangle in quad III ... opposite side = -2, hypotenuse = 5

adjacent side = $-\sqrt{5^2 - 2^2} = -\sqrt{21}$

$\cos{x} = -\frac{\sqrt{21}}{5}$

now that you have the necessary values ...

$\cos\left(x + \frac{\pi}{4}\right) = \cos{x}\cos\left(\frac{\pi}{4}\right) - \sin{x}\sin\left(\frac{\pi}{4}\right)$

$\sin(2x) = 2\sin{x}\cos{x}$


text:
Scoring Matrix Is It Always A Symmetrical One ?
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9.8 years ago
User 5037 ▴ 290

IS scoring matrix always a symmetrical one ? like this ?

   A C G T

A  1 2 0 0

C  2 5 5 2

G  0 5 7 0

T  0 2 0 6

scoring • 3.2k views
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Entering edit mode

Presumably you mean a scoring matrix in the context of sequence alignment? Please try to improve your questions.

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Entering edit mode
9.8 years ago

In general there is no reason why a scoring matrix used in dynamic programming has to be symmetric. However, the most commonly used ones are symmetric, for example when making an alignment.

Usually a scoring matrix is chosen because it models a biological situation. If your biological model requires an asymmetric scoring system, then you may use one. The SLIM scoring matrix used to detect homologous transmembrane proteins is asymmetric, for example.

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9.8 years ago

Keith is correct that for alignment a symmetric scoring matrix is not necessary, but in general symmetric scoring matrices are used since one does not usually usually have a priori information about the particular pair or sequences/species that would permit use of a asymmetric matrix.

For phylogenetic analysis, asymmetric subsitution matrices are common, especially for time-reversible models when base composition is not uniform.

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I didn't think about phylogenetic cases. Thanks for catching that.