## CHECK THESE SAMPLES OF N1

...? Explaining How Prahalad and Krishnan’s R=G, N Formula Applies To Operations Introduction In present days, the nature of association between customers and organizations has transformed radically. Traditionally, organizations developed the products and services by considering undifferentiated customers. However, in modern days, organizations have progressed through several levels of marketplace segmentation regarding different customer groups with diverse choices. Today, organizations have reached to that point where convergence of internet, digitization, and the merging of industry and technology limitations are generating new demand forces in the market. Previously, it was anticipated that organizations build value... Explaining How...

3 Pages(750 words)Essay

...) to get correct estimates.
Memo
Introduction:
Since the higher education market is clearly grouped we have to use stratified random sampling
So that each strata (group) gets properly represented by the sample selected and give a more precise estimate of the entire population.
In stratified random sampling the entire population is divided into strata and samples are picked from each strata. There can be three types of stratified random sampling
Proportionate
Disproportionate
Optimal
Proportionate stratified random sampling: the ratio of-the number of elements chosen from a particular stratum (**n1**) to the total number of elements in that particular stratum(**N1**) is equal to the ratio...

4 Pages(1000 words)Assignment

...examples
Value for the modulus (**n1**)
257
Value for Chung's secret key (kC)
234
Value for Lilly's secret key (kL)
127
Value of message V1
126
Value of message V2
37
Value of message V3
72
Table A2 Calculation of matching session key
Step 1
Chung operates on V3 by exponentiation modulo **n1** with kC to give V3 kC mod **n1** and sends the results to Lilly.
V3 kC mod **n1**
72234 mod 257
235 mod 257
Step 2
Lilly operates on V3 by exponentiation modulo **n1** with kL to give V3 kLkC mod **n1** and sends the result to Chung.
V3 kL mod **n1**
72127 mod 257
25 mod 257
Step 3
Chung takes V3 kL mod **n1** received from Lilly and...

6 Pages(1500 words)Essay

...is consistent with the policy.
3.
Given:
Victorian: Federal:
Sample size (**n1**): 1600 Sample size (n2): 2000
Proportion: 642 Proportion: 861
p1 = 642/1600 = 0.40 p2 = 861/2000 = 0.43
a. The confidence interval for the sample proportions are determined as:
+/- 1.96 (((p1 (1-p1)/**n1**) + (p2- (1-p2)/n2))1/2
(-0.03)+/- 196 (0.00015 + 0.00012) 1/2
The confidence interval lies from -0.062 to 0.002.
b. In order to see if there is a significant difference in the proportion of the populations tested, we till employ a t-test, where:
Ho = p1- p2 = 0
Ha = p1- p2 ≠ 0
We will employ a z-test, where z = p1- p2 / (((p1 (1-p1)/**n1**) + (p2- (1-p2)/n2))1/2. Plugging in the appropriate...

7 Pages(1750 words)Essay

...)
>> subplot(2,1,2),plot(t,q),xlabel(time)
Comparison:
The signal is compared both in frequency and time domain, before and after filtering. As seen in the graphs the amplitude of the signal after filtering is 1. The amplitude thus decreases from 2 to 1, showing 3dB attenuation. In frequency domain the higher frequency coefficients are zero. They cover almost half the total bandwidth. Thus the half band filter has made half of the coefficients zero with 3dB attenuation.
Functions:
function[y,n] = sigadd(x1,**n1**,x2,n2)
n= min(min(**n1**),min(n2)):0.001:max(max(**n1**),max(n2));
y1=zeros(1,length(n));
y2=y1;
y1(find((n>=min(**n1**))&(n<=max(**n1**))==1))=x1;
y2(find((n>=min(n2))&(n<=max(n2))==1))=x2;
y=y1+y2... ;
--------------------
function [y,n] =...

15 Pages(3750 words)Essay

...Objective: "You used the wrong numbers. You should use the following numbers: x1 = 990, **n1** = 7356, x2 = 999, n2 = 7356." Above is the Revision comments for the below order
You can use StatCrunch or the TI to help with this test. Needed information for both tools include:
x1 = number of red
**n1** = total number of candies
x2 = number of brown
n2 = total number of candies
For the TI, you will want 2-PropZTest. Then select the appropriate alternative (not equal), and Calculate then enter. The output will have the test statistic (z), p-value (p), sample p values, weighted p (), then repeat of sample sizes."You used the wrong numbers. You should use the following numbers:
x1 = 990, **n1** =...

2 Pages(500 words)Speech or Presentation

...Diderot Questions Word Count: 257 page N2’s judgment seems to show disappointment. What are the indications of this? What would N2 have preferred in order to satisfy his expectations of a good story? And how does **N1** disagree, or at least defend himself?
The indications that N2 feels disappointment are demonstrated in his manner of speech. N2 would have preferred that a good story would make sense. **N1** defends himself by rationalizing that it is a good story.
2. In a similar way, **N1** accuses N2 of having listened despite his objections. In this riposte, he sets forth a definition of the good listener much the same way N2 had attempted to define the good story-teller. Write down these two definitions, structuring your formulation... of the...

1 Pages(250 words)Assignment

...is not controlled directly by supervisory control unit, which calculates the value of the main control parameter (**N1** or EPR) corresponding to the demand thrust to obtain **N1** command or EPR command. This function is called ‘power management’ and the calculating unit is called ‘power management control’ (PMC) (Diesinger 89). Supervisory engine management is used in proven hydro-mechanical fuel control to obtain complete control over all engine functions and is characterized by a mechanical backup which monitors engine control in case of failure of the electronic control. Supervisory engine management systems, thus, perform functions necessary for engine operation and protection enabling reduction of crew...

7 Pages(1750 words)Coursework

...Standard deviation=0.28808
Standard deviation=0.553894
c) Interpretation of the descriptive statistics.
The mean CAL in controls is higher than the mean CAL in the cases. The standard deviations are large implying that the values are highly distributed in the population and are far from the means.
d) The mean CAL values are normally distributed in the two groups. Most of the data points are close to the mean, where as few data points are far from the mean. Therefore, there is a high likelihood that the data follows a normal distribution.
e) The difference in the mean CAL between case and control groups with the corresponding 95% confidence interval.
The confidence interval is given by the formula M1-M2 ± t α/2, df ×...

4 Pages(1000 words)Essay

...The Scheme’s Properties
Given two nodes, **n1**, n2, with level1, level2 as their levels (the level refers to the level of the element node in the XML tree), and with labels A and B, where global labels are a1.a2…am , ith and b1.b2…bn, , jth respectively and their local labels are La1. La2 and Lb1. Lb2, the label properties can be defined as in the following:
Node Level:
The level information of each node can be derived from its global label as follows:
The level is the number of components in the first part of the node’s global label plus 1 if the second part of the global label exists; i.e. if the SG equals null, the level is the number of component in FG.
e.g. As shown in Fig. 4.1 and Table 4.1, the level of node ‘B’...

12 Pages(3000 words)Essay