How to create learning objectives that are clear and concise, containing the required elements and components, resulting in effective learning for students.  Please give me an example for to the following for better understanding.

• The trainer will explain how to create spreadsheets.
• After attending the course, participants will be better communicators.
• The purpose of this training program is to help managers deal with conflict.
• The trainees will be more efficient upon completion of this training program.

 

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##abs(vec a) = 5 m/s^2 ##

## theta = arctan (4/3)##

by definition average acceleration over time ##t## is:

##vec a = (Delta vec v)/(t) = 1/t (vec v_2 – vec v_1)##

using W-E as positive Cartesian x-direction and S-N as positive y direction, we have

##vec v_1 = ((0),(-8)), qquad vec v_2 = ((6),(0))##

##implies vec a = 1/2( ((6),(0)) – ((0),(-8)) )= 1/2 ((6),(8)) = ((3),(4))##

##implies vec a = ((3),(4))##

from that we can calculate the magnitude of the average acceleration as

##abs(vec a) = sqrt(3^3 + 4^2) = 5 m/s^2 ##

it’s direction, where ##theta## is measured positive anti clockwise from the x-axis is

##tan theta = 4/3, theta = arctan( 4/3)##

 

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How to explain the following questions:

1. Risk and return for Sm Stock, LG Stock, Bonds….

2. Systematic and unsystematic risk, names etc

3. Beta for risk free and the market and what is usually used to reference each

4. Someone tells you there is a reward for taking on risk so they say you are correctly rewarded when you buy a single stock. How do you respond?

 

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##lim_(x->1) [1/(1-x) – 3/(1-x^3)] = -1## and ##lim_(x->1) (x^n-1)/(x-1) = n##

First Limit

##lim_(x->1) [1/(1-x) – 3/(1-x^3)]## We cannot use the for limits, because the denominator has a limit of ##0##.

Write as a single ratio.I will use ##(1-x^3) = (1-x)(1+x+x^2)## (difference of cubes).

##lim_(x->1) [1/(1-x) – 3/(1-x^3)] = lim_(x->1) [(1+x+x^2)/((1-x)(1+x+x^2)) – 3/(1-x^3)]##

## = lim_(x->1) ((1+x+x^2) – 3)/(1-x^3)##

## = lim_(x->1) [-(x^2+x-2)/(x^3-1)]##

## = lim_(x->1) [-((x+2)(x-1))/((x-1)(x^2+x+1))]##

## = lim_(x->1) [-(x+2)/(x^2+x+1)]##

## = -3/3 = -1##

Second Limit

##lim_(x->1) (x^n-1)/(x-1)##We cannot use the quotient rule for limits, because the denominator has a limit of ##0##.

Because ##1## is a zero of the numerator, ##x-1## is a factor of the numerator. Either recall the factorization or use division to get:##x^n-1 = (x-1)(x^(n-1)+x^(n-2)+ * * * +x^2+x+1)##

So,

##lim_(x->1) (x^n-1)/(x-1) =lim_(x->1) (x^(n-1)+x^(n-2)+ * * * +x^2+x+1)##

There are ##n## terms in the expression, each evaluating to ##1##, so, we get

## = underbrace(1+1+1+ * * * +1)_(n” terms”)##

## = n##

 

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How to do the following in SPSS software, the File is Lesson 34 Exercise File1 : Question is the following:

Assumption regarding the Absence of Multicollinearity among the Predictor Variables: If a predictor variable (x) is highly correlated with another predictor variable (x), they essentially provide the same information about the criterion variable. If the Variance Inflation Factor (VIF) is too high (greater than 10), you have multicollinearity and have violated the assumption. Acceptable values are between 1and 5. You are looking for the Absence of Multicollinearity.

I just need steps how can apply this in SPSS software, and which one to choose from for Acceptable values

 

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