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Friday, January 30, 2015


MATH E 115  ASSIGNMENT FOR FEBRUARY 6, 2015


1.  The base of a right prism is a rhombus with a side of 12 cm and an altitude of  15 cm. Determine the diagonals of the prism if the if the height of the prism is 60 cm.

2.  The total area of a regular square prism is 122 dm2 and its altitude is 240 mm, Determine the volume of the prism.

3.  The volume of a rectangular prism  is 6,840 cm3. Determine its dimensions if they are in the ratio of in the ratio of 2:3:5.

4.  The base of an aquarium has dimensions of  60 cm by  150 cm.  If the height of the water inside is 45 cm, how many kilograms of water does it hold if density of water is  1000 kg/m3 ?

5.Find the volume of the largest cylinder with a circular base  that can be inscribed in a cube whose volume is  14,137.16 cm3.

6.  A circular log of radius 60 cm and  600 cm long is to be cut to form a lumber with
a square cross section. Find the largest volume of the lumber that can be obtained.  

7.  Determine the surface area of a circular log of radius 45 cm and  550 cm  long.  What is the base area and volume of the log ?

8.  A rectangular tank with base dimension of  220 cm by 3.6 m is containing water with height of 1.86 m.  If this water is transferred into  cylindrical tank with diameter of  200 cm,  what is the height of the water. How many kilogram of water does the tank hold ?

E N D



Thursday, January 29, 2015

F Elect 3 assignment due on February 3, 2015


  F Elect 3 assignment due on February 3, 2015

    I.  Use quadratic functions to solve the problems.  

       1. A rectangular field is to be enclosed, and divided into three lots by fences parallel
           to one of the sides. Determine the dimensions of the largest field that can be
           enclosed by 2,400 m of fencing material.

       2. A rectangular lot  is to be fenced off along a highway. If the fence along the
           highway cost  P 150  per meter, and on the other sides  P 100 per meter,
           determine the dimensions of the largest lot that can be enclosed  for a budget
           of  P 100,000.

       3. A rectangular field is to be fenced off along the bank of a river. If no fence
           is needed along the river, what is the shape of the  field requiring 400 m of fence if
           the area is to be maximum? 

      4. A triangle has a base of  12 cm  and an altitude of  8 cm. Find the area of the
          largest rectangle that can be inscribed in the triangle so that the base of the 
          rectangle falls on the base of the triangle.
         
      5. A right triangle has a hypotenuse of  30 cm.  Show that the area is maximum
          when the triangle is isosceles.

      II.  Find the critical points and determine whether it is maximum or minimum.      
           1.   y  =  6  –  5x  –  x2  
           2.   y  =  x2  –  7x  + 12  
           3.  3x2 – 6x +  9 +  3y = 0
           4.  4y = 8x2 +  12x
           5.  4x2 – 2y +  6 = 0 



CSE 323 Assignment for February 2, 2015




  CSE  323 – CALCULUS  ASSIGNMENT FOR  FEBRUARY  2,  2015

  1.  Find the volume of the largest box that can be made by cutting
      equal squares of side x out of the corners of a piece of cardboard  
      of dimension 28 cm by 42 cm, and then turning up the sides.  

   2. What is the most economical proportions of a quart can?
                ( Hint : a quart can is close on both ends )

   3. What is the most economical proportions of cylindrical cup?
               ( Hint : a cup is close on one end only ) 

   4. A closed box whose length is twice its width, is to have a
       total surface of 192 cm2. Find the dimensions of the box 
       when its volume is maximum.  

   5.  If the three sides of a trapezoid are each 8 cm long, how
        wide must the 4th side be if the area is to be maximum ?

   6.  Find two positive numbers whose product is 64 and whose
        sum is a minimum.

   7.  if the hypotenuse of  right triangle is given, show that the area
        Is maximum when the triangle is isosceles.

   8.  The strength of a rectangular beam is proportional to the breadth
        and the square of the depth. Find the shape of the strongest beam
        that can be cut from a log of given size.

                                       E N D


LESSON 5 : PARABOLA

PARABOLA

Parabola is the locus of a point which moves so that its distance from a fixed point and a fixed line are equal.

The fixed point is called the focus of the parabola and the  line is called the directrix of the parabola.


      DD’  –  directrix
      E, E’  –  end points of the latus rectum
      V  –  vertex of the parabola
       F – focus of the parabola 
      VF = a   –  focal distance
       DD’V =  VF =  a 
      EF = FE’  =  2a
  The axis of the parabola is horizontal ( OX ).

From the above figure and by definition of parabola :

 PF  =  PQ  

Ö ( x – a )2 + ( y – 0 )2    =  ( x + a ) 

Square both sides

( x – a )2 + y2  = ( x + a )2

X2 – 2ax + a2  + y2  =  x2  + 2ax  + a2

X2 – x2 + a2 – a2 + y2 = 2ax + 2ax

y2 =  4ax 
 
Standard equation of a parabola having horizontal axis and opens to the right.        
                                                          
From the figure above and by definition of parabola :

PF  =  PQ

Ö( x – 0)2 + ( y – a )2   =  y + a

Square both sides

X2 + y2 – 2ay + a2  =    y2 + 2ay + a2

X2 + y2 – y2 + a2 – a2  =  2ay  + 2ay

X2 = 4ay
Standard equation of a parabola with vertical axis (OY) and opens upward.

General Equation of a Parabola

1.  x2 + Dx + Ey + F = 0 ,  E ¹ 0,   with vertical axis ( OY )

2.  y2 + Dx + Ey + F = 0,   D ¹ 0,  with horizontal axis (Ox )

Standard Equation of the Parabola

A. with vertex at V( 0, 0 )

      1.  y2 =  4ax  ----   opens to the right

      2.  y2 =  – 4ax  ----  opens to the left

      3.   x2 = 4ay  ----  opens upward

      4.   x2 =  – 4ay  ----  opens downward  
     
B. with vertex at  V( h, k )

      1.  ( y – k )2 =  4a( x – h )    ----   opens to the right

      2.  ( y – k )2 =  – 4a( x – h)  ----  opens to the left

      3.   ( x – h )2 = 4a( y – k )   ----  opens upward

      4.  ( x – h )2 =  – 4a( y – k   ----  opens downward  

Examples.

Draw the graph of the following parabola.
  1.  2x2 – 8y = 0                                   3.  x2 + 2x – 6y = 2
  2.  3y2 + 18x = 0                     4.  y2 + 2y – 6x = 2 
 5.  x2 + 4y – 4 = 0                                 8.  y2 + 8x  + 16 = 0
 6.  y2 – 4x + 4 = 0                                 9.  x2 = 4x + 4y 
 7.  y2 + 2x + 6y + 17 = 0                     10.  x2 – 2x + 2y + 7 = 0


QUADRATIC FUNCTIONS

        A function of the form y = ax2 + bx + c,  where  a ¹ 0 is a quadratic function. Since the equation has the form of the general equation of a parabola  x2 + Dx + Ey + F = 0, the graph of every quadratic function is a parabola with a vertical axis. It is concave upward if a is positive or concave downward if a is negative.

           At the vertex of the parabola,  x = – b/2a. This value of x gives the function its minimum value if a is positive or concave downward if a is negative.    The   equation   of the   axis of the   parabola is x = – b/2a.   

        As x increases, many functions increase to a maximum value, decreasing thereafter, or decrease to a minimum and begin to increase. In such cases, it is a problem of prime importance to determine the maximum or minimum value. To solve problems for functions in general, differential calculus is required. For the quadratic function it may evidently be solve by merely finding the vertex of the parabola, since the ordinate of the vertex is the greatest or least ( algebraically ) of all the ordinates, according as the parabola opens downward or upward.


STEPS IN GRAPHING A QUADRATIC FUNCTION
  1. Compute x = – b/2a, and then the corresponding value  of y
      to obtain the coordinates of the vertex of the parabola.
       
  2. If the value of  x = – b/2a is convenient, form a table by using
      pairs of values of x where each pair, the values are
      equidistant from the vertex on either side. The value of y
      corresponding to any pair will be equal.

  3. If the value of  x = – b/2a  is inconvenient, form the table by
      choosing values of x arbitrarily on either side of  x = – b/2a.

Example.  Determine the critical points whether it is maximum or minimum.
                1.  y  =  x2 – 10x + 25 
                2.  y =  4x –  x2                                               
                3.  y =  2x2 – 10x + 3
                4.  y = 15x – 3x2  
                                                          
Application problems.

Use quadratic function to evaluate the following problems. 
 1. Find the dimensions of the largest rectangular lot that can be
     enclosed by 100 m of fencing material.
      
 2. A rectangular field is to be fenced off along the bank of a
   river. If no fence is needed along the river, what is the shape
  of the field requiring 400 m of fence if the area is to be
  maximum?

   1. Solution
       Let  x and y be the dimensions (length and width) of the lot       
            Area         :   A = xy
               Perimeter :    2x + 2y = 100                                      
                                x +  y = 50                                                                            
                                y = 50 – x
                            x(50 – x ) = A                                                                                             
                             50x – x2  =  A                                              
                           Standardize the equation :
                               X2 – 50x = – A                                                                
                            x =  – b/2a  = – (–50) / 2 ( 1 )
                            x =  25 m  
                            y = 50 – x
                            y = 50 – 25
                            y = 25 m

                  Alternately :       

                                          By completing the square :
                              x2 – 50x + 625 = – A + 625  
                             ( x – 25 )2 = – ( A – 625 )
                             x – 25 = 0        |   A – 625 = 0
                             x = 25 m          |   A = 625 m2   
                            y = 50 – 25
                              = 25 m
               The dimensions are 25 m by 25 m.  
               The maximum area is  625 m2                  
  B.  Use quadratic functions to solve the problems. 
     1. A rectangular field is to be enclosed, and divided into three
        lots by fences parallel to one of the sides. Determine the
      dimensions of the largest field that can be enclosed by
      2,400 m of fencing material.
  
    2.  A rectangular lot  is to be fenced off along a highway. If the
         fence along the highway cost  P 150 per meter, and on the
         other sides  P 100 per meter, determine the dimensions of
         the largest lot that can be enclosed  for a budget of
         P 100,000.
  
    3. A rectangular field is to be fenced off along the bank of a
        river. If no fence is needed along the river, what is the
        shape of the  field requiring 400 m of fence if the area is to
        be maximum?

   4. A triangle has a base of  12 cm  and an altitude of  8 cm. 
       Find the area of the largest rectangle that can be inscribed
       in the triangle so that the base of the rectangle falls on the              
       base of the triangle.  

   5. A right triangle has a hypotenuse of  30 cm.  Show that the
       area is maximum when the triangle is isosceles.

                                   


Saturday, October 11, 2014

STATISTICS


COLLECTION AND PRESENTATION OF  DATA
 Types of data :
1.      Qualitative data are categorical data taking the form of attributes or categories such as sex, course, race, religion, blood type, etc.  This type of data can only be categorized but cannot be quantified.

2.      Quantitative data or numerical data are obtained from measurements like weights, ages, heights, temperatures, score and other measurable quantities.


Measurement is a process to convert qualitative data to quantitative data. By measurement, numbers are used to code objects or items so that they can be treated statistically.

Measurement scales
1.      Nominal measurement is used merely for identification or classification purposes.
Example. A group of students maybe classified according to their courses:
    1 – BSED                              4 – BSBA                   7 – BSIS 
    2 – BEED                              5 – BSIT                     8 – BSA
    3 – BSCS                              6 – BSN                      9 – BSCpE

2.      Ordinal measurement. The ordinal measurement does not only classify items but also give the order or rank of classes, items or objects.
          Example.  a) Ranks as  1st,  2nd,  3rd  in an oratorical contest
                           b) Ranks as  1st, 2nd,  3rd  in a poster making contest

3.      Interval measurement. In the interval measurement, numbers assigned to the items or objects not only identify and rank the objects but also measure the degree of differences between two classes.
Example.  Height, mass, temperature, score
a)      If  Jay has a mass of 48 kg and Claire is  54 kg, the mass difference of 6 kg implies that Claire is 6 kg heavier than Jay.
b)      If water from the faucet is 22 degrees Celcius, and that coming from the well is  27 degrees Celcius, the temperature difference of 5 degrees implies that water from the well is warmer than the water from the faucet.
c)      If  John has a height of  5’ 8” and  Jim stands  5’ 4”, then the difference of 4 inches implies that John is taller than Jim by 4 inches.
4.      Ratio measurement.  For ratio measurement, the ratio of the numbers assigned in the measurement reflects the ratio in the amount of property being measured. Multiplication and division have meaning in ratio measurements.
          Thus, if Rezzie is 18 years old and Stephen is 24 years old, then their ages can expressed              
          in the ratio of  3 : 4.
          If a motorcycle moves 80 km/hr and a car moves with a speed of  100 km/hr, then their 
          speeds can be expressed in the ratio of  4 : 5. 

SAMPLING TECHNIQUES

Sampling techniques are use to ensure the validity of conclusions or inferences from the sample to the population.
 Common types of sampling techniques:
    1.  Simple random sampling
    2.  Stratified random sampling
    3.  Systematic random sampling
    4.  Cluster sampling
    5.  Multi-stage sampling

Samples can either be a Judgment sampleConvenience sample or a Random sample.
   
* Judgment sample is one in which the individual selecting the sample items uses experience as the basis for choosing the items to be included in the sample with the objective to make the sample as representative of the population as possible. An example is an auditor who chooses only few records necessary for sample audit base on the judgment that these records are representative of the records in general. 

* Convenience sample includes the most easily accessible measurement or observations. An example is taking an opinion poll from students passing by the Plaza Madonna entrance gate.

* Random sample is a sampling procedure by which every element in the population has a known and usually equal chance of being chosen for inclusion in the sample.

 *  Simple random sampling.
         A random sample has limited number of individuals chosen from the population. Before the selection is done every individual has an equal chance of being selected in the sample.
        Lottery is the simplest method of random sampling. To illustrate the process, consider a population of 356 CITE students. Obtain a coded or numbered list of all the students. Write the numbers or codes in a small slip of paper, roll the papers and place in a container. Jumble it thoroughly; and without looking at the slip of papers draw, i.e. 160 slips representing 160 students, the desired sample size. The process ensures a representative sample because each item has the same chance of getting into the sample.

        If the population is large, the table of random numbers can be used. This allows us to select the desired sample on a purely chance basis.
1.      Identify the population and determine the sample size.
2.      Make a list of all members of the population and number them consecutively from zero to the last number of the population.
3.      Using the table of random number, point at any number without looking at the table. Since the code numbers is composed of 3 digits, just look at the last three digits of the number pointed out in the table. If the last three digits is a number in the list, then the member corresponding to that number is selected to be included in the sample.
4.      Using this number as reference, read the table horizontally whether to the left or to the right; or vertically up or down or diagonally. Continue recording those numbers found in the population ignoring those numbers which are not found. Discard duplicate numbers and continue until the required sample size is completed.

*  Stratified Random sampling
           - is used to avoid biasedness. This sampling technique is done by dividing the population into categories or strata and drawing the members to be included in the sample at random proportionate to its stratum or subgroup. T select a stratified random sample, the researcher must possess a clear understanding of the problem in order to determine the strata into which the population should be divided.
           Example. a) A college has a population of 2000 students of which 1200 are females and 800 are males. Here, the strata is the male and the female group of students.  After the categories or strata have been determined, a random sample is selected from each stratum which is proportional to its sample size. This avoids the error of selecting too few or too many members from each stratum. A sample of 400 students is required.  To obtain a sample proportional to the given members in each stratum, compute first for the proportion of the sample to the population. Then multiply the result to the number of males and to the number of females.

            400/2000 = 0.20

            ( 0.20 ) * ( 1200 ) =  240 females in the sample
            ( 0.20 ) * ( 800 ) =  160  males in the sample

The last step is to draw the required random sample size of 400 students either by lottery or by the use of table of random numbers from the 240 females and 160 males.

*  Systematic random sampling
          - is a process of selecting every nth element of the population until the desired sample size is obtained. The members or elements maybe arranged alphabetically or in any systematic fashion. To obtain n, divide the population size by the sample size. In the above example, 2000/400 = 5, which implies that every 5th element of the given population is chosen to be included in the sample.
           To know where to start from the population list, use the table of random numbers. Without looking, point at any number in the table and get the last digit. Supposed that the last digit in the number pointed is 3, it means that the 3rd member in the list is included. From the number 3, take every 5th element in the population as belonging to the sample. This means in the list 3, 8, 13, 18, 23, 28, 33, 38, 43, 48, 53, etc. are included in the sample.  

*  Cluster sampling
        Cluster is an intact group possessing common characteristics.
        Cluster sampling is an advantageous procedure if the population is spread out over a wide geographical area. It is also a practical sampling technique used if the complete list of members of the population is not available. An example is a population consisting of 600 teachers in the 30 public schools of Midsayap. To illustrate how the desired sample size of 140 teachers could be selected from the given population,  the steps are as follows :
       1.  Prepare a list of the cluster comprising the population and determine the sample size. The logical cluster is by school. Prepare a list of all the 30 public schools.
       2.  Estimate the average number of members per cluster in the population. Assume that the average number of teachers per school is 20.
       3.  Divide the required sample size by the average number of teacher per cluster to get the number of clusters to be selected. Since, the sample size is 140 and the average teacher per school is 20, the number of clusters shall be :   140/ 20 = 7                                          
       4.  Select the needed number of clusters. By using the table of random numbers, select the 7 public schools from the 30 population schools. Include all the members in the selected clusters. Since there is an average of 20 teachers per school, and 7 clustered school, the sample size of 140 teachers is completed.
                        Summary :
                        Number of public schools                               :  N = 30
                        Average number of teachers per school          :  x = 20
                        Desired sample size                                         :  n = 140
                        Required cluster                                              :  y = n/x = 140/20 = 7
Hence, there are 7 clusters. 

*  Multi stage sampling
             This is a more complex sampling technique which involves the following steps :
   1.  Divide the population into strata.
   2.  Divide each stratum into clusters.
   3.   Draw a sample from each cluster using the simple random sampling technique.
To illustrate, take a random sampling of all first year high schools in the island of Mindanao. Mindanao is divided into 6 regions which forms our stratum. From each stratum, we choose a private and a public high school which again forms a stratum. From these public and private high school, we select first year high school classes forming our clusters. From these clusters, we select first year students from each class by using the simple random sampling technique.


METHODS OF COLLECTING DATA
There are several methods of collecting data. There is however, no best method to obtain the desired information under investigation. The choice of appropriate methods to be used depends on the following factors :
1.      Nature of the problem.
2.      The population under investigation, and
3.      The time and the material factors.

The methods of collecting data are :
1.      The direct interview method
2.      The indirect or questionnaire method.
3.      The registration method, and
4.      Other methods such as;
a)      Observation
b)      Telephone interview
c)      Experiments

*  The direct or interview method
          This is considered as one of the most effective methods of collecting original data. To obtain accurate responses, the maybe conducted by well-trained interviewers. The interviewers maybe of great help to the respondents in answering questions which the respondents could not understand.

Advantages of the interview method
1.      It can give complete information needed in the study.
2.      It can yield precise and consistent information since the interview can immediately clarify any misinterpretation made by the respondents.

Limitations of the interview
      1.  It is more expensive and time consuming.       
      2.  It may yield inaccurate information since the interviewer  can influence the respondent’s    
           through his facial expression, tone of voice or wording the question.
    3. The interviewer may cheat by turning in dishonest responses if their expected or desired responses 
         are  not obtained. 
*  Indirect or Questionnaire method
           - is one of the earliest method of data gathering. It takes time to prepare because questionnaires need to be attractive. It can include illustrations, pictures and sketches. Its content especially the instructions, must be precise, clear, and self-explanatory.

Advantages of the questionnaire method       
1.      It is less expensive since questionnaires can be distributed personally or by mail.
2.      It is less-time-consuming since it can be distributed over a wider geographical area in a shorter time.
3.      It can give confidential responses since the respondents can answer the questionnaire privately.
4.      The answers obtained are free from any influence from the interviewer.

Limitations of the questionnaire method
1.      It cannot be accomplished by illiterates.
2.      It has a high proportion of non-response or non-return.
3.      It tends to yield wrong information since answers cannot be corrected right away.
4.      It tends to give incomplete information.

*   The Registration method
       By registration method, the respondents give information in compliance with certain laws, policies, rules, regulations, decrees or standard practices. Data which can be collected by registration method are as follows : marriage contracts, birth certificates, motor vehicle registration, license of firearms, registration of corporations, real estates, voters, etc.

Other methods :  
 *  Observation method is used to gather data regarding attitudes, behaviour, values, and cultural pattern of the sample under investigation.

*  Telephone interview are employed if the questions to be asked are brief and few. An example is the checks made on listeners to certain radio programs, like asking what programs his radio is turned on to. This method is used to find the most popular Radio or TV programs.

*  Experiment is employed to collect data if the investigator wishes to control the factors affecting  the variables being studied.  An example is when the researcher wants to determine the different factors affecting the academic performance of the students such as IQ methods or approaches used in teaching.

METHODS OF PRESENTING DATA
1.      Textual or Textular method
2.      Tabular method
3.      Semi-tabular method
4.      Graphical method

*  Textual presentation – the data is presented in paragraph form. The reader gets information by reading the gathered data in the paragraph.
*  Tabular presentation – the data is presented in rows and columns. This is more effective way of showing relationships or comparison of numerical data. It gives more precise, systematic and orderly presentation of data in rows and columns. This method makes comparison of figures easy and comprehensible.

*  Semi-tabular presentation – employs both the textual and the tabular methods. This method is used only if there are a few figures to be tabulated. The tables are followed by narrative explanations to make he facts more understandable.

*  Graphical presentation. The use of graphs is the most effective method of presenting statistical results or findings. It gives the relationship of data in pictorial form. Presentation of facts are made attractive and meaningful especially if colors are made and pictures re used, making it easy for important information to be grasped by the reader.

   **  Advantages of Graphical presentation
1.      Graphs enable students, readers and the busy executive to easily understand the essential facts that numerical data intend to convey. Private and government agencies use charts and graphs in their reports.
2.      They can easily attract attention and are more readily understood. It is easier to go through graphs than through quantitative data.
3.      Graphs simplify concepts that would otherwise have been expressed in so many words.

   **  Limitations of graph
      1.  Graphs are not as precise as tables.
      2.  Graphs require more skill and time to prepare.
      3.  Graphs can only be made after the data have been presented in tabular form.

Parts of a Statistical Table
  1.  Table heading – consist of a table number and the title.
  2.  Stub – are found at the left side of the body of the table which are categories or
       classifications.
  3.  Box head – The box head identifies what are contained in the column. Included in the Box
        head are the stub head, master caption and column caption.
  4.  Body – main part of the table. This contains the substance or the figures of one’s data.  

Table number
Title

Stub
Head
Master  Caption
Column Caption
Column Caption
Column Caption
Row Caption



Row Caption



Row Caption
B O D Y
of  the
Table
Row Caption



Row Caption




               Footnote
               Source of data

Example
Table 1.1
College Student Population of XYZ University
S. Y. 2007 – 2012

School Year
BSCS*
BSIT*
BSED*
BEED*
BSBA*
AS*
BSN*
2007 – 2008 
75
150
45
60
290
130
56
2008 – 2009 
70
140
 34
50
285
108
45
2009 – 2010 
50
120
30
45
270
62
34
2010 – 2011 
48
85
28
43
225
48
23
2011 – 2012 
30
75
22
38
213
44
18

 *  Summer enrolment is not included
     Source of data : Registrar’s Office

TYPES OF GRAPHS COMMONLY USE

1.      Linear graph ( Line graph )
2.      Bar graph
3.      Hundred percent chart ( Pie chart or Circle graph )
4.      Pictogram ( Pictograph )
5.      Statistical maps

*  Linear graph is a practical device uniquely suited to portray changes in values effectively over successive periods of time. Variations in the data are indicated by the variation in the movement of linear curves.
   **  Advantages of  linear graph
1.      The curves shows data as a continuous line; hence, it is continuous in its effect.
2.      The wandering line of the curve tells the whole story. At a glance one can see just what the situation is and what is likely to happen.
3.      Its preparation requires less time and skill. 

*  Bar graph – is one of the most common and widely used graphical devices. This consists of bars of equal width either all vertical or all horizontal. The length of each bar represents the frequency of each class. The bars must be of the same width and is arbitrary. The space between bars is about one-half of the width of the bars itself.

*  Pie chart ( circle graph ) – is use to represent quantities that make up a whole. It consists of a circle subdivided into sectors which look like pieces of a pie whose sizes are proportional to the magnitudes or percentages they represent. The pie chart aims to show per cent distribution of a whole into its component parts.

     **  Construction of a Pie chart    
1.      Express each component part as a certain per cent of the whole and multiply the result by 3.6 since 1 %  of the circumference is equal to 3.6 degrees.
2.      Mark off the desired number of degrees on the circumference of the circle using a protractor; connect this points to the center of the circle in order to produce pie-shaped areas which make-up the whole circle.
3.      Label the various segments horizontally whenever possible with the percentages indicated.
4.      Arrange the sector clockwise according to size.
5.      Use cross-hatching, coloring, shading, etc. to increase the effectiveness of a pie chart.
6.      Avoid overloading the chart by showing too many categories; simplicity is a fundamental characteristics of all charts.

*   Pictogram ( pictograph )
        An excellent device for portraying data is by means of pictures or symbols. Its chief purpose is to catch the reader’s attention to convey to him in a vivid manner basic numerical facts. It does not attempt to show details; it simply tries to facilitate comparison of approximate quantities.
        The symbol or picture should suggest the nature of the data being presented. For instance      
              a) corn production can be represented by a picture of a sack of corn;
b) truck importation can be represented by a picture of a truck;
c) population in one’s place can be represented by a person, etc.
    
*  Statistical maps
        One of the best ways to present data is through statistical maps. A  map can be drawn and divided into desired regions. Each region maybe distinguished from the other by use of shades, dots or cross-hatching. A statistical map is always accompanied by a legend which tells the meaning of the lines, colors or other symbols used.