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Absolute Value
Absolute Value:
Expression in distance, and the result is either 0 or positive.

|x|=a then a0


  • Theorem 1: If |x|=a then x=a or x=-a
  • Theorem 2: If |x|>a then x>a or x>-a
  • Theorem 3: If |x|<a then -a<x<a or x<a and x>-a
  • Midpoint: You can make a line segment using the absolute value equations (like the picture shown below)


    And with the absolute value equation, you are able to solve for the midpoint of the line segment. By using the equation x+y=0 in |x+y|. In this case, the y would represent a random number and the x would represent the midpoint. For instance, if there is the equation |x+1|, then the midpoint would be x=-1 since -1+1=0



  •  ■ Example Question for finding the value of x:
     -2|x+1|-3=-5

     →-2|x+1|=-2
     →|x+1|=1
     →x+1=-1 or x+1=1 using the theorem 1
      →x=-2 or x=0


     ■ Example Question for finding the value of the midpoint:
     |x+6|

     →We must find the midpoint and we can do that by using the equation |x+y|, x+y=0
     →Then we would get x+6=0
     →If we simplify this we would get x=-6 as the midpoint of the line segment of the absolute value.

    Probability 1
    Permutation and Combination

  • Permutation:
    Each of several possible ways in which a set or number of things can be ordered or arranged.
  • When using permutation, you can use the equation nPr=n!(n-r)!

  • Combination:
    Each of several possible ways, in which the order does not matter.
  • When using combination, you can use the equation nCr=n!r!(n-r)!


     ■ Example Question for permutation:
     If there are 12 contestants on a race that gives medals to first, second, and third place, how many possible  ways are there to get the top three runners?

     →First, you would write the equation for this question which would be 12P3 since there are 12 contestants in   total and we are looking for the top three.
     →Then you can simplify this to 12!(12-3)!
     →Which would equal to 1,320 possible ways.



     ■ Example Question for combination:
     If you were picking 3 people for President, Vice President, and Treasury out of 15 canidates, find all the   different possible ways the candinates be chosen for these roles.

     →The first thing you would do here would be to write an equation from this word problem and you would   get 15C3.
     →Then you would simplify it and get 15!3!(15-3)!.
     →Which would equal 455 possible ways.

    Probability 2
    Tree Diagram, Venn Diagram, and the Bell Curve

  • Tree Diagram:
    A tree diagram is a display of all possible outcomes of an event and each branch shows the possible outcomes of that event.


  • Venn Diagram:
    A diagram representing mathematical or logical sets pictorially as circles or closed curves.


  • Bell Curve:
    A graph of a normal distribution, with a large rounded peak tapering away at each end. When using the Bell Curve, this symbol µ represents the mean of the curve and the symbol σ represents the standard deviation.


  • The image on the left shows the different possibilities of each deviation starting from 34%, 13.5%, and to 2.5%. Each deviation can be calculated by using the equation mean+xσ (the x represents the deviation number).









     ■ Example Question for the Bell Curve:
     A sample of 1,000 newborn babies have a mean birthweight of 3.3kg with a standard deviation of 0.2kg. The data is distributed normally. Calculate the following probabilities.

    P(3.1≤x≤3.5)


    P(2.9≤x≤3.7)



     →First, you draw the Bell Curve and label it with the information you are given.
     →Then you can start solving for the probabilities
     →Using the probabilities of each deviations, you can see that P(3.1≤x≤3.5) would equal about 0.68 because of the 1st standard deviation is 0.34 then you would add another 1st deviation since P(3.1≤x≤3.5) is the within that area.
     →For the second probability P(2.9≤x≤3.7) it can be solved in a similar way the first probability was solved. Since both 2.9 and 3.7 are in the 2nd deviation, you can use the probability of the 2nd deviations which is about 0.475 (first deviation+second deviation) then you would add another 0.475 since P(2.9≤x≤3.7) covers the two areas.

    Advanced Probability
    Probability that may be more advanced using permutation, combination, etc.

  • Finding the probability of getting an exact value:
    When looking for a probability of an exact value (for instace drawing exactly 5 red cards out of 10 cards), you can use the equation nCr[p(x)]r[1-p(x)]n-r

    The n would be the total number of the event and the r would represent the wanted exact number from that event.
    [p(x)]rshows the probability of getting the wanted exact number from an event and [1-p(x)]n-rrepresents the probability of getting events other than the wanted one from the total number of events.

  •  ■ Example Question for exact value:
     A major league baseball player has a batting average of 0.35. Calculate the probability of exactly 3 hits from the next 5 at bats.

     →First, you would write down the equation which would be 5C3[0.35]3[1-0.35]5-3.
     →Then you could simplify it and solve for it to get about 0.181.
     →However, another easy way to do this using a GDC is using the function [BIONOMPDF]. Then all you have   to do is to plug in the values into the calculator and you would be able to get the answer.

  • Conditional Probability:
    The probability of one event given that another event had happened. Conditional probability uses the equation P(B|A) which means the probability of event A given that event B has occured.

  •  ■ Example for Conditional Probability:
     In a card game, suppose a player needs to draw two cards of the same suit in order to win. Of the 52 cards,  there are 13 cards in each suit. Suppose first the player draws a heart. Now the player wishes to draw a   second heart. Since one heart has already been chosen, there are now 12 hearts remaining in a deck of 51  cards. So the conditional probability P(Draw second heart|First card a heart) = 12/51.