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Praxis II Preparation
Praxis II Preparation
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Students will tend to cause disruptions in class for a number of reasons. In most cases, it may be due to natural loss of concentration. In Adam’s case, according to Mr. Potter, it was a way of getting attention, a cry for help. However, he chose to ignore his behavior in the hope that he may tire himself out or just get bored. Thompson et al (2010) explains that a need to be pleasing to others is necessary in order to avoid rejection and that one’s survival depends on whether people love them. This is the reason why Adam is always entertaining the class so that his approval rating is always high.
The teacher’s resolution to ignore Adam could be justifiable. Johnson (2011) says that if one has a short temper, the students will take advantage of one’s short fuse. Since his misbehavior occurred repeatedly, it was a signal that he needed help. The teacher should have inquired about Adam’s situation at home. This could be cause of his poor performance and stubborn behavior. If Mr. Potter felt that he was not a qualified counselor to help Adam, he could have sent him to the school’s counselor or psychologist.
The act of sending Adam to the office should have been the last resort. The office was busy, and the principal may not have given his case the full attention it deserved. Johnson (2011) argues that principals do not always support their teachers. One can assume that the teacher just slapped him on the wrist and sent him back to class. That is why he was quiet for just a day and was back to his disruptive behavior the next day because to him the punishment seemed light.
What normally occurs with sending a student routinely to the office is that it forms a cycle whereby the student sees the teacher as the cause of the problem. The student does the punishment given by the principal, comes back to class angry or amused and ready to start the cycle again. This will end in Adam missing valuable hours or even days from your class or other subjects.
Once one has understood the cause for misbehaving, one can put it in perspective. One can also identify a way for corrective action to prevent future recurrences. Adam’s problem seems to sprout during a writing assignment. He may be having a problem and it would be better if the teacher stands next to him to see if he needs any assistance. Nilson (2010) argues that the classiest way to handle a student in a situation like this is to handle it with humor, even asking the student to share the conversation with the entire class.
Instead of reprimanding Adam publicly, he should consider calling him aside privately. Most students do not like embarrassment in front of their classmates and may still be rebellious even after receiving their punishment (Nilson, 2010). Johnson (2011) says that a more subtle way is dropping a behavior card during class so that one does not disrupt the others. Using other non-verbal messages such as staring may work since we all respond to body language more than words. Trying to shout over a misbehaving or rowdy student will only make them more excited or get even louder. Adam’s temerity to make a paper plane and fly it in class exemplifies this fact. If left unchecked such behavior may spread to the other students.
To stop this, Mr. Potter needs to draft a classroom conduct contract at the beginning of the semester in which students read and sign showing that they will adhere to the set rules. The students can also add to the contract listing what the norms in the classroom should be. This way, they also feel responsible and part of the system. Finally, positive feedback will help improve Adam’s behavior. When he does behave appropriately, one should appreciate that there was notable improvement (Johnson, 2011).
References
Johnson, L., (2011). Teaching Outside the Box: How to Grab Your Students By Their Brains. (2nd ed.). San Francisco: John Wiley and Sons.Nilson, L. B., (2010). Teaching At Its Best: A Research-Based Resource for College Instructors. (3rd ed.). San Francisco: John Wiley and Sons.Thompson, C. L., & Henderson, D. A., (2010). Counseling Children. (8th ed.). Belmont, CA: Cengage Learning
Take Home test on Probability) N@
Take Home test on Probability) N@
Use the chart of sums of rolling 2 dice.
234567
345678
456789
5678910
67891011
789101112
Find the probability of the following sums
P (sum of 4)
Psum of 4=336=112P (sum 3) or P (sum of 5)
Psum 3=236and Psum of 5=436Psum 3 or Psum of 5=236+436=636=16P (sum greater than 8)
Psum greater than 8=1036=518P (sum less than 6)
Psum less than 6=1036=518Use the following chart. Suppose you encounter a delegate at random at a convention.
Women Men
Republicans 20 29
Democrats 24 17
Independents 7 3
Find the following probabilities:
You meet a woman
Pwoman=51100You meet an Independent
Pindependent=10100=110You do not meet a Democrat
Pnot a democrat=1-Pdemocrat=1-41100=59100You meet a female Republican
Pfemale republican=20100=15You meet someone who is not a male Democrat
Pnot a male democrat=1-Pmale democrat=1-17100=83100Use the following deck of cards
Spades Black A K Q J 10 9 8 7 6 5 4 3 2
Hearts Red A K Q J 10 9 8 7 6 5 4 3 2
Clubs Black A K Q J 10 9 8 7 6 5 4 3 2
Diamonds Red A K Q J 10 9 8 7 6 5 4 3 2
Find the following probabilities when selecting one card at a time from a deck of 52 playing cards with the following 26 red, 26 black, 4 of a kind, 13 in each suit
P (K)
PK=452=113P (red card)
Pred card=1352=14P (spade)
Pspade=1352=14P (of at least one Queen if you select a card from the deck 5 times)
Pat least one queen=1-Pnon-queen card=1-48C552C5=1-0.6588=0.34P (10 or a spade)
P10 or spade=P10+Pspade-P10 and spade=452+1352-152=413You have a jar of jelly beans.
5 Red 6 blue 5 yellow 8 orange 2 green and 3 purple
Find the following probabilities if you replace the jelly bean before making another selection
Total number of beans = 5+6+5+8+2+3=29P (green and red)
Pgreen and red=Pgreen×Pred=229529=10841P (blue and blue)
Pblue and blue=629629=36841P (purple and orange and yellow)
Ppurple and orange and yellow=329829529=12024389P (white)
Pwhite=029=0Find the following probabilities if you eat each selected jelly bean before making another selection.
P (yellow and orange and blue
Pyellow and orange and blue=529828627=201827P (red and red and red)
Pred and red and red=529428327=51827 P (black)
Pblack=0Flip a coin three times
These are all the possibilities
H H H
H H T
H T H
H T T
T H H
T H T
T T H
T T T
Find the following probabilities
P (3 tails)
P3 tails=18P (3 heads)
P3 heads=18P (exactly 2 Heads)
Pexactly 2 heads=38P (at least 2 Heads)
Pat least 2 heads=48=12P (more than 1 Tail)
Pmore than 1 tail=1-Pno tail+P1 tail=1-18+38=1-48=12P (4 tails)
P4 tails=0How many telephone numbers can be created if all numbers can be repeated and there are no restrictions.
_ _ __ _ __ _ _ _
If we see the phone number as having three “slots,” one for the area code, one for the first digit, and the third for the remaining six digits, there are 1 way to fill the first slot, 9 ways to fill the second slot, and 10*10*10*10*10*10 = 1000000 ways to fill the last slot, for a total of 1*9*106 = 954 possibilities.
How many 8 letter passwords can be created from the following? 26 Uppercase and 26 lower case letters
Numbers: 0 – 9 Symbols: % $ # * &
All numbers, letters and symbols can be repeated and there are no restrictions
_ _ _ _ _ _ _ _
Only lowercase: 268Only uppercase: 268Only digits: 108Only special: 338Grand total: 95^8−69^8−69^8−85^8−62^8+43^8+59^8+36^8+59^8+36^8+52^8−26^8−26^8−10^8−33^8 = 3025989069143040≈3.026×101
How many outfits can you create from 8 pairs of foot ware, 12 shirts, 9 pairs of slacks, 5 pairs of shorts?
8*12*9*5=4320 outfitsHow many ways can 9 people line up for a bus?
9x8x7x6x5x4x3x2x1, or 362,880 ways of arranging the peopleHow many passwords of 6 digits can be made using the following symbols?
@,#,$,%,&,* Numbers from 0 – 9
All upper and lower case letters
266*266*106*66=4.45*1027 passwordsA restaurant offers 5 vegetables, 6 main dishes, 10 desserts, 8 beverages, and 3 appetizers. How many different meals can be formed?
5*6*10*8*3=7200 meals
Roger.
& Webster
