Loren Timpone
Class notes from 10/28/08
MAE 301

Review of questions from the Regents Exam

  1. The first question that we went over from the Regents is

a)Since January 1980, the population of the city of Brownville has grownaccording to the mathematical model y = 720,500(1.022)x, where x isthe number of years since January 1980.Explain what the numbers 720,500 and 1.022 represent in this model.If this trend continues, use this model to predict the year duringwhich the population of Brownville will reach 1,548,800.

  • 720,500 1.022 represents a growth rate of 2.2% added to the current population
  • 1.022 represents the growth rate of the population is 2.2% per year
    OR
  • Population increase multiplied by 1.022 each year
  • It is not sufficient to say that the rate of the equation is 1.022 because the population does not increase by 1.022%

Rate of change was defined in class as

  • It is also defined as the change in the value of a quantity divided by the elapsed time (mathwords 2008).
  • Also if 1.022 was the rate of change how is it represented in the equation?
  • It was also mentioned that a derivative could be a possible definition to rate of change. However this definition is more sufficient to describe the instantaneous rate of change or the rate of change at a particular moment. Same as the value of the derivative at a particular point (mathwords 2008).
  1. Another question on the Regents asked to simplify the following expression.
  • This answer shows two equivalent expressions but not equality. These expressions can only be considered equal if it is given that.
  • It is often arbitrarily stated on regents that. Even though that information is not necessary to solve the problem it is good for students to understand its importance to the expressions.
  1. Given that the AB=8, BC=13 and m. How many distinct triangles can you make?

What this question is asking is there another angle that fit within the dimensions given.

Using the Law of Sine’s

C= sin-1 (

C

Therefore one angle for C is 21. This would make

To see if there is another triangle that would fit the criteria given find an equivalent angle. In this case 159. However 159 plus 36is greater then 180 and therefore cannot be anangle in the triangle.

Counting

Last class we were presented with the problem. Given 4 white rooks and 4 black rooks how many ways can you place them on the chessboard so that no 2 can be in the same horizontal or vertical line?

B
W

The first rook can be placed anywhere on the chessboard. So there are 64 possible spots to place it. If the first rook, black, was placed in the top corner then no other rook, no matter what color, can be placed in the first column or the first row. Therefore there are only 49 possible spots for the second rook to be placed. If a white rook was placed in the second space of the second columnthen no other rook can be placed in the second row or second column. Continuing this for all the columns the next rook would have 36 spots available then 25, 26, 9, 4, and 1 respectively.

However, multiplying all these values together would give you a value to big because it doesn’t take into account the number of repetitions of the white and black rooks.

Therefore you have to divide by the number of repetitions of the black and white rooks. So by dividing by 4!, to account for the shuffle of white rooks, and 4!, for the shuffle of black rooks, the equation you get is.

One student explained the repetition by using the word MISSISSIPPI. To find the number of different ways the letters can be written one must divide by the number of repetitions because it is the same combination even if some S’s exchange places.

Another counting example was presented. It stated how many integers 100-999 inclusive have distinct digits?

  • In this context distinct means integers 100-999 that do not contain repeated digits. Some examples would be 102 and 345 whereas 101 and 454 do not have distinct digits.

One way to solve this problem would be to notice that there are 9 possible choices for the first number. It can be 1, 2, 3, 4, 5, 6, 7, 8, or 9. The only restriction is that the first number can’t be zero because then it will not be a 3 digit number.

The second number can be 0, 1, 2, 3, 4, 5, 6, 7, 8 or 9 but since one number was already chosen for the first term there are only 9 possible chooses left.

The last number can also be 0, 1, 2, 3, 4, 5, 6, 7, 8 or 9 but since two numbers were already chosen for the first and second terms there are only 8 possible chooses left.

Therefore the number of integers is

Another way that was presented to solve the problem is by using a chart

0-9 / 10-19 / 20-29 / 30-39 / … / 90-99
100’s / 8 possibilities / 0 possibilities / 8 possibilities / 8 possibilities / 8 possibilities
200’s / 8 possibilities / 8 possibilities / 0 possibilities / 8 possibilities / 8 possibilities
300’s / 8 possibilities / 8 possibilities / 8 possibilities / 0 possibilities / 8 possibilities
400’s / 8 / 8 / 8 / 8 / 8 possibilities
500’s / 8 possibilities / 8 possibilities / 8 possibilities / 8 possibilities / …
… / … / … / … / … / 8 possibilities
900’s / 8 possibilities / 8 possibilities / 8 possibilities / 8 possibilities / 0 possibilities

In the first row and column there are 8 possibilities out of a possible 10 numbers to choose from. 101 and 100 are excluded because they are not distinct.In the second column in the 100’s row all values between 110 and 119 are excluded because they all have a repeating one. Similarly to the first column the rest of the ranges have 8 possibilities because two numbers out of the possible ten contain repeating digits.

In the proceeding columns and rows the same pattern occurs. Therefore multiplying

The second part of this question was: how many of those are odd?

  • One possible solution is to use the counting principle. Some students may try to start with the first number and then subtract possibilities as you go on, like the first half of the problem. However, it becomes difficult when getting to the last number.

The first number can be 1, 2, 3, 4, 5, 6, 7, 8, and 9. And the second number can be 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 minus the number that was used in the first. So there are also nine choices. However the third number can be either 5 numbers, if neither of the first two are odd, 4 if only one of them is odd or 3 different numbers if they are both odd.

However it becomes easier if you start with the third number because it is the most restricted. The last number can have 5 possibilities 1, 3, 5, 7 and 9. Then the next number that is restricted is the first one because it can’t be zero. So there are 8 possibilities for the first number, 1, 2, 3, 4, 5, 6, 7, 8 and 9. The second number can also have only eight possibilities because it can be 0-9 minus two numbers which were already used. Therefore,

  • Another method that was used to solve this problem was to use the 100’s and 200’s group to simulate what happens in all other groups (300’s - 900’s).

In the group 100-199 we know that 72 numbers do not repeat and 28 do.

Out of the 28 that do repeat 18 of those are odd (101, 111, 113, 115, 117, 119, 121, 131, 133, 141, 151, 155, 161, 171, 177, 181, 191, 199) and 10 are even (100,110, 112, 114, 116, 118, 122, 144, 166, 188)

In the group 200-299 there are also 28 numbers that repeat. Out of those there are 18 even numbers and 10 odd numbers.

Then by taking the 5 odd sets of numbers (100’s, 300’s, 500’s, 700’s and 900’s) and multiplying it by the number of odd numbers in each odd set you get

Then if you take the 4 even sets of numbers (200’s, 400’s, 600’s, and 800’s) and multiply it by the number of odd numbers in each even set you get

By adding 90 + 40 = 130, you find that there are 130 odd numbers that repeat.

Since we know that 100-999 inclusively there are 450 odd numbers. By subtracting the number of odd numbers that repeat from the total number of odd numbers we get the total number of distinct odd digits.

Another example of a counting problem

  • In a certain country telephone numbers have 9 digits. The first two digits are the area code (01) and are the same within a given area. The last 7 digits are the local number and cannot begin with 0. How many different telephone numbers are possible within a given area code in this country?
  • The diagram below shows the number of choices for each digit. The first digit of the area code is 0, no choice. The second digit of the area code is 1, no choice. The first digit of the local code can be any digit except 0, so 9 choices. The 2nd, 3rd, 4th, 5th, 6th and 7th digits of the local code can be any digit, therefore 10 choices each.

(analyzemath.com)

Probability

The theoretical definition of probability from last class:

Given 3 fair coins find the following probabilities given and find an answer in terms of the definition.

  • Using the definition, the probability of getting 2 heads is:

=

There are three possible combinations that give two heads and eight total possibilities.

Therefore: P (2 Heads) =

  • Using the definition, the probability of getting at least 2 heads is:

=

Thereare four possible combinations that give two or more heads and eight total possibilities.

Therefore: P (at least 2 Heads) =

  • Using the definition, the probability of getting no heads is:

=

There is one possible combination that gives no heads and eight total possibilities.

Therefore: P (no heads) =

Given 3 weighted coins and that and find the following probabilities.

  • P( 2 Heads)= P(HHT) + P(HTH) + P(THH)

=

  • P( at least 2 Heads)= P(THH) + P(HHT) + P(HTH) + P(HHH)

=

  • P(no heads)= P(TTT)

=

The question that was posed for next class was to describe how these answers fit the current definition and how they do not.