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BUSINESS MATHEMATICS AND STATISTICS

Jul 13, 2014

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BUSINESS MATHEMATICS AND STATISTICS. A PRESENTATION ON. THE ADDITION AND THE MULTIPLICATION THEOREM OF PROBABILITY. INTRODUCTION.

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BUSINESS MATHEMATICS AND STATISTICS A PRESENTATION ON THE ADDITION AND THE MULTIPLICATION THEOREM OF PROBABILITY

INTRODUCTION • In our daily life, we all make the use of the word ‘probability’. In simple terms “Probability means the measurement of chance factor. It helps in quantifying uncertainty.” For instance, the probability it may rain today is half, i.e., it may or may not rain. Thus, the happening of the event is not certain. The value of probability of any event lies between 0 and 1. • Probability may be defined as the ratio of the favourable cases to the total number of equally likely cases. • Symbolically, • P (A) = p = = • Where, P (A) = Probability if occurrence of an event A • m = Number of favourable cases • n = Total number of equally likely cases • Similarly, • P ( ) = q = 1 – P (A) = 1 =

THEOREMS OF PROBABILITY • There are three main theorems of probability are : • ADDITION THEOREM • MULTIPLICATION THEOREM • COMBINED USE OF ADDITION AND MULTIPLICATION THEOREM • BERNOULLI’S THEOREM • BAYES’ THEOREM

THE ADDITION THEOREM • The Addition Theorem of Probability is studied under two headings: • Addition Theorem for Mutually Exclusive Events • Addition Theorem for Not Mutually Exclusive Events THE ADDITION THEOREM FOR MUTUALLY EXCLUSIVE EVENTS The Addition Theorem states that if A and B are two mutually exclusive events, then the probability of occurrence of either A or B is the sum of the individual probabilities of A and B. Symbolically, P (A or B) = P (A) + P (B) OR P (A + B) = P (A) + P (B) It is also known as the ‘Theorem of Total Probability.’

PROOF OF THE THEOREM Let n be the total number of exhaustive and equally likely cases of an experiment. Further let and be the number of cases favourableto the happening of the event A and B respectively. Then, P (A) = P (B) = Since, the events A and B are mutually exclusive, the total number of ways in which event A or B can happen is + , then P (A or B) = = + = P (A) + P (B) Hence, the theorem is proved. Generalisation The theorem can be extended to three or more mutually exclusive events. If A, B and C are three mutually exclusive events, then P (A + B + C) = P (A) + P (B) + P (C)

EXAMPLES ILLUSTRATING THE APPLICATION OF THE ADDITION THEOREM Question: A card is drawn from a pack of 52 cards. What is the probability of getting either a king or a queen? Solution: There are 4kings and 4 queens in a pack of 52 cards. The probability of drawing a king card is P (K) = and the probability of drawing a queen card is P (Q) = Since, both the events are mutually exclusive, the probability that the card drawn is either a king or a queen is P (K or Q) = P (K) + P (Q) = + = =

Question : A perfect die is tossed. What is the probability of throwing 3 or 5? Solution : The probability of throwing 3 is P (A) = The probability of throwing 5 is P (B) = P (A or B) = P (A) + P (B) = + = = Question :A card is drawn at random from a pack of cards. Find the probability that the drawn card is either a club or an ace of diamond. Solution :Probability of drawing a club P (A) = Probability of drawing an ace of diamond P (B) = P (A or B) = P (A) + P (B) + = =

Question: An investment consultant predicts that the odds against the price of a certain stock will go up during the next week are 2 : 1 and the odds in favour of the price remaining the same are 1 : 3. What is the probability that the price of the stock will go down during the next week. Solution: Let A denote the event ‘stock price will go up’, and B be the event ‘stock price will remain same’. Then P (A) = , and P (B) = P (stock price will either go up or remain same) = P (A U B) P (A) + P (B) = + = Now, P (stock price will go down) = 1 – P (A U B) = 1 - =

ADDITION THEOREM FOR NOT MUTUALLY EXCLUSIVE EVENTS Two or more events are known as partially overlapping if part of one event and part of another event occur together. Thus, when the events are not mutually exclusive the addition theorem has to be modified. Modified Addition Theorem states that if A and B are not mutually exclusive events, the probability of occurrence of either A or B or both is equal to the probability of that event A occurs, plus the probability that event B occurs minus the probability that events common to both A and B occur simultaneously. Symbolically, P (A or B or Both ) = P (A) + P (B) – P (AB) The following figure illustrates this point: NOT MUTUALLY EXCLUSIVE EVENTS A B AB Overlapping Events

Generalisation The theorem can be extended to three or more events. If A, B and C are not mutually exclusive events, then P (Either A or B or C) = P (A) + P (B) + P (C) – P (AB) – P (AC) – P (BC) + P (ABC) EXAMPLES ILLUSTRATING THE APPLICATION OF THE MODIFIED ADDITION THEOREM Question: A card is drawn at random from a well shuffled pack of cards. What is the probability that it is either a spade or a king ? Solution: The Probability of drawing a spade P (A) = The Probability of drawing a King P (B) = The Probability of drawing a King of Spade P (AB) = P ( A or B or Both) = P (A) + P (B) – P (AB) = + - = =

Question: A bag contains 30 balls number from 1 to 30. One ball is drawn at random. Find the probability that the number of ball is a multiple of 5 or 6. Solution: Probability of the ball being a multiple of 5 P (A) = Probability of the ball being a multiple of 6 P (B) = Since, 30 is a multiple of 5 as well as 6, therefore the events are not mutually exclusive. P (A and B) = (common multiple 30) P (A or B) = P (A) + P (B) – P (AB) = + - =

MULTIPLICATION THEOREM MULTIPLICATION THEOREM FOR INDEPENDENT EVENTS The Multiplication Theorem states that if A and B are two independent events, then the probability of the simultaneous occurrence of A and B is equal to the product of their individual probabilities. Symbolically, P (AB) = P (A) X P (B) Let be the number of cases favourable to the happening of the event A out of exhaustive and equally likely cases. P (A) = Let be the number of cases favourable to the happening of the event B out of exhaustive and equally likely cases. P (B) = Now, by the Fundamental Principle of counting , the number of cases favourable to the happening of the event AB is out of P (AB) = P (AB) = P (A).P (B) Hence, the theorem is proved.

Generalisation The theorem can be extended to three or more independent events. If A, B and C are three independent events, then P (ABC) = P (A) x P (B) x P (C) EXAMPLES ILLUSTRATING THE APPLICATION OF THE MULTIPLE THEOREM Question: A coin is tossed three times. What is the probability of getting all the three heads? Solution: Probability of head in the first toss P (A) = Probability of head on the second toss P(B) = Probability of head on the third toss P(C) = P (ABC) = P (A) x P (B) x P(C) = x x =

Question: From a pack of 52 cards, two cards are drawn at random one after the another with replacement. What is the probability that both cards are kings? Solution: The probability of drawing a King P (A) = The probability of drawing again a king after replacement P (B) = P (AB) = P (A) x P (B) = x =

Probability of happening of atleast one event in case of n independent events P (happening of atleast one of the events) = 1 – P (happening of none of the events) EXAMPLES ILLUSTRATING THE APPLICATION OF THIS THEOREM Question: A problem in statistics is given to three students. A, B and C whose chances od solving it are ½, 1/3 and ¼. What is the probability that the problems will be solved? Solution: Probability that A will solve the problem = P (A) = Probability that B will solve the problem = P (B) = Probability that C will solve the problem = P (C) = Probability that A will not solve the problem = P ( ) = 1 - = Probability that B will not solve the problem = P ( ) = 1 - = Probability that C will not solve the problem = P ( ) = 1 - =

P (that none will solve the problem) = P ( ).P ( ).P ( ) = x x = P (that problem will be solved) = 1 – P (that none will solve) = 1 - =

Question: A candidate (Mr. X) is interviewed for 3 posts. For the first post, there are 3 candidates, for the second post, there are 4 and for the third there are 2. What are the chances of Mr. X being selected ? Solution : Probability of selection for 1st post = P (A) = Probability of selection for 2nd post = P (B) = Probability of selection for 3rd post = P (C) = Probability of not getting selected for 1st post = P () = 1 - = Probability of not getting selected for 2nd post = P () = 1 - = Probability of not getting selected for 3rd post = P () = 1 - = P () = P ( ).P ( ).P ( ) = x x = Probability of selection for at least 1 post = 1 – P (not selected at all) = 1 - =

CONDITIONAL PROBABILITY Dependent events are those in which the occurrence of one event affects the probability of other events. The probability of the event B given that A has occurred is called conditional probability of B. It is denoted by P (b/A). Similarly, the conditional probability of A given that B has occurred is denoted by P (A/B). If A and B are two dependent events, then the conditional probability of B is given A is defined and given by: P (B/A) = provided P (A) > 0 Similarly, the conditional probability of A given B is defined and given by: P (A/B) = provided P (B) > 0

MULTIPLICATION THEOREM IN CASE OF DPENDENT VARIABLES When the events are not independent, i.e., they are dependent events, then the multiplication theorem has to be modified. The Modified Multiplication Theorem states that if A and B are two dependent events, then the probability of their simultaneous occurrence is equal to the probability of one event multiplied by the conditional probability of the other. P (AB) = P (A) . P (B/A) OR P (AB) = P (B) . P (A/B) Where, P (B/A) = Conditional Probability of B given A. P (A/B) = Conditional Probability of A given B.

EXAMPLES ILLUSTRATING THE APPLICATION OF THE MODIFIED MULTIPLICATION THEOREM Question: A bag contains 10 white and 5 black balls. Two balls are drawn at random one after the other without replacement. Find the probability that both balls drawn are black. Solution: The probability of drawing a black ball in the first attempt is: P (A) = = The probability of drawing the second black ball given that the first drawn is black and not replaced is: P (B/A) = = Since, the events are dependent, so the probability that both balls drawn are black is: x =

Question: Find the probability of drawing a king, a queen and a knave in that order from a pack of cards in three consecutive draws, the cards drawn not being replaced. Solution: The probability of drawing a king = P (A) = The probability of drawing a queen after a king has been drawn P (B/A) = The probability of drawing a knave after a king and a queen have been drawn P (C/AB) = P (ABC) = x x =

COMBINED USE OF ADDITION AND MULTIPLICATION THEOREM Question: A speaks truth in 80% cases, B in 90% cases. IN what percentage of cases are they likely to contradict each other in stating the same fact. Solution: Let P (A) and P (B) denote the probability that A and B speak the truth. Then, P (A) = = P ( ) = 1 – P (A) = 1 - = P (B) = = P ( ) = 1 – P (B) = 1 - = They will contradict each other only when one of them speaks the truth and the other speaks a lie. Thus, there are two possibilities: (i) A Speaks the truth and B tells a lie (ii) B Speaks the truth and A tells a lie.

Since, the events are independent, so by using the multiplication theorem, we have: Probability in the 1st case = x = Probability in the 2nd case = x = Since, these cases are mutually exclusive, so by using the addition theorem. We have the required probability = + = =26%

Question: A bag contains 5 white and 3 red balls and four balls are successively drawn out and not replaced. What is the chance that they are alternatively f different colours? Solution: 4 balls of alternative colours can be either White, Red, White, Red or Red, White, Red, White Beginning with White Ball: The probability of drawing a white ball = The probability of drawing a red ball = The probability of drawing a white ball = The probability of drawing a red ball = P (1W 1R 1W 1R) = x x x = Beginning with Red Ball: The probability of drawing a red ball = The probability of drawing a white ball = The probability of drawing a red ball = The probability of drawing a white ball =

P (1R 1 W 1R 1W) = x x x = Required Probability = + =

BERNOULLI’S THEOREM IN THE THEORY OF PROBABILITY Bernoulli’s theorem is very useful in working out various probability problems. This theorem states that if the probability of happening of an event on one trial or experiment is known, then the probability of its happening exactly, 1,2,3,…r times in n trials can be determined by using the formula: P (r) = nCrpr . qn-r r = 1,2,3,…n Where, P (r) = Probability of r successes in n trials. p = Probability of success or happening of an event in one trial. q = Probability of failure or not happening of the event in one trial. n = Total number of trials.

EXAMPLES ILLUSTRATING THE APPLICATION OF BERNOULLI’S THEOREM Question: Three coins are tossed simultaneously. What is the probability that there will be exactly two heads? Solution: P (r) = = nCrpr . qn-r Given, n = 3, r = 2, p = probability of head in throw of a coin = q = 1 - = P (2H) = 3C2 () 2 . 3-2 = x = 3 x =

BAYE’S THEOREM Baye’s Theorem is named after the British Mathematician Thomas Bayes and it was published in the year 1763. With the help of Baye’s Theorem, prior probability are revised in the light of some sample information and posterior probabilities are obtained. This theorem is also called Theorem of Inverse Probability. STATEMENT OF BAYE’S THEOREM If A1 and A2 are mutually exclusive and exhaustive events and B be an event which can occur in combination with A1 and A2, then the conditional probability for event A1 and A2 given the event B is given by: P (A1/B) = Similarly, P (A2/B) =

PROOF OF THE THEOREM A1 A2 B Since, A1and A2 are mutually exclusive events and since the event B occurs with only one of them, so that B = BA1 + BA2 or B = A1B + A2B By the addition theorem of probability, we have P (B) = P (A1B) + P(A2B) …(i)

Now, by the multiplication theorem, we have P (A1B) = P (A1) . P (B/A1) …(ii) P (A2B) = P (A2) . P (B/A2) …(iii) Substituting the values of P (A1B) and P (A2B) in (i), we get P (B) = P (A1) . P (B/A1) + P (A2) . P (B/A2) …(iv) Hence, P (B) = P (Ai) . P (B/Ai) …(v) Again by the theorem of conditional probability, we have P (A1/B) = …(vi) Substituting the values of P (A1B) and P (B) from (ii) and (iv) in equation (vi), we get P (A1/B) = Similarly, P (A2/B) = The probabilities P (A1) and P (A2) are called posterior probabilities and probabilities P (A1/B) and P (A2/B) are called posterior probabilities.

The theorem can be expressed by means of the following figure: P (B/A1) P (A1) . P (B/A1) First Branch P (A1) P (A2) Second Branch P (B/A2) P (A2) . P (B/A2) Prior Probability Conditional Probability Joint Probability Of A1 andA2 of B given A1 and A2

Now, P (A1/B) = = Similarly, P (A2/B) = =

EXAMPLES ILLUSTRATING THE APPLICATION OF BAYE’S THEOREM Question: In a bolt factory machine A, B and C manufacture respectively 25%, 35% and 40% of the total. Of their output 5, 4, 2 per cent are defective bolts. A bolt is drawn at random from the product and is fount to be defective. What is the probability that it was manufactured by machine C? Solution: Let A, B and C be the events of drawing a bolt produced by machine A, B and C respectively and let D be the event that the bolt is defective. The prior probabilities are: The conditional probabilities are: P (A) = 25% = = 0.25 P (D/A) = 5% = = 0.05 P (B) = 35% = = 0.35 P (D/B) = 4% = = 0.04 P (C) = 40% = = 0.40 P (D/C) = 2% = = 0.02

P (C/D) = = = = = 0.2318 or 23.18%

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Beginning with the ones place on the right, the place values are grouped in ... use a caret ( ^ ) to show the new position of the decimal point. ... – powerpoint ppt presentation.

  • Whole Numbers and Decimals
  • Read decimals
  • Write decimals
  • Round decimals
  • Add decimals
  • Subtract decimals
  • Multiply decimals
  • Divide decimals
  • Read whole numbers
  • Write whole numbers
  • Round whole numbers
  • Add whole numbers
  • Subtract whole numbers
  • Multiply whole numbers
  • Divide whole numbers
  • Our system of numbers, the decimal number system uses 10 symbols called digits 0,1, 2, 3, 4, 5, 6, 7, 8, and 9.
  • Place-value system a number system that determines the value of a digit by its position in a number.
  • Beginning with the ones place on the right, the place values are grouped in digits of three places.
  • For example 145,874,322
  • Each group is called a period.
  • Each period has a name and a ones place, a tens place and a hundreds place
  • In a number, the first period from the left may have fewer than three digits.
  • In many cultures, the periods are separated by commas.
  • Identify the period name of the leftmost group.
  • Read the three digit number from left to right.
  • Name the period.
  • 34,786,654 would read thirty four million seven hundred eighty six thousand six hundred fifty four.
  • Do not read or name a period that is all zeros.
  • 34,000,892 would read thirty four million eight hundred ninety two.
  • Do not name the units period (892).
  • The period name will be read at each comma.
  • Period names are read in the singular (thousand not thousands).
  • Hundreds is not a period name.
  • Do not say the word and when reading whole numbers.
  • Calculator displays ordinarily do not show commas insert when writing the number.
  • Begin recording digits from left to right.
  • Insert a comma at each period name.
  • Every period after the first period must have three digits.
  • Insert zeros as necessary.
  • Seven million, three hundred three thousand, nine hundred twenty eight.
  • 303, thousand
  • 928 (units)
  • is written 7,303,928.
  • Rounding to a specific place
  • Identify the place
  • (nearest hundred, for example)
  • Look at the number immediately to the right.
  • Is it 5 or higher? Round up.
  • Is it 4 or lower? It stays the same.
  • All digits to the right of the specified place become zeros.
  • Round to the nearest hundred
  • 2,345 12,517 234,567 12,345,078
  • And the answers are
  • 2,300 12,500 234,600 12,345,100
  • Write the numbers in a vertical column, aligning digits according to their places.
  • Beginning with the ones column, add the place digits.
  • Add, if necessary, to the tens column.
  • Repeat the operation, adding to the hundreds column, if necessary until you have reached the farthest column of digits to the left.
  • Addends numbers being added
  • Sum or total The answer or result of addition.
  • Commutative property of addition two or more numbers can be added in either order without changing the sum
  • Associative property of addition When more than two numbers are being added, the addends can be grouped by two at a time in any way.
  • Add the ones column
  • Place the 8 the bottom of the ones column
  • Carry the 2 to the tens column
  • Place the 4 in the tens column.
  • Carry the 2.
  • Finish the operation
  • Answer 64,948
  • Estimate to find a reasonable approximate answer for a calculation.
  • Use estimating as a quick tool when an exact number is not required.
  • Round whole numbers to the place desired for an estimate.
  • What was the weeks total to the nearest hundred?
  • Answer 3,200
  • Sales for last weeks concession stand
  • Tuesday 877
  • Wednesday 455
  • Thursday 614
  • The order of the numbers is important so therefore, subtraction is not commutative.
  • Grouping in subtraction is important. Subtraction is not associative.
  • (8 - 3) -1 5 1 4 but
  • 8 - (3 -1) 8 - 2 6
  • Minuend the beginning amount or number that a second number is being subtracted from.
  • Subtrahend the number being subtracted.
  • Difference the answer or result of subtracting
  • Borrow regroup digits in the minuend by borrowing 1 from the digit to the left of the specified place and adding 10 to the specified place.
  • Subtract 128 from 1,345 by rounding each number to the nearest hundred to estimate the difference.
  • 128 would become 100.
  • 1,345 would become 1,300.
  • The estimated difference would be 1,200.
  • Borrow 1 from the tens column.
  • Subtract 8 from 13.
  • Borrow 1 from the hundreds column
  • Subtract 9 from 18
  • Borrow 1 from the thousands column
  • Subtract 5 from 11
  • Numbers can be multiplied in any order without affecting the result.
  • 8 x 3 x 4 4 x 3 x 8
  • Multiplicand the number being multiplied
  • Multiplier the number multiplied by
  • Factor each number involved in multiplication
  • Product the answer or result of multiplication
  • Partial product the product of one digit of the multiplier and the entire multiplicand
  • Identify each
  • 123 x 466 ?
  • Answer 57,318
  • Answer 8,040
  • Answer 9,396
  • Division is used to find the number of equal parts a whole quantity can be separated into.
  • A 40 tip is shared equally among 5 servers. How much does each server receive?
  • 40 5 servers 8 each
  • Dividend the number being divided or the total quantity
  • Divisor The number to divide by
  • Quotient The answer or result of the operation
  • Whole-number part of the quotient the quotient without regard to its remainder
  • (continued on the next slide)
  • Remainder of quotient a number that is smaller than the divisor that remains after division is complete.
  • Partial dividend the part of the dividend that is being considered at a given step of the process.
  • Partial quotient the quotient of the partial dividend and the divisor.
  • There will be a remainder if an amount is too small to be further divided by the divisor.
  • For example 152 3 50 R 2
  • That amount may be expressed as a remainder (R 2), a fraction, or a decimal.
  • 1. Beginning with its leftmost digit, identify the first group of digits of the dividend that is larger than or equal to the divisor.
  • Is it 1? No.
  • Is it 12? Yes.
  • 5 goes into 12 two times. Place the 2 above the 2 in the dividend.
  • (Go on to next slide)
  • 2. Multiply 2 by the divisor. Place 10 under the 12 and subtract. The result is 2.
  • 3. Bring down the following digit which is 3 and divide 5 into 23. The result is 4.
  • 4. Place the 4 directly above the 3 in the dividend. Multiply 4 by the divisor.
  • 5. Place 20 under the 23 and subtract. The result is 3.
  • 6. Bring down the last digit which is 5 and divide 5 into 35. The result is 7. Place 7 directly above the 5.
  • 7. You have finished and the answer is 247.
  • 6,750 cases of detergent will be distributed evenly to 25 local stores. How many will each receive?
  • 420 bottles of fabric softener in the warehouse are packed a dozen to case. How many cases are there in the warehouse?
  • Read and write decimals
  • 1.2345 rounded to the nearest tenth is 1.2
  • Our money system, based on the dollar, uses the decimal system.
  • Moving one place from right to left increases the value ten times.
  • Moving one place from left to right, causes the value of the digit to become ten times smaller.
  • It is one part of a 10-part whole.
  • 0.1 is read one tenth
  • If this chart represented a dollar, the white segment would be equal to 0.10.
  • Separates the whole number part from the decimal part, as the number extends from left to right.
  • 34.7 is read thirty four and seven tenths
  • or 34 point 7.
  • The first place to the right of the decimal point is tenths. (0.1)
  • Second place is hundredths. (0.01)
  • Third place is thousandths. (0.001)
  • Fourth place is ten-thousandths. (0.0001)
  • 3.12 Three and twelve hundredths
  • 9.067 Nine and sixty-seven thousandths.
  • 4.5 Four and five tenths.
  • Read the whole number part first, saying and to indicate the beginning of the decimal part of the number.
  • When reading numbers that represent money amounts, read whole numbers as dollars.
  • Decimal amounts are read as cents.
  • 35.98 is read thirtyfive dollars and 98 cents.
  • 1. Find the digit in the specified place.
  • 2. Look at the next digit to the right.
  • If this digit is less than 5, eliminate it and all digits to its right.
  • If the digit is 5 or more, add 1 to the digit in the specified place, and eliminate all digits to its right.
  • Round to the nearest tenth
  • Add and subtract decimals
  • 3.234 6.8 ?
  • Attach extra zeros to the right end of each number so each number has the same quantity of digits.
  • Add or subtract as though the numbers are whole numbers.
  • Place the decimal point in the sum or difference to align with the decimal point in the respective operation.
  • 7.098 2.6 0.8 13.999
  • .976 - .04217
  • Multiply the decimal numbers as though they are whole numbers.
  • Count the digits in the decimal parts of both decimal numbers.
  • Place the decimal point in the product so that there are as many digits in its decimal part as there are digits you counted in the previous step.
  • If necessary, attach zeros to the left end of the product to place the decimal point accurately.
  • 3.45 x 4.082
  • How many places are there to the right of the decimal point?
  • Five so, the answer will have five places to the right of the decimal.
  • The answer is 14.08290
  • The last zero can be dropped and the answer would be 14.0829.
  • 3.07 x 8.008
  • .01 x 1.001
  • Divide a decimal by a whole number
  • Place a decimal point for the quotient directly above the decimal point in the dividend.
  • Divide as though the decimal points are whole numbers.
  • 3.4 divided by 3 ?
  • 2.06 (repeating)
  • Jill wants to buy a bottle of detergent. If a 100-ounce bottle costs 6.49 and a 50- ounce bottle costs 3.99, which would be the better buy on cost per ounce basis? What are those amounts?
  • Answer The 50 - ounce bottle has a cost of .0798 per ounce while the 100-ounce bottle has a cost of .0649 per ounce. The bigger bottle is a better buy.
  • Change the divisor to a whole number by moving the decimal point to the right, counting the places as you go.
  • Use a caret ( ) to show the new position of the decimal point.
  • Move the decimal point in the dividend to the right as many places as you moved the divisor.
  • Place the decimal point for the quotient directly above the new decimal point for the dividend.
  • Divide as you would divide a whole number.
  • Bill Sullivan has an hourly rate of 14.32 and his gross weekly pay was 572.80. How many hours did he work?
  • Jan Stevens has an hourly rate of 7.75 per hour and her gross weekly pay was 193.75. How many hours did she work last week?
  • All resources can be found on
  • The resource drive (Resource\BM\Math)
  • Online labs can be found at
  • http//linux.herzing.ca/kim
  • (as of July 30, this link only works from school network support is to have it up online by the end of this week)
  • Online Lab 1
  • Homework due at the end of the week
  • Chapter 3 Exercises Set A

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