basic math conversions math for water technology mth 082 fall 08 chapters 1, 2, 4, and 7 lecture 1...

Basic Math ConversionsBasic Math Conversions

Math for Water TechnologyMTH 082

Fall 08Chapters 1, 2, 4, and 7

Lecture 1

Math for Water TechnologyMTH 082

Fall 08Chapters 1, 2, 4, and 7

Lecture 1

Why are you here?

Lai

d of

f/ret

ra...

Mili

tary

Just

ent

erin

g ...

Dec

ided

to g

o ...

25% 25%25%25%

1. Laid off/retraining

2. Military

3. Just entering college

4. Decided to go back to school to complete college/degree

1. Laid off/retraining

2. Military

3. Just entering college

4. Decided to go back to school to complete college/degree

What is your education level?

Hig

h Sch

ool o

r GED

Ass

ociat

es D

egre

e

BS o

r BA/C

olle

ge D...

Gra

duate

Level

33%

0%

67%

0%

1. High School or GED

2. Associates Degree

3. BS or BA/College Degree

4. Graduate Level

1. High School or GED

2. Associates Degree

3. BS or BA/College Degree

4. Graduate Level

How much math have you taken?

Hig

h Sch

ool M

a...

Colle

ge M

ath (.

..

Colle

ge A

lgeb

r...

Inte

rmed

iate

A...

Cal

culu

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20% 20% 20%20%20%

1. High School Math2. College Math (MTH10-

MTH50)3. College Algebra I (MTH

060)4. Intermediate Algebra II

(MTH 065)5. Calculus or higher (MTH

251+)

1. High School Math2. College Math (MTH10-

MTH50)3. College Algebra I (MTH

060)4. Intermediate Algebra II

(MTH 065)5. Calculus or higher (MTH

251+)

Have you taken the Math Placement Exam for incoming students in the testing center?

Have you taken the Math Placement Exam for incoming students in the testing center?

1. Yes

2. No

1. Yes

2. No

I have had MTH 065 (Intermediate College Algebra II) and thus

completed the prerequisite for this course?

I have had MTH 065 (Intermediate College Algebra II) and thus

completed the prerequisite for this course?

Tru

e

Fal

se

0%0%

1. True

2. False

1. True

2. False

Although I have not completed the prerequisite for this course, I am

willing to work _________% harder than my teammates?

Although I have not completed the prerequisite for this course, I am

willing to work _________% harder than my teammates?

100% 50

%25

% 0%

25% 25%25%25%

1. 100%

2. 50%

3. 25%

4. 0%

1. 100%

2. 50%

3. 25%

4. 0%

ObjectivesObjectives

Review and demonstrate proficiency in math problems that include:

1. manipulation of fractions and decimals2. percent and unit conversions


1. manipulation of fractions and decimals2. percent and unit conversions

RULES TO SOLVING MATH PROBLEMS

RULES TO SOLVING MATH PROBLEMS

1.READ THE PROBLEM FIRST (AND PUT IT INTO YOUR OWN WORDS)

2.LAY OUT THE PROBLEM=DRAW A DIAGRAM3.DETERMINE WHAT YOU HAVE AND WHAT YOU

NEED (YOU MAY HAVE EXTRA)4.PERFORM CONVERSIONS5.ARTICULATE THE REASON FOR USING AN

EQUATION 6.DO DIMENSIONAL ANALYSIS FIRST7.APPLY THE EQUATION---DO NOT PLUG AND

CHUG8.SOLVE THE PROBLEM9.CHECK YOUR WORK

1.READ THE PROBLEM FIRST (AND PUT IT INTO YOUR OWN WORDS)

2.LAY OUT THE PROBLEM=DRAW A DIAGRAM3.DETERMINE WHAT YOU HAVE AND WHAT YOU

NEED (YOU MAY HAVE EXTRA)4.PERFORM CONVERSIONS5.ARTICULATE THE REASON FOR USING AN

EQUATION 6.DO DIMENSIONAL ANALYSIS FIRST7.APPLY THE EQUATION---DO NOT PLUG AND

CHUG8.SOLVE THE PROBLEM9.CHECK YOUR WORK

Decimal PlacesDecimal Places

http://www.gomath.com/htdocs/lesson/decimal_lesson1.htmhttp://www.gomath.com/htdocs/lesson/decimal_lesson1.htm

Greater than 1Greater than 1 Less than 1Less than 1

10

1100

1

1000

1110100


Chapter 1Power and Scientific Notation

Chapter 1Power and Scientific Notation

Rules of Power and Scientific Notation

Rules of Power and Scientific Notation

Rule 1 = when a number is TAKEN out of scientific notation

•Positive exponent value move decimal point to the right

•Negative exponent value move decimal point to the left!

Rule 1 = when a number is TAKEN out of scientific notation

•Positive exponent value move decimal point to the right

•Negative exponent value move decimal point to the left!

Rule 2 = when a number is PUT into scientific notation

•Decimal point to the left indicates a positive exponent

•Decimal point move to the right indicates negative exponent values!

Rule 2 = when a number is PUT into scientific notation

•Decimal point to the left indicates a positive exponent

•Decimal point move to the right indicates negative exponent values!

Rules of Scientific NotationRules of Scientific Notation

Rule 4 = when you divide the numbers in scientific notation, divide the numbers but subtract the exponents.


Rule 3 = when you multiply the numbers in scientific notation, multiply the numbers but add the exponents.Rule 3 = when you multiply the numbers in scientific notation, multiply the numbers but add the exponents.

POWERPOWERNumeric20=121 =222= 2 X 2 = 423 = 2 X 2 X 2= 8

Numeric20=121 =222= 2 X 2 = 423 = 2 X 2 X 2= 8

Englishft2= ft X ftm3= meter X meter X meter

Englishft2= ft X ftm3= meter X meter X meter

33=?

9 27 1 0

7%0%0%

93%

3 X 3 X 3 = 273 X 3 X 3 = 271. 9

2. 27

3. 1

4. 0

1. 9

2. 27

3. 1

4. 0

POWERPOWER

Numeric Expanded and Exponential FormNumeric Expanded and Exponential Form

)3

2)(

3

2()

3

2( 2

)3

1(3

22

))8)(8(

1()

8

1(8

22

))5)(5)(5(

)4)(4(()

5

4(

3

2

English Expanded and Exponential FormEnglish Expanded and Exponential Form

)) )( )( (

) )( (()(

3

2

m

km22

2

)()))((

))((()(

ft

in

ftft

inin

ft

in

Your TurnYour Turn

Scientific NotationScientific Notation

Scientific Notation = number multiplied by power of tenScientific Notation = number multiplied by power of ten

54)104.5(104.5 1

200,1))10)(10)(10(2.1(102.1 3

0362.0)10

62.3())

10

1)(

10

1(62.3(1062.3

22

Your Turn (Write It All out!!!)Your Turn (Write It All out!!!)

10800.4 3 10350 3


Scientific Notation = Taken out!Scientific Notation = Taken out!

Rule 1 = when a number is taken out of scientific notation a positive exponent value indicates a move of the decimal point to the right and a negative exponent value indicates a decimal point move to the left!

Rule 1 = when a number is taken out of scientific notation a positive exponent value indicates a move of the decimal point to the right and a negative exponent value indicates a decimal point move to the left!

10039.0 2

Your TurnYour Turn

105200.0 3

890,977890.910789.9 4 17890.01789010890,17 5

Positive four places to rightPositive four places to right Negative five places to leftNegative five places to left

25 X 10-4 = ?

25.0

025

2500

00

None

of the

ab.

..

0%

13%13%

73%25 X 10-4=

Move decimal four spots to left.0025

25 X 10-4=Move decimal four spots to left

.0025

1. 25

2. .0025

3. 250,000

4. None of the above

1. 25

2. .0025

3. 250,000

4. None of the above


Scientific Notation = Put Into!Scientific Notation = Put Into!

Rule 2 = when a number is PUT into scientific notation a decimal point to the left indicates a positive exponent and a decimal point move to the right indicates a negative exponent values!

Rule 2 = when a number is PUT into scientific notation a decimal point to the left indicates a positive exponent and a decimal point move to the right indicates a negative exponent values!

110 5.775

420

Your TurnYour Turn

0041.0

310 2.11200

-210 62.30362.

0.0058 = ?

5.8

X 1

0-3

58

X 10-

4

0.5

8 X 1

0-2

All

of the

abo...

47%

40%

0%

13%

1. 5.8 X 10-3

2. 58 X 10-4

3. 0.58 X 10-2

4. All of the above

1. 5.8 X 10-3

2. 58 X 10-4

3. 0.58 X 10-2

4. All of the above

Multiplying in Scientific Notation

Multiplying in Scientific Notation

Rule 3 = when you multiply the numbers in scientific notation, multiply the numbers but add the exponents.Rule 3 = when you multiply the numbers in scientific notation, multiply the numbers but add the exponents.

7)34(34 1012)10()43()104()10 3(

7)34(3-4 1030)10()65()106()10 5(

)106()102()103()0006.0()2.0()003(. 413

8)413( 1036)10()36()62 3( Your TurnYour Turn

)002.0()2.0()02(.

Dividing in Scientific NotationDividing in Scientific Notation



1)34(3

4

102)10()2

4(

)102(

)10 4(

3)1(21

2

104)102

8()

102

108(

)2.0(

)800(

Your TurnYour Turn

)3.0(

)006.0(


Chapter 2Dimensional Analysis

Chapter 2Dimensional Analysis

MATT’S RULE

ALWAYS USE DIMENSIONAL ANALYSIS BEFORE YOU PLUG AND

CHUG!

MATT’S RULE

ALWAYS USE DIMENSIONAL ANALYSIS BEFORE YOU PLUG AND

CHUG!

Dimensional AnalysisDimensional Analysis

33

ft

galor

(ft) (ft) ft)(

galftgal

3

3

ftgalmingal

t)(ft)(ft)(fgalmingal

gal/ft

gal/min

3

232

mm

kmor

(mm) (mm) (mm)

(km)(km)mmkm

m

(sec)(sec)

(mm) (mm) (mm)

(km)(km)

(sec)(sec)

(mm) (mm) (mm)(km)(km)

sec/

mmkm2

32

mm

Dividing is the same as multiplying by the INVERSEDividing is the same as multiplying by the INVERSEYour TurnYour Turn

sec/

sec/3

ft

ft

Dimensional AnalysisMultiplication and Division


gal1

gal

1

t)(ft)(ft)(f

(ft) (ft) (ft)

gal))(ftft(gal 33

Need answer in gallonsNeed answer in gallons

Need answer in square feetNeed answer in square feet

WRONG!(sec)

ft

(sec)

(ft)

(sec)

t)(ft)(ft)(fft/sec) (3.5)secft (80

2

43

23 ft(ft)

(sec)

(sec)

t)(ft)(ft)(f

sec

(sec)t)(ft)(ft)(f

ft/sec) )/(3.5secft (80 ft



Need answer in cubic meters per secondNeed answer in cubic meters per second

YEP!(sec)

m

(1)

)(m

(sec)

(m))m (5)secm (100

322

UGLY!(sec)m

m

m

1

(sec)

(m)

1(m)(m)(sec)(m)

)m )/(5secm (10022

2

WORD PROBLEMWORD PROBLEMThe flow rate in a water line is 2.3 ft3/sec. What is the flow rate as gallons per minute? The flow rate in a water line is 2.3 ft3/sec. What is the flow rate as gallons per minute?

Step 1: Use your own words. Got a pipe with a known flow rate, need to convert that value from one unit to another. This is a simple conversion problemStep 1: Use your own words. Got a pipe with a known flow rate, need to convert that value from one unit to another. This is a simple conversion problem

Step 2: Draw a diagramStep 2: Draw a diagram2.3 ft3/sec2.3 ft3/sec gal/min?gal/min?

Step 3: Conversions? GIVEN: 2.3 ft3/sec NEED: gal/minCONVERSIONS:

7.48 ft3/gal 60 sec/min

Step 3: Conversions? GIVEN: 2.3 ft3/sec NEED: gal/minCONVERSIONS:

7.48 ft3/gal 60 sec/min

WORD PROBLEMWORD PROBLEMThe flow rate in a water line is 2.3 ft3/sec. What is the flow rate as gallons per minute? The flow rate in a water line is 2.3 ft3/sec. What is the flow rate as gallons per minute?

Step 4: Convert ft3/sec to gal min. Dimensional Analysis First. To multiply or divide?

Step 4: Convert ft3/sec to gal min. Dimensional Analysis First. To multiply or divide?

2.3 ft3/sec2.3 ft3/sec gal/min?gal/min?

)(min

gal

min

sec

t)(ft)(ft)(f

gal

sec

t)(ft)(ft)(fsec/min 60 )

ft

gal (7.48 )secft (2.3

33 Yes

Step 5: Solve the problem.Step 5: Solve the problem.

min

gal1032sec/min 60 )

ft

gal (7.48 )secft (2.3

33

WORD PROBLEMWORD PROBLEMA channel is 3 ft wide with water flowing to a depth of 2 ft. The velocity in the channel is found to be 1.8 ft/sec. What is the flow rate in the channel in cubic feet per second?

A channel is 3 ft wide with water flowing to a depth of 2 ft. The velocity in the channel is found to be 1.8 ft/sec. What is the flow rate in the channel in cubic feet per second?

Step 1: Use your own words. Got a channel with known dimensions and a flow rate, need to convert that value from one unit to another. This is a simple conversion problem

Step 1: Use your own words. Got a channel with known dimensions and a flow rate, need to convert that value from one unit to another. This is a simple conversion problem

Step 2: Draw a diagramStep 2: Draw a diagram

Step 3: Conversions? GIVEN: 1.8 ft/sec , 3ft, 2 ft NEED: ft3/secCONVERSIONS:

None necessary

Step 3: Conversions? GIVEN: 1.8 ft/sec , 3ft, 2 ft NEED: ft3/secCONVERSIONS:

None necessary

1.8 ft/sec1.8 ft/sec3 ft

2 ftft3/sec?ft3/sec?



Step 3: Conversions? GIVEN: f=1.8 ft/sec, w=3ft, d=2 ft NEED: ft3/secCONVERSIONS: None necessary

Step 4 Equation : flow in channel (FC) = f X w X d

Step 5: Solve Dimensional Analysis First!

Step 3: Conversions? GIVEN: f=1.8 ft/sec, w=3ft, d=2 ft NEED: ft3/secCONVERSIONS: None necessary

Step 4 Equation : flow in channel (FC) = f X w X d

Step 5: Solve Dimensional Analysis First!



YES!sec

ft

1

ft

1

ft

sec

ftft) (2 ft) (3 )secft (1.8

3



Step 6: Solve Problem

Equation : flow in channel (FC) = f X w X d where f = flow w = width of channel d = depth of channel

Step 6: Solve Problem

Equation : flow in channel (FC) = f X w X d where f = flow w = width of channel d = depth of channel



sec

ft8.10ft) (2 ft) (3 )secft (1.8

3


Chapter 3Rounding and Estimating

Chapter 3Rounding and Estimating

Decimal PlacesDecimal Places

http://www.gomath.com/htdocs/lesson/decimal_lesson1.htmhttp://www.gomath.com/htdocs/lesson/decimal_lesson1.htm

Greater than 1Greater than 1 Less than 1Less than 1

10

1100

1

1000

1110100

Basic Rulesof RoundingBasic Rulesof Rounding

A ≈ indicates a number or answer has been rounded

Rule 1: When rounding to any desired place if the digit directly to the right of that place is less then 5 replace all digits to the right with zeros.

Rule 2: When rounding to any desired place if the digit directly to the right of that place is greater then 5, increase the digit in the rounding place by 1 and replace all digits to the right of the increase with zeros.

Rule 3: When rounding decimal numbers to the right of the decimal point, drop the rounded digits

A ≈ indicates a number or answer has been rounded



Rule 3: When rounding decimal numbers to the right of the decimal point, drop the rounded digits

RoundingRounding

Round 342,427 to the nearest thousands342,427 ≈ 342,000

Round 342,427 to the nearest thousands342,427 ≈ 342,000

Rounding place (less then 5 everything to right =0)Rounding place (less then 5 everything to right =0)

hundreds placehundreds place



Round 1,342,427 to the nearest hundred thousands placeRound 1,342,427 to the nearest hundred thousands place

Your TurnYour Turn

1,342,427 ≈

Round 37,926 to the nearest tens 37,926 ≈ 37,930

Round 37,926 to the nearest tens 37,926 ≈ 37,930

Rounding place (greater then 5 increase value by 1)Rounding place (greater then 5 increase value by 1)

tens placetens place

RoundingRoundingRule 2: When rounding to any desired place if the digit directly to the right of that place is greater then 5, increase the digit in the rounding place by 1 and replace all digits to the right of the increase with zeros.


Round 248,722 to the nearest thousands placeRound 248,722 to the nearest thousands place

Your TurnYour Turn

248,722 ≈

Round 5.654 to the nearest tenth 5.654 ≈ 5.7

Round 5.654 to the nearest tenth 5.654 ≈ 5.7

Rounding place (greater then 5 increase value by 1)Rounding place (greater then 5 increase value by 1)

tenths placetenths place

RoundingRoundingRule 3: When rounding decimal numbers to the right of the decimal point, drop the rounded digits.Rule 3: When rounding decimal numbers to the right of the decimal point, drop the rounded digits.

Round 483.16 to the nearest unitRound 483.16 to the nearest unit

Your TurnYour Turn

483.16 ≈

Round 549,012 to the nearest ten thousands

5500

00

5490

12

5490

00

Not e

nough in

f...

94%

0%6%

0%

1. 550,000

2. 549,012

3. 549,000

4. Not enough info given

1. 550,000

2. 549,012

3. 549,000

4. Not enough info given

Estimate the value of 20 X 30 = (2 X 3) = 6 with two zeros at the end =600Estimate the value of 20 X 30 = (2 X 3) = 6 with two zeros at the end =600

EstimatingEstimatingFactoid: Estimating indicates the approximate size of a calculated answer.Factoid: Estimating indicates the approximate size of a calculated answer.

Estimate the value of 40 X 600 = (4 X 6) = 24 with three zeros at the end =24,000Estimate the value of 40 X 600 = (4 X 6) = 24 with three zeros at the end =24,000

Estimate the value of 40 X 20 X 500 = (4 X 2 X 5) = 40 with four zeros at the end =400,000Estimate the value of 40 X 20 X 500 = (4 X 2 X 5) = 40 with four zeros at the end =400,000

Estimate the value of 9 X 700 X 60 X 70 = 2800 with four zeros at the end =28,000,000Estimate the value of 9 X 700 X 60 X 70 = 2800 with four zeros at the end =28,000,000

63 ≈ 60 X 6 360 ≈ 400 X 7

63 ≈ 60 X 6 360 ≈ 400 X 7

9 X 79 X 7

EstimatingEstimatingFactoid: Estimating indicates the approximate size of a calculated answer.Factoid: Estimating indicates the approximate size of a calculated answer.

Estimate the value of 40,000/200 = cancel zeros 400/2=200Estimate the value of 40,000/200 = cancel zeros 400/2=200

Estimate the value of 700/6,000 = cancel zeros 7/60=Estimate the value of 700/6,000 = cancel zeros 7/60= 1.01.00.760

Estimate the value of (20)(400)/(50)(80) = cancel zeros =(20)(4)/(5)(8)= 80/40= cancel zeros =8/4=2Estimate the value of (20)(400)/(50)(80) = cancel zeros =(20)(4)/(5)(8)= 80/40= cancel zeros =8/4=2


Chapter 4Solving for the Unknown

“Basic Algebra”

Chapter 4Solving for the Unknown

“Basic Algebra”

Solving For The UnknownSolving For The Unknown

12X2

24X

)24((X) (2)

Rule 1 = ISOLATE THE X TO THE NUMERATOR AND/OR ONE SIDE OF THE PROBLEM!!

WHATEVER YOU DO TO ONE SIDE DUE TO THE OTHER!

Rule 1 = ISOLATE THE X TO THE NUMERATOR AND/OR ONE SIDE OF THE PROBLEM!!

WHATEVER YOU DO TO ONE SIDE DUE TO THE OTHER!

A

VAQ

V

Q

A THE ISOLATE 1 RULE

A FOR SOLVE

208642

8X

64220X

XFor Solve 20X642

WORKCHECK

2016642

16X

64220X

XFor Solve 20X642

WORKCHECK

Solving For The UnknownSolving For The UnknownRule 2 = In multiplication or division equations with unknown in numerator CROSS MULTIPLY AND THEN ISOLATE/SOLVE THE X

Go from the top of one side to the bottom of the other

Rule 2 = In multiplication or division equations with unknown in numerator CROSS MULTIPLY AND THEN ISOLATE/SOLVE THE X

Go from the top of one side to the bottom of the other

2X

2412X

SIDE ONE TO X ISOLATE :1 RULE

8)(3)((6)(x)(2)

MULTIPLY CROSS :2 ULE

X FOR SOLVE )6

3()

8

)2)(((

R

x

Solving For The UnknownSolving For The Unknown

1

)32

1(322)3

32

1(

23(32)

(8)(4)(X) (16)(2)

SIDE ONE TO X ISOLATE :1 RULE

1

(8)(X))4(

(8)(X)

(16)(2)

1

(8)(X)

other) the todue side one todoyou (Whatever NUMERATOR TO X ISOLATE :1 ULE

))(8(

X FOR SOLVE )4()2)(16(

X

X

XFORSOLVE

X

R

X

Chlorine DosageChlorine DosageChlorine Dosage = Chlorine Demand +Chlorine Residual

The residual in a distribution system is measured to be 0.2 mg/L using a HACH DPD Colorimeter. If the original dose was 7.0 mg/L what is the chlorine demand for the system?

Chlorine Dosage = Chlorine Demand +Chlorine Residual

The residual in a distribution system is measured to be 0.2 mg/L using a HACH DPD Colorimeter. If the original dose was 7.0 mg/L what is the chlorine demand for the system?

DemandChlorinelmg

Xlmglmg

LmgXlmg

Dose

/8.6

/2.0/7

/2.0/7

Residual Demand

A well system was dosed with a slug of 50 mg/L chlorine for 24 hours. The residual in a distribution system is

measured to be 0.5 mg/L using a HACH DPD Colorimeter. How much chlorine

was gobbled up by organics and inorganics (i.e., chlorine demand) in the

water?

A well system was dosed with a slug of 50 mg/L chlorine for 24 hours. The residual in a distribution system is

measured to be 0.5 mg/L using a HACH DPD Colorimeter. How much chlorine

was gobbled up by organics and inorganics (i.e., chlorine demand) in the

water?

55.

5 m

g/L

49.

5 m

g/l

50.

5 m

g/l

0%8%

92%

1. 55.5 mg/L

2. 49.5 mg/l

3. 50.5 mg/l

1. 55.5 mg/L

2. 49.5 mg/l

3. 50.5 mg/l

(X) – 12 = 6 Solve for X? 3

X=2

7

X=3

0

X=5

4

X =

21

10%0%

90%

0%

1. X=27

2. X=30

3. X=54

4. X =21

1. X=27

2. X=30

3. X=54

4. X =21

(X) - 12= 6 3

(X) = 6+12 3

(X) = 18 3

(X) = 18 * 3

(X)= 54----------------(54) - 12= 6 3 18-12=6

6=6

(X) - 12= 6 3

(X) = 6+12 3

(X) = 18 3

(X) = 18 * 3

(X)= 54----------------(54) - 12= 6 3 18-12=6

6=6

FORMULA:

SOLVED:

FORMULA:

SOLVED:

20 ft2 = (15 ft X H) Solve for H? 2

27.

5 ft

2.6

7 ft

1.4

7 ft

0% 0%

100%A= (B X H) 2 2A=(B)(H)2A= H B

2(20ft2)=(15 ft)(H)40 ft2 =(15 ft)(H)40ft2 = (H)15 ft

2.67 ft =H

A= (B X H) 2 2A=(B)(H)2A= H B

2(20ft2)=(15 ft)(H)40 ft2 =(15 ft)(H)40ft2 = (H)15 ft

2.67 ft =H

1. 27.5 ft

2. 2.67 ft

3. 1.47 ft

1. 27.5 ft

2. 2.67 ft

3. 1.47 ft


Chapter 7PercentsChapter 7Percents

Basic Rulesof PercentsBasic Rulesof Percents

100WHOLE

PARTPERCENT

FACTOID. The term efficacy refers to a percentFACTOID. The term efficacy refers to a percent

Rule 1. In calculations greater than 100 percent, the numerator of the percent equation must always be larger than the denominator.


Percents Fractions DecimalsPercents Fractions Decimals

2000

1

10000

5

100

1

100

5

100100

5

or 0005.0100

05.%05.

250

1

1000

4

100

1

10

4

100104

or 004.0100

4.0%4.0

02.0100

2%2

%9595.0100

95or 95.0

100

95%95

%2020.0100

20or 20.0

100

20%20

PercentsPercents

08.0100

8%8

:StateYork Newfor Tax Sales

005.0100

05.%05.

95.0100

95%95

%2020.0100

20or 20.0

100

20%20

Percent Word ProblemsPercent Word ProblemsA certain piece of equipment is having mechanical difficulties. If the equipment fails 6 times out of 25 tests, what percent failure does this represent?

A certain piece of equipment is having mechanical difficulties. If the equipment fails 6 times out of 25 tests, what percent failure does this represent?

100WHOLE

PARTPERCENT

failure 24% 100 0.24 10025

6PERCENT

Percent Word ProblemsPercent Word ProblemsThe raw water entering a treatment plant has a turbidity of 10 ntu. If the turbidity of the finished water is 0.5 ntu, what is the turbidity removal efficacy of the treatment plant.

The raw water entering a treatment plant has a turbidity of 10 ntu. If the turbidity of the finished water is 0.5 ntu, what is the turbidity removal efficacy of the treatment plant.

100WHOLE

PARTPERCENT

efficacy removal turbidity95% 10010

9.5PERCENT

Percent is unknown and 10 ntu = whole. However,0.5 ntu is not the part removed. It is the turbidity still in the water. Thus, 10 ntu-0.5 ntu= 9.5 ntu

Percent is unknown and 10 ntu = whole. However,0.5 ntu is not the part removed. It is the turbidity still in the water. Thus, 10 ntu-0.5 ntu= 9.5 ntu

Percent Word ProblemsPercent Word Problems

A treatment plant was designed to treat 60 Mgd. One day it treated 66 Mgd. What % of the design capacity does this represent.

A treatment plant was designed to treat 60 Mgd. One day it treated 66 Mgd. What % of the design capacity does this represent.

100WHOLE

PARTPERCENT

110%1001.1 10060

66PERCENT



Percent Word ProblemsPercent Word Problems16 is 80% of what?16 is 80% of what?

200.8

16W

160.8W

)1

W(

W

16(W) 0.8

160.8

8.0 100

80%80

W

P%

W

Find 90% of 5?Find 90% of 5?

P4.5

)1

5(

5

P(5) 0.9

5

P0.9

9.0 100

90%90

W

P%

High test hypochlorite or HTH has 32.5 lbs of active chlorine in a 50 lb container. What is the % active in

the container?

High test hypochlorite or HTH has 32.5 lbs of active chlorine in a 50 lb container. What is the % active in

the container?

25% 10 50

65%

0%

100%

0%0%

1. 25%

2. 10

3. 50

4. 65%

1. 25%

2. 10

3. 50

4. 65%

P/W * 100 = %

32.5 lbs X 100 = %50 lbs

0.65 * 100 = %

65%

P/W * 100 = %

32.5 lbs X 100 = %50 lbs

0.65 * 100 = %

65%


Chapter 5Ratios and Proportions

Chapter 5Ratios and Proportions

Rules of Ratios and Proportions

Rules of Ratios and Proportions

Rule 1 = If the unknown is expected to be smaller than the known value, put an x in the numerator of the first fraction, and put the known value of the same unit in the denominator.


Rule 2 = If the unknown is expected to be larger than the known value, put an x in the denominator of the first fraction, and put the known value of the same unit in the numerator.


Rule 3 = Make the two remaining values of the problem into the second fraction. (smaller in numerator, larger in denominator)

Rule 3 = Make the two remaining values of the problem into the second fraction. (smaller in numerator, larger in denominator)

Ratios and ProportionsRatios and Proportions

uelarger val

luesmaller va

uelarger val

luesmaller va

ey...........mon ...........lbs



Problem = If 3 men can do a certain job in 10 hours, how long would it take 5 men to do the same job?

What is the unknown? Time and it will be smaller…so

Problem = If 3 men can do a certain job in 10 hours, how long would it take 5 men to do the same job?


hrsx

x

hrsx

65

30

3055men

3men

hr 10

xhr


uelarger val

luesmaller va

uelarger val

luesmaller va

ey...........mon ...........lbs



Problem = If 5 lb of chemical are mixed with 2,000 gallons of water to obtain a desired solution, how many pounds of chemical would be mixed with 10,000 gallons of water to obtain a solution of the same concentration?

What is the unknown? lbs…so

Problem = If 5 lb of chemical are mixed with 2,000 gallons of water to obtain a desired solution, how many pounds of chemical would be mixed with 10,000 gallons of water to obtain a solution of the same concentration?

What is the unknown? lbs…so

lbgal

gallbx

gal

allbx

allbx

252

)10)(5(

000,2

)g 000,0(1 )5(

)g 000,0(1 )5(000,2

gal 10,000

gal 2,000

x

lb 5


yes! so 4 (4)(1) and 4)2)(2(

?4

2

2

1 Are

alproportionand

yes! so 294 (7)(42) and 294)98)(3(

?98

42

7

3 Are

alproportionand

x

x

x

xand

46.1813

(6)(40)

))(13()40)(6(

for x. solve 4013

6


uelarger val

luesmaller va

uelarger val

luesmaller va

ey...........mon ...........lbs



Problem = If three men can do a certain job in 10 hours, how long would it take five men to do the same job?


Problem = If three men can do a certain job in 10 hours, how long would it take five men to do the same job?


hrsx

x

hrsx

65

30

3055

3

hr 10

x

If a pump will fill a tank in 13 hours at 6 gpm, how long will it take a 15

gpm pump to fill the same tank?

If a pump will fill a tank in 13 hours at 6 gpm, how long will it take a 15

gpm pump to fill the same tank?

1. 5.2 hrs

2. 2.16 hrs

3. 2.5 hrs

4. 32.5 hrs

1. 5.2 hrs

2. 2.16 hrs

3. 2.5 hrs

4. 32.5 hrs

5.2

hrs

2.1

6 hrs

2.5

hrs

32.

5 hrs

75%

25%

0%0%

uelarger val

luesmaller va

uelarger val

luesmaller va

...........gpm ...........hrs

X hrs = 6 gpm13 Hrs 15 gpm

(15 gpm)(X HRS) = (6 gpm)(13 hrs)

( X hrs) = (6 gpm)(13 hrs) (15 gpm)Hrs = 5.2

X hrs = 6 gpm13 Hrs 15 gpm

(15 gpm)(X HRS) = (6 gpm)(13 hrs)

( X hrs) = (6 gpm)(13 hrs) (15 gpm)Hrs = 5.2

Mixed numbersMixed numbersMixed Numbers as Fractions uses Circles to demonstrate how a fraction can

be renamed from mixed form to fraction form. The circles below show the mixed number 2 2/5. You are to write 2 2/5 in fraction

form with only a numerator and denominator.

To write the example, you can think of each whole number as 5/5. So in the above example you would have:

On the pretest, you can think of 13/8 .

Mixed Numbers as Fractions uses Circles to demonstrate how a fraction can be renamed from mixed form to fraction form.

The circles below show the mixed number 2 2/5. You are to write 2 2/5 in fraction form with only a numerator and denominator.

To write the example, you can think of each whole number as 5/5. So in the above example you would have:

On the pretest, you can think of 13/8 .

8

51

8

51

8

5

8

8

8

13

http://www.visualfractions.com/MixtoFrCircle.html

42/5 is what mixed number? 42/5 is what mixed number?

4 2/

51

3/5

1 1/

5 0

...

88%

0%0%

13%

1. 22/5

2. 8/5

3. 6/5

4. 2/20

1. 22/5

2. 8/5

3. 6/5

4. 2/20

4(2/5) =?

(4)(5)+2= 22

22 5

4(2/5) =?

(4)(5)+2= 22

22 5

Basic MathBasic Math

Significant FiguresSignificant Figures

Sig FigsSig Figs

1. Non-zero digits are always significant

2. Any zeros between two significant digits are significant.

3. A final zero or trailing zeros in the decimal portion ONLY are significant.

1. Non-zero digits are always significant

2. Any zeros between two significant digits are significant.

3. A final zero or trailing zeros in the decimal portion ONLY are significant.

http://www.sciencebyjones.com/multiply_sig_figs.htm

Sig FigsSig FigsRule 1: All non-zero digits are significant.12.83 cm [4] 16935 g [5]

Rule 2: Zeros between other significant figures are significant.12,038 cm [5] 169.04 g [5] 70,304 g [ ] 395.01 kg [ ]

Rule 3: Zeros to the right of a decimal point and to the right of a number are significant.12.380 cm [5] 169.00 m [5] 3.010 mL [4] 1.30 kg [ ] 1691.100 cm [ ]

Rule 4: A zero standing alone to the left of a decimal point is not significant.0.421 g [3] 0.5 m [ ]

Rule 5: Zeros to the right of the decimal and to the left of a number are not significant.0.000421 mg [3] 0.00180 cm [3] 0.010 kg [ ] 0.01010 m [ ]

Rule 1: All non-zero digits are significant.12.83 cm [4] 16935 g [5]

Rule 2: Zeros between other significant figures are significant.12,038 cm [5] 169.04 g [5] 70,304 g [ ] 395.01 kg [ ]

Rule 3: Zeros to the right of a decimal point and to the right of a number are significant.12.380 cm [5] 169.00 m [5] 3.010 mL [4] 1.30 kg [ ] 1691.100 cm [ ]

Rule 4: A zero standing alone to the left of a decimal point is not significant.0.421 g [3] 0.5 m [ ]

Rule 5: Zeros to the right of the decimal and to the left of a number are not significant.0.000421 mg [3] 0.00180 cm [3] 0.010 kg [ ] 0.01010 m [ ]


Sig FigsSig FigsRule: When adding and subtracting numbers that come from measurements, arrange the numbers in columnar form. The final answer can contain only as many decimal places as found in the measurement with the fewest number of decimal places.

Example: 134.050 m + 1.23 m =

134.050 m+ 1.23 m135.28 m (2 decimal places)

Rule: When adding and subtracting numbers that come from measurements, arrange the numbers in columnar form. The final answer can contain only as many decimal places as found in the measurement with the fewest number of decimal places.

Example: 134.050 m + 1.23 m =

134.050 m+ 1.23 m135.28 m (2 decimal places)


Sig FigsSig FigsRule: In multiplication and division, the result may have no more significant figures than the factor with the fewest number of significant figures.

Example: 2.52 m x 1.0004243 m = 2.521069236 m2

but must be recorded as 2.52 m2 (3 sig figs)

Rule: In multiplication and division, the result may have no more significant figures than the factor with the fewest number of significant figures.

Example: 2.52 m x 1.0004243 m = 2.521069236 m2

but must be recorded as 2.52 m2 (3 sig figs)


How many Sig Figs are in 108,602?

How many Sig Figs are in 108,602?

4 6 1 3

0% 0%0%

100%

1. 4

2. 6

3. 1

4. 3

1. 4

2. 6

3. 1

4. 3

108,602

All numbers are significant

108,602


How many Sig Figs are in 108.00108?

How many Sig Figs are in 108.00108?

3 8 4 3

0% 0%0%

100%

1. 3

2. 8

3. 4

4. 3

1. 3

2. 8

3. 4

4. 3

108.00108


108.00108



Unit ConversionsMathematics Chapter 2 Dimensional Analysis

Unit ConversionsMathematics Chapter 2 Dimensional Analysis

RULES FOR CONVERSIONSRULES FOR CONVERSIONS

1.SHOW ALL WORK

2.CARRY YOUR UNITS TILL THE END

3.FOLLOW PROPER ORDER OF OPERATIONS

4.CARRY OUT ALL SQUARING OR CUBING ACTIONS

5.DO NOT JUST WRITE DOWN ANSWERS WITHOUT WORK

6.USE YOUR UNITS TO GUIDE YOU

1.SHOW ALL WORK

2.CARRY YOUR UNITS TILL THE END

3.FOLLOW PROPER ORDER OF OPERATIONS

4.CARRY OUT ALL SQUARING OR CUBING ACTIONS

5.DO NOT JUST WRITE DOWN ANSWERS WITHOUT WORK

6.USE YOUR UNITS TO GUIDE YOU

Unit ConversionsUnit ConversionsExample 1. Convert 4,000 cu. Inches to cu. yardsExample 1. Convert 4,000 cu. Inches to cu. yards

)!(8.0)27

1)(

1728

1)(000,4(

)!(........)3

1)(

12

1)(000,4(

)3

1)(

12

1)(000,4(

)3

1()

12

1)(000,4(

33

3

3

33

33

3

3

33

3

3

3

33

333

yesydft

yd

in

ftin

yesydft

yd

in

ftin

ft

yd

in

ftin

ft

yd

in

ftin

Step 1. Set up conversion

Step 2. Carry out unit order of operations (cube)Step 3. Cancel units (do you have the right answer??)Step 4. perform numerical order of operations (square and cube numbers)

Final 2 Steps. Multiply denominator together and then divide (2 steps=less likely for mistake with the TI calculator)




Step 1

Step 2

Step 4

Step 3

Unit ConversionsUnit ConversionsExample 2. Convert 5000 gallons to cu. yardsExample 2. Convert 5000 gallons to cu. yards

)!(7.24)27

1)(

48.7

1)(000,5(

)!(........)27

1)(

48.7

1)(000,5(

)3

1)(

48.7

1)(000,5(

)3

1)(

48.7

1)(000,5(

33

33

33

33

3

33

33

yesydft

yd

gal

ftgal

yesydft

yd

gal

ftgal

ft

yd

gal

ftgal

ft

yd

gal

ftgal







Step 1

Step 2

Step 4

Step 3

How many gallons are there in 82 ft3?

10.

9 g

613

gal

I don’t

know

18%

0%

82%

7.48g (82 ft3) =613 or 610 g rounded 1 ft3 7.48g (82 ft3) =613 or 610 g rounded 1 ft3

1. 10.9 g

2. 613 gal

3. I don’t know

1. 10.9 g

2. 613 gal

3. I don’t know

Convert 3.2 ft3/sec to million gallons per day?

3.2

mgd

2.1

mgd

5.0

mgd

I don’t

know

0% 0%0%

100%

3.2 ft3 60 sec 1,440 min 7.48 gal 1 million gallon sec 1 min 1d 1 ft3 1,000,000 gallons

2.1 mgd

3.2 ft3 60 sec 1,440 min 7.48 gal 1 million gallon sec 1 min 1d 1 ft3 1,000,000 gallons

2.1 mgd

1. 3.2 mgd

2. 2.1 mgd

3. 5.0 mgd

4. I don’t know

1. 3.2 mgd

2. 2.1 mgd

3. 5.0 mgd

4. I don’t know


% to mg/L

REMEMBER: 1 ppm = 1mg/L% to mg/L

REMEMBER: 1 ppm = 1mg/L

% to Mg/L Word Problems% to Mg/L Word ProblemsYou can memorize or set up a ratio. Its your choiceYou can memorize or set up a ratio. Its your choice

Rule 1. to convert mg/L (ppm) to % multiply by 0.0001Rule 1. to convert mg/L (ppm) to % multiply by 0.0001

Rule 2. to convert % to mg/L (ppm) multiply by 10,000 Rule 2. to convert % to mg/L (ppm) multiply by 10,000

Rule 3. Ratio for percent to mg/L:Rule 3. Ratio for percent to mg/L:

l

mgx

l

mg%

000,10

%1

% to Mg/L Word Problems% to Mg/L Word ProblemsExample 1. Convert 0.55% to mg/LExample 1. Convert 0.55% to mg/L

l

mg

l

mgC

l

mg

l

mgC

l

mgC

l

mg

l

mgx

l

mg

500,5)(

%)55.0)(000,10()%)(1(

%55.0

000,10

%1

%

000,10

%1

Step 1. Show formula

Step 2. Set up ratio

Step 3. Cross multiply

Step 4. Solve for variable

Final Step . Are units correct?





Final Step . Are units correct?

Step 1

Step 2

Step 3

Step 4

% to Mg/L Word Problems% to Mg/L Word ProblemsExample 2. Convert 2,000 mg/L to percentExample 2. Convert 2,000 mg/L to percent

%)(%2.0

%)()10(

)2(

%)()000,10(

)000,2%)(1(

%))(000,10()000,2%)(1(

000,2

%

000,10

%1

%

000,10

%1

P

P

l

mgl

mg

P

l

mgl

mg

Pl

mg

l

mgl

mgP

l

mg

l

mgP

l

mg





Step 5. Reduce Fraction

Final Step. Solve….Are units correct?





Step 5. Reduce Fraction

Final Step. Solve….Are units correct?

Step 1

Step 2

Step 3

Step 4

Step 5

A solution was found to be 1.3% alum. How many milligrams per liter of alum are in the solution?

13,

000

mg/L

1.3

mg/L

130

,000

mg/L

I don’t

know

86%

0%7%7%

10,000 mg/L = X 1% 1.3%

10,000 mg/L (1.3) = X

13000 mg/L = x

10,000 mg/L = X 1% 1.3%

10,000 mg/L (1.3) = X

13000 mg/L = x

1. 13,000 mg/L

2. 1.3 mg/L

3. 130,000 mg/L

4. I don’t know

1. 13,000 mg/L

2. 1.3 mg/L

3. 130,000 mg/L

4. I don’t know


Temperature ConversionsTemperature Conversions

Temperature ConversionsTemperature Conversions

oF= (9 * oC) + 32 5

oF= (9 * oC) + 32 5

oC= 5 * (oF – 32) 9

oC= 5 * (oF – 32) 9

Convert 17oC to Fahrenheit Convert 17oC to Fahrenheit

Convert 451oF to degrees CelsiusConvert 451oF to degrees Celsius

oF= (9 *17)+32=62.6oF= 63oF 5

oF= (9 *17)+32=62.6oF= 63oF 5

Celsius to Fahrenheit 1. Begin by multiplying the Celsius temperature by 9. 2. Divide the answer by 5. 3. Now add 32.

Celsius to Fahrenheit 1. Begin by multiplying the Celsius temperature by 9. 2. Divide the answer by 5. 3. Now add 32.

Fahrenheit to Celsius1. Begin by subtracting 32 from the Fahrenheit #. 2. Divide the answer by 9. 4. Then multiply that answer by 5.

Fahrenheit to Celsius1. Begin by subtracting 32 from the Fahrenheit #. 2. Divide the answer by 9. 4. Then multiply that answer by 5.

oC= 5* (oF -32)=232.7oC= 233oC 9

oC= 5* (oF -32)=232.7oC= 233oC 9

Convert 75oF to degrees Celsius?

24

oC

107

oC

I don’t

know

50%

0%

50%

1. 24 oC

2. 107 oC

3. I don’t know

1. 24 oC

2. 107 oC

3. I don’t know

oC=5/9 (oF - 32)oC=5/9 (o75 - 32)oC=0.55 (43)oC = 24

oC=5/9 (oF - 32)oC=5/9 (o75 - 32)oC=0.55 (43)oC = 24

The objectives for this week were met with the assignment

and lecture?

Stro

ngly A

gree

Agre

e

Dis

agre

e

Stro

ngly D

isag

ree

46%

0%

8%

46%

1. Strongly Agree

2. Agree

3. Disagree

4. Strongly Disagree

1. Strongly Agree

2. Agree

3. Disagree

4. Strongly Disagree


1. Manipulation of fractions and decimals2. Percent and unit conversions


1. Manipulation of fractions and decimals2. Percent and unit conversions

basic math conversions math for water technology mth 082 fall 08 chapters 1, 2, 4, and 7 lecture 1...

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