# 110_-_7_-_Homework_R.. - Skilled Trades Math On-line

Πολεοδομικά Έργα

29 Νοε 2013 (πριν από 4 χρόνια και 5 μήνες)

96 εμφανίσεις

-

Michael Bush

1

Industrial Application of Basic

Mathematical Principles

Session 7

Homework Assignments

-

Michael Bush

2

Industrial Application of Basic

Mathematical Principles Session 7 Homework

Unit 27 Problem 1 Page 216

Compute the Celsius melting point for Chromium

a.

5
C = F - 32
9

5
C = 2,740 - 32
9
  
5
C = 2,708 =
9
 
1504 C

-

Michael Bush

3

Industrial Application of Basic

Mathematical Principles Session 7 Homework

Unit 27 Problem 1 Page 216

Compute the Celsius melting point for Cast Iron

a.

5
C = F - 32
9

5
C = 2,300 - 32
9
  
5
C = 2,268 =
9
 
1260 C

-

Michael Bush

4

Industrial Application of Basic

Mathematical Principles Session 7 Homework

Unit 27 Problem 1 Page 216

Compute the Celsius melting point for Copper

a.

5
C = F - 32
9

5
C = 1,940 - 32
9
  
5
C = 1,908 =
9
 
1060 C

-

Michael Bush

5

Industrial Application of Basic

Mathematical Principles Session 7 Homework

Unit 27 Problem 1 Page 216

Compute the Celsius melting point for Aluminum

a.

5
C = F - 32
9

5
C = 1,200 - 32
9
  
5
C = 1,168 =
9
 
649 C

-

Michael Bush

6

Industrial Application of Basic

Mathematical Principles Session 7 Homework

Unit 27 Problem 1 Page 216

Compute the Celsius melting point for Lead

a.

5
C = F - 32
9

5
C = 620 - 32
9
  
5
C = 588 =
9
 
327 C

-

Michael Bush

7

Industrial Application of Basic

Mathematical Principles Session 7 Homework

Unit 27 Problem 2 Page 217

8
10
12
14
16
18
20
22
24
26
28
30
Degrees Fahrenheit (hundreds)
6
CHROME
CAST
COPPER
ALUMINUM
2
4
6
12
14
16
18
10
8
Degrees Celsius (hundreds)
20
-

Michael Bush

8

Industrial Application of Basic

Mathematical Principles Session 7 Homework

Unit 27 Problem 3 Page 217

Determine difference between Chromium and Lead

a.

Chromium = 1,504 C

1,504 C
327 C

 
1,177 C

-

Michael Bush

9

Industrial Application of Basic

Mathematical Principles Session 7 Homework

Unit 27 Problem 3 Page 217

Determine difference between Cast Iron and Aluminum

a.

Cast Iron = 1,260 C

Aluminum = 649°C
1,260 C
649 C

 
611 C

-

Michael Bush

10

Industrial Application of Basic

Mathematical Principles Session 7 Homework

Unit 27 Problem 3 Page 217

Determine difference between Copper and Lead

a.

Copper = 1,060 C

1,060 C
327 C

 
733 C

-

Michael Bush

11

Industrial Application of Basic

Mathematical Principles Session 7 Homework

Unit 28 Problem 1 Page 217

Generate a line graph

10
12
14
16
18
20
22
24
26
28
30
Revolutions Per Minute (hundreds)
TEMPERATURE (Fahrenheit)
700
800
900
1000
1100
1200
1300
32
34
36
-

Michael Bush

12

Industrial Application of Basic

Mathematical Principles Session 7 Homework

Unit 28 Problem 2 Page 217

Determine the average difference in exhaust
temperatures

Medium Compression engine

825 920 1,000 1,080 1,150 1,220
     
Low Compression engine

780 870 950 1,030 1,100 1,160
     
1032.5 F - 981.7 F=
 
50.8 F

6195°F
6195°F 6 =

1032.5°F
5890°F
5890°F 6 =

981.7°F
-

Michael Bush

13

Industrial Application of Basic

Mathematical Principles Session 7 Homework

Unit 28 Problem 2 Page 217

Determine the difference in exhaust temperatures at
3,250 RPM

High Compression engine

3,250 RPM 1,240 F
 
Low Compression engine

3,250 RPM 1,130 F
 
1240 C - 1130 C=
 
110 F

-

Michael Bush

14

Industrial Application of Basic

Mathematical Principles Session 7 Homework

Unit 29 Problem 1 Page 218

Draw a Circle Graph

All
Others
USA
France
Great
Britain
-

Michael Bush

15

Industrial Application of Basic

Mathematical Principles Session 7 Homework

Unit 29 Problem 2 Page 218

Determine the Percentages

175,000
90,000
125,000
100,000
+ 60,000
550,000
175,000 550,000=

.3182=
31.82%
125,000 550,000=

.2273=
22.73%
90,000 550,000=

.1636=
16.36%
100,000 550,000=

60,000 550,000=

.1818=
.1091=
18.18%
10.91%
-

Michael Bush

16

Industrial Application of Basic

Mathematical Principles Session 7 Homework

Unit 29 Problem 3 Page 218

Determine the Percentages

175,000
90,000
125,000
100,000
+ 60,000
550,000
175,000 550,000=

.3182=
31.82%
125,000 550,000=

.2273=
22.73%
31.82%
+ 22.73%
54.55%
-

Michael Bush

17

Industrial Application of Basic

Mathematical Principles Session 7 Homework

Unit 29 Problem 4 Page 218

All
Others
USA
and
Great
Britain
and
France
Draw a Circle Graph

-

Michael Bush

18

Industrial Application of Basic

Mathematical Principles Session 7 Homework

Unit 30 Problem 1 Page 218

a.

Sampling provides an economical method of
establishing the acceptability of a manufactured lot
by a limited sampling verses complete inspection of
all manufactured parts.

b.

In single sampling, a random, single sample of a
specified number of parts is inspected. A larger
number of parts is required verses the first samples
in sequential sampling. Sequential requires greater
numbers of samples but fewer parts with each
sample.

-

Michael Bush

19

Industrial Application of Basic

Mathematical Principles Session 7 Homework

Unit 30 Problem 2a Page 218

Tempering Temperature Fahrenheit
220
240
300
Tempering Temperature Celsius
430
450
470
490
510
530
550
570
590
610
630
650
A
B
C
D
E
F
G
H
CARBON
STEEL TOOLS
320
340
280
260
Construct a Line Graph

-

Michael Bush

20

Industrial Application of Basic

Mathematical Principles Session 7 Homework

Unit 30 Problem 2b Page 218

Plain

Carbon Steel Tools

Tempering Temperatures

°
F

°
C

A

Roughing Mills

430

221

B

Counterbores

460

238

C

Knurls

485

251

D

Tube Cutters

485

251

E

Taps

500

260

F

530

277

G

Pneumatic Tools

580

302

H

Non
-
Cutting Tools

640

338

4110
°
F

2138
°
C

4110°F ÷ 8 =
513.75°F
2138°C ÷ 8 =
267.25°C
-

Michael Bush

21

Industrial Application of Basic

Mathematical Principles Session 7 Homework

Unit 30 Problem 2c Page 218

Plain

Carbon Steel Tools

Tempering Temperatures

°
F

°
C

A

Roughing Mills

430

221

B

Counterbores

460

238

C

Knurls

485

251

D

Tube Cutters

485

251

E

Taps

500

260

F

530

277

G

Pneumatic Tools

580

302

H

Non
-
Cutting Tools

640

338

485°F + 500 F =

985°F
251°C + 260 C =

511 C

Median

985°F ÷ 2 =
492.5°F
511°C ÷ 2 =
255.5°C