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Maths Examples and Topics

Subject: Maths

Pages: 12 Words: 3600

Subject: Maths

Pages: 10 Words: 3000

Math Assignment

[Author Name(s), First M. Last, Omit Titles and Degrees]

[Institutional Affiliation(s)]

Author Note

Math Assignment

Online Homework 2.1

1)

{A, B} and {a, b}

A→ a, B→ b

A→ b, B→ a

2)

{a, b, c} and {1, 2, 3}

a→ 1, b→ 2, c→ 3

a→ 1, b→ 3, c→ 2

a→ 2, b→ 1, c→ 3

a→ 2, b→ 3, c→ 1

a→ 3, b→ 1, c→ 2

a→ 3, b→ 2, c→ 1

3)

a)

b)

4)

a)

b)

5)

LXIII

6)

DCCLVIII

7)

MDCCCLXVII

8)

MMCMLXXXIX

9)

MCCXXXXIV

10)

65

11)

782

12)

1392

13)

3434

14)

3451

15)

CDLVI

16)

DXCVI

17)

a)

900

b)

CM

c)

d)

Online Homework 2.2

1)

937

2)

362

3)

3 + (14x20) + (0x20x20) + (5x20x20x20) = 40283

4)

18 + (12x20) + (17x20x20) + (8x20x20x20) = 71058

5)

6)

Online Homework 2.3

1)

5,631 = 5000 + 600 + 30 + 1

2)

8,327 = 8000 + 300 + 20 + 7

3)

1,232 = 1000 + 200 + 30 + 2

4)

32

5)

59

6)

59

7)

63

8)

6588

9)

20378

Online Homework 2.4

1)

2202

2)

3312

3)

2033

4)

1001100

5)

1110001

6)

1111010

7)

1111100

Module 1 and 2 Test Practice

1)

1110111

2)

(A ꓵ B)c = {1,2,3,5,6,7,8,9,10}

(Ac U Bc) = {1,2,3,5,6,7,8,9,10}

3)

4

4)

(A U B U C) = {1,7,8,11,15,18,20,21,22,23,24,26,29,30}

(A ꓵ B ꓵ C) = {8}

5)

(a), (b), (d) and (f) i.e. {5,7,11,13,17,19}, {17}, {} and {5,7}

6)

7)

47

8)

9)

10)

(Ac U ϕ) = {3,7,8,11}

(A ꓵ C) = {2}

(Ac U B) = {1,3,5,7,8,11}

(B ꓵ C) ꓵ Ac = {3,7}

11)

(A U B) = {a,b,c,f,g,h,i,j,k,m}

(A ꓵ B) = {f,h}

(A ꓵ B)c = {a,b,c,g,i,j,k,m}

k € A, (A U B), (A ꓵ B)c and U

{f,k} ⸦ U, (A U B)

12)

5,493 = 5000 + 400 + 90 + 3

13)

575

14)

MMDXXXVII

15)

33523

16)

a)

(N U S) = {4D,4N,4Q,5N,6N,7N,8N}

b)

(Q U W) = {4Q,5Q,6D,6H,6N,6P,6Q,7Q,8Q,9Q}

c)

(D U P) = {4D,5D,5P,6D,6P,7D,7P,8D,8P,9D}

17)

n(Ac ꓵ Bc) = 1

18)

DCLIX

19)

4242

20)

1913

21)

n(A) = 4

n(A ꓵ C) = 2

n(A ꓵ B ꓵ Cc) = 1

22)

(A ꓵ C)c U (B ꓵ A)c = {b,c,d,e,f,g,i,j}

23)

n(A ꓵ C) = 2

24)

n(P) = 4

n(N) = 5

n(Q) = 6

n(D) = 6

n(Z) = 3

n(H) = 4

n(Y) = 4

n(T) = 0

n(S) = 3

n(W) = 5

n(X) = 5

n(V) = 5

n(C) = 25

n(A) = 0

25)

56326

Subject: Maths

Pages: 11 Words: 3300

Your Name

Instructor Name

Course Number

Date

Title: Math Homework:

Car 1 make: Mercedes-Benz-E-ClassModel:2017Price: $53,500

Car 2 make: BMW 5 SeriesModel:2017Price: $ 53,400

Q1: What does it mean to be upside down in a loan? Look it up, the write the definition in your own words.

When the amount of loan borrowed using a collateral is more than the value of the item that you had bought, an upside loan situation occurs. In this collateral could be anything from car to home. Being Upside down in a loan happens when market value of the product bought has depreciated faster in respect to the decrease in loan balances.

Q2: Rate of depreciation of two new 2017 vehicles and their value in ten years.

Using the values of the cars listed above, their depreciation is calculated using reducing balance method.

Car 1

Depreciation for 1st year = 1337553500*100=25%

Depreciation for 2nd year = 18190-1337553500-13375*100=12%

Depreciation for 3rd year = 22427-1819053500-18190*100=12%

Depreciation for 4rth year = 26777-2242753500-22427*100=14%

Depreciation for 5th year = 30786-2677753500-26777*100=15%

Depreciation for 6th year = 34193-3078653500-30786*100=15%

Depreciation for 7th year = 36896-3419353500-34193*100=14%

Depreciation for 8th year = 39054-3689653500-36896*100=13%

Depreciation for 9thyear = 40392-3905453500-39054*100=9%

Depreciation for 10th year = 42440-4039253500-40392*100=16%

Car 2:

Depreciation for 1st year = 1335053400*100=25%

Depreciation for 2nd year = 18156-1335053400-13350*100=12%

Depreciation for 3rd year = 22385-1815653400-18156*100=12%

Depreciation for 4rth year = 26727-2238553400-22385*100=14%

Depreciation for 5th year = 30728-2672753400-26727*100=15%

Depreciation for 6th year = 34129-3072853400-30728*100=15%

Depreciation for 7th year = 36827-3412953400-34129*100=14%

Depreciation for 8th year = 38981-3682753400-36827*100=13%

Depreciation for 9thyear = 40856-3898153400-38981*100=13%

Depreciation for 10th year = 42361-4085653500-40856*100=12%

Calculation:

Year

Depreciation rate for Car 1($)

Depreciation rate for Car 2 ($)

0

53500

53400

1

13375

13350

2

18190

18156

3

22427

22385

4

26777

26727

5

30786

30728

6

34193

34129

7

36896

36827

8

39054

38981

9

40392

40856

10

42440

42361

Q3: Graphs for the depreciation of each car

Graph for Cost of Depreciation for car 1

Graph for Cost of Depreciation for car 2

Q4: Using 5.25% as APR, calculate the total loan payments that you would have to make for year 3, 5, 7 and 10 years.

Year

Beginning Balance car 1 ($)

Total Loan Payment car 1 ($)

Year

Beginning Balance car 2 ($)

Total loan payment car 2 ($)

3

44917.28

4,640.58

3

44833.32

4631.93

5

35386.55

5153.14

5

35320.40

5143.55

7

24083.08

5722.36

7

24756.72

5711.68

10

6696.18

6696.16

10

6683.66

6683.71

Calculation:

Number of Loan Years

Total Payments for Car1 ($)

Total Payments for Car 2 ($)

3

4,640.58

4631.93

5

5153.14

5143.55

7

5722.36

5711.68

10

6696.16

6683.71

Compare these results, why some are higher than others from the same car

The table illustrates an increase in the value of loan payments for both cars in each of the 4 periods. This difference in values is because the interest paid lowers and the principle increases over till the loan’s maturity. This leads to increasing loan payments as the loan period increases further from 3 to 10 years for both cars.

Q5: Write the total amount of interest that you have to pay for each of the 4 situations described above

Number of Loan Years

Total amount of interest for car 1 ($)

Total amount of interest for car 2 ($)

3

2247.54

2243.35

5

1734.98

1731.73

7

1165.76

1163.60

10

191.96

191.57

year

Beginning Balance car 1 ($)

Interest car 1($)

Year

Beginning Balance car 2 ($)

Interest car 2 ($)

3

44917.28

44917*5.25%=2247.54

3

44833.32

44833.32*5.25%=2243.35

5

35386.55

35386.55*5.25%=1734.98

5

35320.40

35230.40*5.25%=5143.55

7

24083.08

24083.08*5.25%=165.76

7

24756.72

24756.72*5.25%=1163.60

10

6696.18

6696.18*5.25%=191.96

10

6683.66

6683.66*5.25%=191.57

Calculation:

Compare these results, why some are higher than others from the same car?

Total amount of interest paid over the years will decrease as the principal increases. At first year interest is high because principal paid on the loan is less. As more and more of the loan is paid off, the interest shrinks and keeps on decreasing till the loans maturity.

Q6: Assuming that you will take a 10 year long loan on your car, complete the following table with the amount of principal left after each year (look at the balance value).

Year

Principle left (Balance for ) Car 1($)

Principal left (Balance for) Car 2 ($)

0

53500

53400

1

53500-2709.13-4178.99 = 49321.02

53400-2704.08-4171.20= 49228.83

2

4931.02-2484.38-4403.75 = 44917.28

49228.83-2479.76-4395.52 = 44833.52

3

44917.28-2247.54-4640.58 = 40276.70

44833.32-2243.35-4631.93= 40201.42

4

40276.70-1997.96-4890.16 = 35386.55

40201.42-1994.23-4881.05= 35320.40

5

35386.55-1734.98-5153.14 = 30233.39

35320.40-1731.73-5143.55= 30176.88

6

30233.39-1457.82-5430.30 = 24803.08

30176.88-1455.08-5420.20= 24756.72

7

24803.08-116576-5722.36 = 19080.72

24756.72-1163.60-5711.68= 19045.06

8

19080.72-858.01-6030.11 = 13050.61

19045.06-856.41-6018.87= 13026.21

9

13050.61-533.69-6354.43 = 6696.18

13026.21-532.71-6342.57= 6683.66

10

6696.18-191.96-6696.16 = 0

6683.66-191.57-6683.71= 0

*To calculate the principle left after every year (not the principle paid every year) interest and principle paid within any given year will be subtracted from the beginning balance of every year. The remaining balance left would be the principle left after every year.

Q7: On the same graphs for Q2, graph the principal left to pay from Question 6 for both cars in RED. Connect the dots in a curve pattern.

In what years is the Red Line above the Blue Line for car 1?

Till year 3 the Red line line, denoting principle left (Balance for) car 1 has stayed above the Blue line, denoting depreciation for Car 1. This can be seen in the graph above.

In what years is the Red Line above the Blue Line for car 2?

The graph above illustrates that the Red line for principle left (Balance for) car 1 has stayed above the Blue line for depreciation of car 2 till 3 years.

Q8: What does it mean when the Red Line is above the Blue Line?

As, the period of the loan increases over time its principal amounts are going to decrease till its maturity. During this time the product bought using the loan, will have an increasing rate of depreciation and will continue to have a diminishing value even after the loan reaches maturity. In this case the value for the two cars bought using the loans will keep on decreasing due to their increase in depreciation value of the cars. In this time, their principal payments will get lower until the whole loan is paid off.

Q9: Based on these two graphs, which car do you think is the best option to purchase?

Explain why, please write your answer as a short paragraph

Based on these graphs, the recommendation would be to buy car 2 (BMW 5 series). This is based on the fact that firstly the price of car 2 ($53400) is bit lower than car 1 ($53500). This makes it a bit cheaper and has a high accessibility chance than car 1. Secondly, car 2 has a lower depreciating rate and depreciation cost than car 1. This will allow car 2 to retain more of its value over its lifetime and would be able to sold off at a higher price in comparison to car 1.Third, its interest payment is lower compared to car 1, so less amount of principle would have to be paid over the course of the loan period. Lastly, car 1 for total 10 years loan payment is $68881.27, whereas, for car 2 it is $68752.52. This gives the idea the it will be cheaper to pay off loan for car 2 than 1.

Subject: Maths

Pages: 10 Words: 3000

[Name of the Writer]

[Name of Instructor]

[Subject]

[Date]

Math hw

5- (cos2θ/ sinθ) + sinθ = cotθ/ secθ

LHS: Left Hand Side

RHS: Right Hand Side

LHS= (Cos2θ/ sinθ) + sinθ

Cos2θ = cos2θ-sin2θ

= ((cos2θ – sin2θ)/ sinθ) + sinθ

= (cos2θ – sin2θ + sin2θ)/ sinθ

= cos2θ/ sinθ

= (cosθ/ sinθ)* cosθ

cosθ/ sinθ=cotθ and cosθ=1/ secθ

= cotθ/ secθ = RHS

Hence proved LHS = RHS

6- tanθ + cotθ = 2/ sin2θ

LHS= tanθ + cotθ

tanθ = sinθ/ cosθ and cotθ = cosθ/ sinθ

= sinθ/ cosθ + cosθ/ sinθ

= (sin2θ + cos2θ)/ sinθcosθ

sin2θ + cos2θ = 1

= 1/ sinθcosθ

Multiply and divide by 2

=2/ 2sinθcosθ

sin2θ = 2sinθcosθ

=2/ sin2θ = RHS

Hence proved LHS = RHS

7- sin2θ = 2tanθ/ (1 + tan2θ)

RHS= 2tanθ/ (1 + tan2θ)

tanθ = sinθ/ cosθ

= 2 (sinθ/ cosθ)/ (1 + sin2θ/ cos2θ)

= 2 (sinθ/ cosθ)/ ((cos2θ + sin2θ)/ cos2θ)

sin2θ + cos2θ = 1

= 2 (sinθ/ cosθ)/ (1/ cos2θ)

=2 (sinθcos2θ / cosθ)

=2sinθcosθ

sin2θ = 2sinθcosθ

= sin2θ = LHS

Hence proved LHS = RHS

8- cos2θ = (1 - tan2θ)/ (1 + tan2θ)

RHS= (1 - tan2θ)/ (1 + tan2θ)

tanθ = sinθ/ cosθ

= (1 - (sin2θ/ cos2θ))/ (1 + (sin2θ/ cos2θ))

= ((cos2θ - sin2θ)/ cos2θ)/ ((cos2θ + sin2θ)/ cos2θ)

cos2θ + sin2θ = 1 & cos2θ - sin2θ= cos2θ

= (cos2θ/ cos2θ)/ (1/ cos2θ)

= cos2θ = LHS

Hence proved LHS = RHS

9- sin2θsec2θ = 2tanθ

LHS= sin2θsec2θ

sin2θ= 2sinθcosθ and sec2θ= 1+ tan2θ

= 2sinθcosθ (1+ tan2θ)

tanθ= sinθ/ cosθ

= 2sinθcosθ + 2sinθcosθ (sin2θ/ cos2θ)

= (2sinθcos3θ + 2 sin3θcosθ)/ cos2θ

= (2sinθcosθ (cos2θ + sin2θ))/ cos2θ

cos2θ + sin2θ = 1

= 2sinθcosθ/ cos2θ

= 2sinθ/ cosθ

tanθ= sinθ/ cosθ

=2tanθ = RHS

Hence proved LHS = RHS

10- 2 - sec2θ = cos2θsec2θ

LHS= 2 - sec2θ

sec2θ = 1 + tan2θ

= 2 – (1 + tan2θ)

= 1 – tan2θ

tanθ = sinθ/ cosθ

= 1 – (sin2θ/ cos2θ)

= (cos2θ - sin2θ)/ cos2θ

cos2θ - sin2θ= cos2θ and sec2θ= 1/ cos2θ

= cos2θsec2θ = RHS

Hence proved LHS = RHS

11- (cosθ – sinθ)2 = 1 - sin2θ

LHS= (cosθ – sinθ)2

= cos2θ + sin2θ – 2sinθcosθ

cos2θ + sin2θ= 1 and sin2θ = 2sinθcosθ

= 1 - sin2θ = RHS

Hence proved LHS = RHS

12- (cosθ + sinθ)2 = 1 + sin2θ

LHS= (cosθ + sinθ)2

= cos2θ + sin2θ + 2sinθcosθ

cos2θ + sin2θ= 1 and sin2θ= 2sinθcosθ

= 1 + sin2θ = RHS

Hence proved LHS = RHS

13- (2sin2θ/ sin2θ) + cotθ = secθcscθ

LHS= (2sin2θ/ sin2θ) + cotθ

sin2θ = 2sinθcosθ and cotθ = cosθ/ sinθ

= (2sin2θ/ 2sinθcosθ) + cosθ/ sinθ

= (sinθ/ cosθ) + (cosθ/ sinθ)

= (sin2θ + cos2θ)/ cosθsinθ

sin2θ + cos2θ = 1

= 1/ cosθsinθ

secθ = 1/ cosθ and cscθ = 1/ sinθ

= secθcscθ = RHS

Hence proved LHS = RHS

14- cos2θ = (cot2θ -1)/ (cot2θ + 1)

RHS= (cot2θ -1)/ (cot2θ + 1)

cotθ = cosθ/ sinθ

= ((cos2θ/ sin2θ) – 1)/ ((cos2θ/ sin2θ) + 1)

= ((cos2θ - sin2θ)/ sin2θ)/ ((cos2θ + sin2θ)/ sin2θ)

cos2θ =cos2θ - sin2θ and cos2θ + sin2θ= 1

= cos2θ = LHS

Hence proved LHS = RHS

Subject: Maths

Pages: 5 Words: 1500

Subject: Maths

Pages: 30 Words: 9000

Mathematical solution

Student’s name

Institution

Date

Problem 1: Sandwich sales

Problem

Double Meat Place intends to launch a product in the market, but it has certain challenges, which it has to address for its sales to be successful. The main problems are the inconsistent or varieties of the sales price of the product. The prices of the product vary according to a region, and this affects the average unit sale per region. The analysis of the market text unit reveals that the average unit sales depend on the price of the Trios. The lowest sales price generates the highest average unit sales and this could and the highest sales price generates the lowest average unit sales. This could mean that the demand for the product depend on the price of the product. The fact that the averages sales depend on the price of the product is a problem because it means that the company would be able to register high sales only when the prices are low and therefore, this problem must be addressed by the marketing and sales team. Again, it is not appropriate for a company to have differences prices of the same product, this could affect the brand and image of the company in the market and therefore, it is important for the marketing and the sales department to work on the modernity to realign prices so that a company can have a common price in the market.

Designing a solving problem

In order to address the problem, regression formula or linear regression would be used to predict the appropriate sales price of Trios which can allow Double Meat Place to sell an average unit of 2300 per location. The regression will involve the use of graphing looking for a pattern that is applied to determine the sales unit of product for a specific region. Regression is used to establish an understanding among the variables and therefore, it will be used to determine how independent variables are related to the dependent variable in order to establish the sales price of a product per region based on the specific average unit sales. The efficient address of the problem will require the drawing of a regression curve or graph, which will provide the exact sales price per region. The data will be transfer to the excel sheet, where it will be analyzed using a data analysis tool. The coefficient of X variable and the significant F are used to draw a graph that helps in predicting the unit price of sales which will give the company an average of sale of 2300. The listed below are the sale prices and average sales per region obtained from the market test.

Table 1: Market test Unit price

The regression (graphing method would be used to solve the problem identified. The graphing method is efficient and simple. Therefore, it would be able to help in obtaining the sales price faster and accurately as well compared to mathematical formula.

Implementation of the plan

The plan would be implemented through intensive marketing strategy focusing on building the image of the company. Image is the brand and therefore, it is important for the company to engage in different strategy to improve its brand and that will allow the company to have a common unit price in all the regions. The social media will be used to highlight the prices of Trios across the regions. The common price or unit price of the Trios based on the tabulations and as illustrated in the graph would be 2.63. Therefore, it means that the sales price of 2.63 of Trios would give Double Meat Place average sales of 2300 per region. Therefore, the marketing strategy will focus on brand building using social media, promotion and advertisement to bring customers on and also to build relationship with customers which will help the company to improve its sales in the market.

Figure 1: Regression analysis of the data

SUMMARY OUTPUT

Regression Statistics

Multiple R

0.999597922

R Square

0.999196005

Adjusted R Square

0.998928007

Standard Error

0.013052246

Observations

5

ANOVA

df

SS

MS

F

Significance F

Regression

1

0.635168917

0.635168917

3728.368

9.67773E-06

Residual

3

0.000511083

0.000170361

Total

4

0.63568

Coefficients

Standard Error

t Stat

P-value

Lower 95%

Upper 95%

Lower 95.0%

Upper 95.0%

Intercept

4.018475341

0.019818954

202.7592031

2.65E-07

3.955402584

4.0815481

3.955402584

4.081548099

X Variable 1

-0.000500292

8.1934E-06

-61.06035981

9.68E-06

-0.000526367

-0.0004742

-0.000526367

-0.000474217

Figure 2: Graph of average unit sales of Trios per region

Evaluation

Based on the analysis of the process, it is evident that nothing has been miscalculated. The method used here graphing which involve the use regression techniques to establish the unit price and the general performance of the company.

Problem 2: Drug Safety

Problem

The main problem is the pain killer drug has a serious side effect. It is observed that the pain killer affects the patient’s reaction time and decision-making process. Therefore, the main problem is the side effects that are caused by the pain killer drugs being manufactured by the company. This side effect is serious because it makes it unsafe for the patient to operate heavy machinery immediately after being admitted, and therefore, the immediate solution is required.

Solving plan

The problem can be solved using graphing where the concept of linear regression is applied. In order to solve the problem, a suitable concentration level and hours to be taken would be determined using the regressions method. The linear graph will be drawn using two variables, independent and depended on variables. These two variables will make it easy to find the suitable time and solution level which is appropriate and cannot affect the decision-making process of any patient. The graphing method would be the best choice for this study because it provides accurate results and using will allow the team to identify the correct solution level and time with no error. Chances of making calculation errors are limited in the graphing method. Therefore, it is the best method that is used to analysis the data to complete the tabulation as well.

Implementation

In order to address the problem, a suitable solution level will be established using the graphing method and linear regressions. In this case, the data collected analyzed using excels worksheet, where the graph of the linear regression represented the data to obtain an appropriate point was done. Below is the illustration of how the suitable solution level was determined using coefficient and significant F, which are the values of the two variables, independent and dependent variables. The illustrated below in table 2 is the data collected from patients, which was analyzed using regression method and graphing.

Table 2: Data collected from patients

Hours after full administration

Concentration mcg

0

19.985

1

18.052

2

16.301

3

14.725

4

13.299

5

11.992

6

10.885

7

9.77

8

8.818

Figure 3: Regression analysis of the data

00

SUMMARY OUTPUT

Regression Statistics

Multiple R

0.993663263

R Square

0.98736668

Adjusted R Square

0.98556192

Standard Error

0.329067775

Observations

9

ANOVA

df

SS

MS

F

Significance F

Regression

1

59.242

59.242

547.0903

6.63E-08

Residual

7

0.757999

0.108286

Total

8

60

Coefficients

Standard Error

t Stat

P-value

Lower 95%

Upper 95%

Lower 95.0%

Upper 95.0%

Intercept

13.81095535

0.433557

31.85502

7.77E-09

12.78576

14.83615

12.78576

14.83615

X Variable 1

-0.713080331

0.030487

-23.39

6.63E-08

-0.78517

-0.64099

-0.78517

-0.64099

Figure 5: Graph of concentration level

However, the unexpected outcome was revealed. The result of the study indicates that there no appropriate time for the solution level. Though it is stated that the solution level should be 5mgc it is established that the time is not constant and therefore, the result does not provide a permanent solution to the problem being experienced by the company.

Evaluation

The final result means that the solution level 5gmc does not time for the result before starting to operate a heavy machine. Therefore, time can still fluctuate, which means serious measures still needed to be taken.

Subject: Maths

Pages: 5 Words: 1500

Subject: Maths

Pages: 9 Words: 2700

Assignment 1.1: Working with Whole Numbers

Complete the Following Problems. Be sure to show all work.

For each whole number, determine the place value of the digit 1.

197

In this number, 1 is at hundreds place as it is third from right.

910,398

In this number, 1 is at ten thousands place as it is fifth from right

1,029,933

In this number, 1 is at millions place as it is seventh from right

203,391

In this number, 1 is at ones place as it is first from right.

3,103,929

In this number, 1 is at hundred thousands place as it is sixth from right.

Write each whole number in expanded form.

549

5*100+4*10+9*1

1,093

1*1000+0*100+9*10+3*1

239,519

2*100000+3*10000+9*1000+5*100+1*10+9*1

Round each whole number to the given place.

732 to the nearest hundred

700 since 32 is less than 50

91,359 to the nearest ten

91360 as 9 is greater than 5

29,499 to the nearest thousand

29000 as 499 is less than 500

Add.

63 + 49

112

6 + 17 + 24

47

45 + 1029 + 769 + 5000

6843

444467

Subtract.

931 – 120

711

657 – 275

382

2593

34083

Multiply.

5(615)

3075

10431

c. 36612

d. 168732

7) Divide.

13

Undefined

0

880

616

5492

Write using exponential notation.

3^5

13^3

5^3*4^2

Evaluate.

64

625

36

Find each square root.

9

13

Simplify.

50

-27

21

-4

13

*The similarity index is high because this is a numerical solution.

Subject: Maths

Pages: 1 Words: 300

Permutation and combination

Student’s name

Course Title

Instructor’s name

Date

AND & OR are mostly used in set theory during computations involved in permutations and combination. There exist differences between AND & OR because they are used in different areas and to get distinctive results. The OR logic can be used in the cases when an object meets at least one criteria with the workflow enrollment or list of filters. On the other hand, the AND logic is only used when the object under consideration meets all the requirements in the group. In other words, the logic AND represents that both conditions under consideration must be met when combining two or more pieces logic. This is in contrast to OR because the combination of the element can be completed when at least one condition is met. We can use the truth table to show the difference between the AND & OR below. We will use 1 to represent 'TRUTH' and 0 to represent 'FALSE.' In this case, we will consider 'q' and 'r' as our elements.

AND

q

r

q AND r

1

1

1

1

0

0

0

1

0

0

0

0

From the above table, q AND r is only true when both q and r is true

OR

q

r

q OR r

1

1

1

1

0

1

0

1

1

0

0

0

The above table shows that q OR r is true when either q or r is true

We can as well use mathematical inequalities to show the difference that exists between AND&OR. For instance, when we say x ≤ 4, it means that x=4 or x < 4. If p: x = 4 and q: x < 4 and x = 2 then p will be false while q will be true. X ≤ 4 will still be true. On the other hand, can use another statement to represent AND. If we take 2 < x > 8, it means that x>2 and x<8. Let p: x>2 and q: x<8. If x=1, then p becomes false and q becomes true. The whole statement becomes false because all conditions are not met.

Finally, AND can be used to show multiplication while OR can be used to represent addition mostly in the calculations of probability. For example;

P (A and B) = P (A). P (B) for AND. For the case of OR it is calculated as P (A or B) = P (A) +P (B).

Bibliography

"Software survey section." Computers & Mathematics with Applications 20, no. 1 (1990): I-II. doi:10.1016/0898-1221(90)90073-s.

Subject: Maths

Pages: 1 Words: 300

The Pick’s Theorem

Student’s Name

Institution

Date

Introduction

Pick’s theorem is described as gems of elementary mathematics. It has the most innocent sounding hypotheses that entail a very surprising conclusion. Pick’s theorem is named after its founder George Alexander Pick. George Alexander Pick was born in 1859 in Vienna, Austria and spent a better part of his life in the academic field teaching at different universities. Pick discovered the first mathematics paper in 1875. And in 1899, he discovered theorem but it was not published until 1969 several years later. Pick has published over 70 different mathematical issues related o calculus, algebra, and geometry and therefore, he was regarded as the father of mathematical formula. The formulas he established were used and currently being used to solve various mathematical problems including calculating the area of a polygon. A lattice polygon is referred to as a simple polygon, which is fixed in or embedded on a grid, whose vertices are integer coordinates and are referred to as lattice point. However, if the lattice point of the polygon is P, then the formula, which is involved, is simply obtained by adding the points located at the boundaries of the polygon which is b, dividing by b by 2 and then adding the number of lattice points available at the interior polygon which is i and then subtracting 1 from i to get the area of the polygon.

Pick’s Theorem provides the simplest and the best method for calculating the area of a polygon. According to Schultz (2015), it is the best method, which is used to determine the area of a polygon, whose vertices are placed on a lattice. It provides a correct spaced array of point and therefore, it an important aspect in mathematics. Though polygon area can be tabulated using other methods such as using surrounding rectangular and partitioning into smaller pieces, pick’s theorem offers the best and alternative way of calculating the area of a polygon. Pick's theorem is, therefore; provide the simple and best alternative formula for calculating the area of a polygon.

Figure 1: Polygon

The pick’s theorem illustrates that the area of a simple lattice polygon, which is S is always given as A (S) =i-1/2b-1= u-1/2b-1. In this case, i, b, and u are the number of interior lattice points. The number of boundary lattice points and the total number of lattice points of S CITATION Dal \l 1033 (Varberg, 2014). However, to understand the formula of polygon it is important to look into the boundary and interior points of the polygon. The boundary point, which is B, is the lattice point of the polygon and it includes all the vertices. The interior point (i) is a point, which exists in the interior region of a polygon. As stated by Varberg (2014) pick’s theorem applies the description to highlights the area of a polygon whose vertices are lattices points. The formula of the Pick's theorem is, therefore, defined as

914400topPick's theorem is, therefore, provides the best description of a polygon and it is the true construction of polygon. The polygon is, therefore, used to construct from n and 1 triangle. Pick theorem is, therefore, an important factor for tabulation of polygon area. According to Pick's Theorem, the calculation of the area of a polygon is done simply by counting the points allocated on the interior and on the shape of the boundary CITATION Jac14 \l 1033 (Kowalski, 2014). Without the use of Pick’s theorem, it might be difficult to tabulate the area of lattice polygon. Larsson and Lofberg (2014) pointed out that the decomposition of lattice polygon to form triangles has provided the best way to solve the problem of a polygon without pick's theorem. This could be done by tabulating the area of each triangle using the sine rule and therefore, it assumes that the area of each triangle gets the lattice polygon. It is also pointed out that the Shoelace theorem operates side by side and it is used to find the area of any figure when the coordinates are given. Larsson and Lofberg (20140 argued that pick's theorem provides the best and simple way to solve a mathematical problem which includes calculating the area of a polygon. Shoelace theorem could not provide the simple method though it is being used for the calculation. According to Larsson and Lofberg (2014), pick's theorem provides the best way of calculating the polygon a simple way compared to Shoelace and other methods used in the calculation the area of a polygon. The Pick's Theorem is, therefore, entails the use of the interesting corollaries as illustrated below:

Proof

Pick’s Theorem illustrates that, let P to be the lattice polygon and the B (P) to be referred as the number of lattice point, which is allocated at the edge of P and I(P), which is to be the number of lattice points, which lies on the interior of the P of the polygon. According to Kowalski (2014), the area of P denoted to A (P) is equal to B (P)/2 + I (P) – 1. Studies have shown that this kind of formula allows the simplest and the best calculation by simply tabulating the area of polygon or lattice polygon with vertices lies on the integer coordinates, with just one simple question. The formula used in the theorem is simple to draw a conclusion on the concept and therefore, it is easy and simpler to use in solving polygon problems. The problem of polygon question can be solved by using only one simple question. It means that the area of a polygon is always half of the integers and therefore, it is easy to prove the lattice using rectangular. The rectangular then provide a clear illustration or the formula for calculating the area of a polygon using rectangular.

Figure 2: Rectangular with holes

The polygon above is an example of a rectangular. In the above polygon i = 7 and b = 8. According to Pick’s Theorem the area = 7+ frac (8) (2) – 1 = 10 and thus provide the correct area of polygon above. The area is calculated by the number of rectangular allocated on the polygon. The Pick’s Theorem is used even to calculate very complex a simple polygons. Some of the polygons are not intersecting themselves and do not have any holes and it can include in the holes for easy calculation. The Pick’s Theorem is used to calculate simple questions of an area of a polygon with holes.

Polygons with Holes of Pick’s Theorem

The Pick’s Theorem indicated above could be solved by focusing on the green color. The theorem would then be applied on the green shape without holes and then subtracting the established area of the hole. The Area of the green shape which does not have holes applying Pick’s Theorem and the area of the rectangular of the hole using Pick's Theorem and therefore, the total area of the shape is established to be CITATION Ram14 \l 1033 (Raman & Ohman, 2014). The approach to solving the problem can be expanded to a polygon, which has any number of holes. However, if it happens to interested in proofing the existence of Pick’s Theorem is could be too long but very hard to comprehend. And therefore, proofing the formula of Pick’s Theorem is quite challenging and it is time-consuming. It means that the Pick's Theorem formula is the best way to solve the area of a polygon and this could help to formulate the formula. It is, therefore, clear that counting holes or dots within the polygon are one of the simplest ways to establish the area of a polygon.

The Equilateral Triangles

It has also been established that Pick's Theorem can apply to indicate that it is difficult to draw an equilateral triangle to a lattice to allow each vertex to exist on a grid point. However, imagine that there is an equilateral triangle, which has a base of b, and height of h. In order to tabulate the height in terms of a base is, therefore, necessary to use Pythagoras Theorem.

1233170184785

Diagram 1: Triangle used in the application of Pick’s Theorem

For instance, we must first calculate the area using the formula to get the accurate Area of the triangular. The area of an equilateral triangle is, therefore, could be obtained using Pick’s Theorem. It is, therefore; prove the importance of theorem is the calculation of the area of various objects CITATION Con15 \l 1033 (Conover, Marlow, Neff, & Spung, 2015). The formula being used in the calculation of triangle should, therefore, be full because a complete drawing is needed for equilateral triangles. The Pick's Theorem, therefore, holds true for any triangle, which shares its enclosing rectangular.

Conclusion

Based on several studies, it is evidence that Pick’s Theorem is an important mathematical theory, which has been used to calculate the area of several objects. It was discovered by George Alexandria Pick and since 1899 has been used to help to address several mathematical problems. The Pick's Theorem is, therefore, used for the calculation area of the polygon through small rectangular, and small colored dots or small hole to give an accurate answer. It is also important to note that the formulation or generating the Theorem formula is hard and its application is simple by using derived formulas.

References

BIBLIOGRAPHY Conover, B., Marlow, C., Neff, J., & Spung, A. (2015). Pick's Theorem.

https://sites.math.washington.edu/~julia/teaching/445_Spring2013/Project_Pick.pdf, 2-34.

Kowalski, J. M. (2014). Recurrent Theme of Pick’s Theorem.

https://arxiv.org/ftp/arxiv/papers/1707/1707.04808.pdf, 2-31.

Larsson, E., & Löfberg, H. (2014). A Proof of Pick’s Theorem. Mathematical Communication,

2-15.

Raman, M., & Ohman, L.-D. (2014). Two Beautiful Proofs Of Pick’s Theorem.

https://pdfs.semanticscholar.org/12b2/234857bc83581fe972820a4d6955b9feb322.pdf, 2-34.

Schultz, K. (2015). An Investigation of Pick's Theorem.

http://jwilson.coe.uga.edu/emat6680fa05/schultz/6690/pick/pick_main.htm, 2-34.

Varberg, D. (2014). Pick Theorem Revisited. https://faculty.math.illinois.edu/~reznick/496-2-6-

17.pdf, 2-34.

Subject: Maths

Pages: 5 Words: 1500

Math Research

[Author’s name]

[Institutional Affiliation(s)]

Author Note

Math Research

Part 1: Inductive and Deductive Reasoning

Do an internet search on Carl Friedrich Gauss. Write a summary of who Gauss was and how some his accomplishments made math what it is today. Add a reference identifying the website you used.

Answer

Carl Friedrich Gauss (1777-1855) is considered to be the greatest German mathematician of the nineteenth century. He contributed immensely to the fields of astronomy, geodesy, physics, number theory, and electromagnetism. His initial discoveries include the discovery that a 17-sided polygon could be drawn using a compass and a straight edge, and the fundamental theorem of Algebra which states that a polynomial will always have at least one root solution. He also helped develop methods by which the orbiting of asteroids could be determined with great precision ADDIN ZOTERO_ITEM CSL_CITATION {"citationID":"Q9vsFIrk","properties":{"formattedCitation":"({\\i{}Gauss}, n.d.)","plainCitation":"(Gauss, n.d.)","noteIndex":0},"citationItems":[{"id":236,"uris":["http://zotero.org/users/local/DTmO0ro3/items/GY5UYJL3"],"uri":["http://zotero.org/users/local/DTmO0ro3/items/GY5UYJL3"],"itemData":{"id":236,"type":"webpage","title":"Gauss","URL":"http://www.math.wichita.edu/history/men/gauss.html","accessed":{"date-parts":[["2020",1,16]]}}}],"schema":"https://github.com/citation-style-language/schema/raw/master/csl-citation.json"} (Gauss, n.d.). He devoted his life to mathematics and is thus considered, along with Newton and Archimedes, as one of the greatest mathematicians to have ever lived.

Part 2: Investigating Number Patterns

Using your own words, explain how to complete the following successive differences problem: 2, 57, 220, 575, 1230, and 2317. Identify the next number in the sequence and describe how you calculated this value.

Answer

The next number in the sequence is 3992. In the beginning, I assumed the unknown number to be ‘x’. Then I took differences between consecutive numbers in the sequence and noted them down. Then I tried to see if there was a pattern, I could not spot any. Then again, I calculated the difference between consecutive values in the emerging series. After repeating this process a couple of times, a series of differences developed amongst which the difference was 24 between consecutive values. From that, I found the value of ‘x’ which came out to be 3992.

Part 3: Problem Solving Strategies

Write about how Polya's four-step process for problem solving can be used in everyday life. Provide specific examples of how you or somebody else might use this process.

Answer

The four steps in Polya’s four step process include understanding the problem, devising a plan, carrying out the plan, and then interpreting the results. Let’s take the real-life example of a broken door handle. The first step will be to understand the problem i.e. if only the handle is broken or the wood of the door is also damaged. The second step will be to devise a plan for the replacement of the door handle i.e. you might plan to call the carpenter for this. The third step will be to call the carpenter and make him change the door handle. The last step will be to see if the handle works fine after the plan has been carried out.

Part 1: Venn Diagram and Subsets

Do an internet search on John Venn. Write a summary of who Venn was and discuss how his invention of Venn diagrams and his work on logic affect mathematics. If you are looking for content/length, discuss real life examples where a Venn diagram could be used. Add a reference identifying the website you used.

Answer

John Venn was an English logician, philosopher, and mathematician who is famously known for his work in the introduction of the Venn diagrams. A Venn diagram is often used in the representation of mathematical and logical concepts, particularly, relationships between variables in various fields of science. In the early part of his life, he had a passion for building machines due to which he invented the world’s first bowling machine. It was so accurate that when the Australians came to visit the place, one of their top batsmen got out four times consecutively. His main field of study, however was logic ADDIN ZOTERO_ITEM CSL_CITATION {"citationID":"UeCDY09V","properties":{"formattedCitation":"({\\i{}John Venn\\uc0\\u8212{}Oxford Reference}, n.d.)","plainCitation":"(John Venn—Oxford Reference, n.d.)","noteIndex":0},"citationItems":[{"id":234,"uris":["http://zotero.org/users/local/DTmO0ro3/items/AK98IHBH"],"uri":["http://zotero.org/users/local/DTmO0ro3/items/AK98IHBH"],"itemData":{"id":234,"type":"webpage","abstract":"(1834–1923)British logician who, in his work Symbolic Logic of 1881, introduced what are now called Venn diagrams.","language":"en","note":"DOI: 10.1093/oi/authority.20110803115435590","title":"John Venn - Oxford Reference","URL":"https://www.oxfordreference.com/view/10.1093/oi/authority.20110803115435590","accessed":{"date-parts":[["2020",1,16]]}}}],"schema":"https://github.com/citation-style-language/schema/raw/master/csl-citation.json"} (John Venn—Oxford Reference, n.d.). He published three books on several topics related to logic and is considered as an authority in the field. He died in 1923.

Part 2: Operations with Sets

Do an internet search on Augustus DeMorgan. Write a summary of who DeMorgan was and discuss what his contribution was to logic and mathematics. If you are looking for content/length, discuss examples of where we use DeMorgan’s logic today. Add a reference identifying the website you used.

Answer

Augustus DeMorgan was a British Mathematician who is very well known for his work on mathematical induction. The process of mathematical induction was being used at that time repeatedly to solve problems but was not clearly defined. His greatest contribution, however, is in the field of mathematical logic where he introduced DeMorgan’s law. DeMorgan’s laws revolve around set theory and the inter-relationship between them ADDIN ZOTERO_ITEM CSL_CITATION {"citationID":"SqQ8t5El","properties":{"formattedCitation":"({\\i{}Augustus De Morgan}, n.d.)","plainCitation":"(Augustus De Morgan, n.d.)","noteIndex":0},"citationItems":[{"id":232,"uris":["http://zotero.org/users/local/DTmO0ro3/items/5PMLCZJN"],"uri":["http://zotero.org/users/local/DTmO0ro3/items/5PMLCZJN"],"itemData":{"id":232,"type":"webpage","title":"Augustus De Morgan","URL":"https://www2.stetson.edu/~efriedma/periodictable/html/Mg.html","accessed":{"date-parts":[["2020",1,16]]}}}],"schema":"https://github.com/citation-style-language/schema/raw/master/csl-citation.json"} (Augustus De Morgan, n.d.). They are, to this day, being used in several fields of science.

Part 3: Cardinal number of infinite sets

Do an internet search on Georg Cantor. Include the website address in your journal entry. Summarize some of his contributions to set theory and mathematics and discuss where we might see his accomplishments in today’s world.

Answer

Georg Cantor was a German Mathematician born in Russia, and is most commonly considered as the father of modern set theory. His discovery that the linear continuum is not countable meant that its points can not be counted using natural numbers. This meant that even though real numbers and natural numbers are both infinite, the number of real numbers will always remain greater than the number of natural numbers. This gave rise to the concepts of different sizes of infinity and set theories.

Part 1: Logical Fallacies

Do an internet search on Logical Fallacies. Identify at least one logical fallacy that we didn’t study in this Unit. Define the fallacy (using letters and logic symbols) and give at least one example of this fallacy. Add a reference identifying the website you used.

Answer

One fallacy that we did not study in the unit goes by the name of false cause. If two events that happen to occur at the same time are given a causal connection without any evidence, then such a fallacy is known as a causal fallacy or false cause fallacy. One classical example of this fallacy is the statement that my going to sleep causes the sun to set ADDIN ZOTERO_ITEM CSL_CITATION {"citationID":"xFuiTXzJ","properties":{"formattedCitation":"({\\i{}Fallacies (Stanford Encyclopedia of Philosophy)}, n.d.)","plainCitation":"(Fallacies (Stanford Encyclopedia of Philosophy), n.d.)","noteIndex":0},"citationItems":[{"id":240,"uris":["http://zotero.org/users/local/DTmO0ro3/items/EZCHHS82"],"uri":["http://zotero.org/users/local/DTmO0ro3/items/EZCHHS82"],"itemData":{"id":240,"type":"webpage","title":"Fallacies (Stanford Encyclopedia of Philosophy)","URL":"https://plato.stanford.edu/entries/fallacies/","accessed":{"date-parts":[["2020",1,16]]}}}],"schema":"https://github.com/citation-style-language/schema/raw/master/csl-citation.json"} (Fallacies (Stanford Encyclopedia of Philosophy), n.d.). It may be the case that both of these events have the same timeline but that does not necessitate a causal connection.

Part 2: History of Logic

Do an internet search on Aristotle. Write a summary of who Aristotle was and some his accomplishments that lead people to believe that he is the father of modern logic. Add a reference identifying the website you used.

Answer

Aristotle is counted in the list of the greatest philosophers who ever walked the Earth. His works in the field of logic form the basis of formal logic. His main works include his theory of syllogism which had an unparalleled influence on the history of Western thought. His works were inherited by the Arabic and the Latin medieval traditions and were thus made the norm of civilization ADDIN ZOTERO_ITEM CSL_CITATION {"citationID":"f00JtU3l","properties":{"formattedCitation":"({\\i{}Aristotle (Stanford Encyclopedia of Philosophy)}, n.d.)","plainCitation":"(Aristotle (Stanford Encyclopedia of Philosophy), n.d.)","noteIndex":0},"citationItems":[{"id":242,"uris":["http://zotero.org/users/local/DTmO0ro3/items/YBXEDJMC"],"uri":["http://zotero.org/users/local/DTmO0ro3/items/YBXEDJMC"],"itemData":{"id":242,"type":"webpage","title":"Aristotle (Stanford Encyclopedia of Philosophy)","URL":"https://plato.stanford.edu/entries/aristotle/","accessed":{"date-parts":[["2020",1,16]]}}}],"schema":"https://github.com/citation-style-language/schema/raw/master/csl-citation.json"} (Aristotle (Stanford Encyclopedia of Philosophy), n.d.). It was also said by philosophers that anyone after Aristotle who ever said anything new was either confused, perverse or stupid.

Part 3: Logic in advertising

Do an internet search for “Logic Fallacies in Advertising”. Study one example and share what you learned. Reference the website (URL) that you used/reviewed. As you write-up what you learned, be sure to use some of the terminology that was taught in this unit and/or discuss how you will be a better independent thinker as a result of what you’ve learned in this unit.

Answer

The first and the most used logical fallacy in advertisement is known as the gross generalization fallacy. It means to suggest that the small sample of observations can be shown to suggest that the same conclusion will be applicable to the larger sample. Another one that is popularly used to lure customers is known as the appeal to authority. It suggests that since a particular actor is advertising a product, therefore, I will buy it ADDIN ZOTERO_ITEM CSL_CITATION {"citationID":"Es3jWGM9","properties":{"formattedCitation":"({\\i{}10 Logical Fallacies in Advertisements \\uc0\\u8211{} Homan Kaur}, n.d.)","plainCitation":"(10 Logical Fallacies in Advertisements – Homan Kaur, n.d.)","noteIndex":0},"citationItems":[{"id":238,"uris":["http://zotero.org/users/local/DTmO0ro3/items/6GR59HWY"],"uri":["http://zotero.org/users/local/DTmO0ro3/items/6GR59HWY"],"itemData":{"id":238,"type":"post-weblog","language":"en-US","title":"10 Logical Fallacies in Advertisements – Homan Kaur","URL":"https://make.sailacademy.ca/hkaur/2017/10/27/10-logical-fallacies-in-advertisements/","accessed":{"date-parts":[["2020",1,16]]}}}],"schema":"https://github.com/citation-style-language/schema/raw/master/csl-citation.json"} (10 Logical Fallacies in Advertisements – Homan Kaur, n.d.). A number of other fallacies are also used in advertising as the main purpose is persuasion due to which logic takes the back seat.

Works Cited:

ADDIN ZOTERO_BIBL {"uncited":[],"omitted":[],"custom":[]} CSL_BIBLIOGRAPHY 10 Logical Fallacies in Advertisements – Homan Kaur. (n.d.). Retrieved January 16, 2020, from https://make.sailacademy.ca/hkaur/2017/10/27/10-logical-fallacies-in-advertisements/

Aristotle (Stanford Encyclopedia of Philosophy). (n.d.). Retrieved January 16, 2020, from https://plato.stanford.edu/entries/aristotle/

Augustus De Morgan. (n.d.). Retrieved January 16, 2020, from https://www2.stetson.edu/~efriedma/periodictable/html/Mg.html

Fallacies (Stanford Encyclopedia of Philosophy). (n.d.). Retrieved January 16, 2020, from https://plato.stanford.edu/entries/fallacies/

Gauss. (n.d.). Retrieved January 16, 2020, from http://www.math.wichita.edu/history/men/gauss.html

John Venn—Oxford Reference. (n.d.). https://doi.org/10.1093/oi/authority.20110803115435590

Subject: Maths

Pages: 2 Words: 600

Mathematics

Student’s Name

Institution

Other numbers system

There are several number systems used in mathematics. The first number system is decimal numbers which is base 10 (ten), binary number system (base 2), the octal number system and the base 16, which is the hexadecimal number system. There are differences among the number system and the differences are based on the numbers and the figures added to the number. The decimal system is regarded as Hindu – Arabic system and it employs 10 (ten) as the base and it requires ten different numbers from 1 to 9. It has tenth as its base that the Roman or Egyptian number system does not have CITATION Phi13 \l 1033 (Herd, 2013). Therefore, the difference between Roman and decimal number system is based on the number on the base. The decimal number is easily readable, easy to manipulate. However, its major drawback is that it wastes a lot of space and time.

Multiplying Numbers

8286752909570A multiply number is a number which is multipliable with another number. The result of the multiplication number is called a product. There are two methods of multiplication, which are, Lattice Method and Russian Peasant method CITATION San14 \l 1033 (Pariyani, 2014). Lattice method calculatesthe product by multiplying the top digit of the column. It can be taught quickly and it does not require any memorization. The lattice method takes less space and it is easy to typeset. It can also be performed using token. However, its drawbacks are long division, addition and multiplication. Lattice method is used sometimes for multiplication.An example of multiply numbers is 23X47 = 1081 CITATION Aja15 \l 1033 (Olajide, 2015). And lattice method is illustrated as indicated on the box below:

Alternate Number-Base Applications

There are several significant aspects of using hexadecimal with computers. Firstly, it can represent 16-digit words and only 8 (eight) bit bytes. However, by using numeration by many symbols, it becomes much simpler to work with, by saving both space and paper CITATION Jod15 \l 1033 (Fasteen, 2015). It becomes probable to understand some of its vast streams of data inside a computer. The system can be used in multiplication of two different numbers. It is also used in the set of digits and in the case where the base part iszero and when there is an absolute number base.

Fasteen, J. I. (2015). An Investigation of the Role of Alternate Numeration Systems in Preservice

Teacher Mathematics Content Courses. Dissertations and Theses , 2-35.

Modulo Arithmetic and Number Base Conversions

The relationship between modulo arithmetic and number base conversion is based on the non-negativenumber CITATION MAk12 \l 1033 (Akhtaruzzaman, 2012). It is expressed based on value of ‘b’ and the unit value is the same to n, mode b,. Modulo arithmetic are performed in a positive integer and number base conversions is done in a negative integer.

Using Abstract Thinking

The abstract thinking is the idea of thinking about thinks which are removed from facts. It is referred to as the concept of being thought about CITATION Jod15 \l 1033 (Fasteen, 2015). The three-dimension object and two-dimension can help in using the abstract of thinking. The idea is coiled from the aspect of freedom and is mostly from the psychological point of view. It is therefore, a non-mathematical aspect of addressing a specific problem.

Amalie Noether and Abstract

Amalie Noether and Abstract is a German mathematician who made a great contribution in the discovery of abstract algebra and several concepts of theoretical physics. He is the genius who discovered a new approach to physics CITATION Aja15 \l 1033 (Olajide, 2015). Therefore, Amalie Noether is one of the brains behind the formation of theoretical physics and algebra in mathematics.

Herd, P. (2013). Imaginary Number Bases. https://arxiv.org/pdf/1701.04506.pdf , 2-35.

Olajide, A. O. (2015). Computer number System . International Journal of Mathematics , 2-35.

Golden Ratio

The golden ratio is when two quantities’ ratio is the same as the sum of the ratio. It is calculated by dividing the line into two parts CITATION MAk12 \l 1033 (Akhtaruzzaman, 2012). It is used to analyze proportion and other related concepts.

Akhtaruzzaman, M. (2012). Golden Ratio, Golden Section, Golden Mean, Golden Spiral . Internal Journal of Mathematics , 2-34.

Pariyani, S. (2014). Why do we need various number base conversions (number system conversions), e.g octal to hexadecimal? Journal of Mathematics and Business , 2-35

Subject: Maths

Pages: 2 Words: 600

Nutritional Values in Breakfast

Student’s Name

Institution

Date

Introduction

The purpose of the study is to determine the range at which male and female takes breakfast daily. Breakfast is an important component of the meal. It is noted that variety of meals are taken during the breakfast to ensure that it is well balanced. Student takes meals rangers from toast, milk, fruits, cooked breakfast, juice, yoghurt and cereals. The study investigates the rate at which male and female students take cooked breakfast. The comparison is made on the way two sexes take place to determine the rate at which each sex takes calories.

The calculation of all relevant statistics (on excel as much as possible)

Male

Female

Mean

0.37563

Mean

0.42684

Standard Error

0.006990343

Standard Error

0.01103996

Median

0.375

Median

0.4375

Mode

0.33

Mode

0.44

Standard Deviation

0.098858373

Standard Deviation

0.156128617

Sample Variance

0.009772978

Sample Variance

0.024376145

Kurtosis

2.827550492

Kurtosis

2.884006361

Skewness

-0.266950352

Skewness

-0.533881685

Range

0.662

Range

0.922

Minimum

0

Minimum

0

Maximum

0.662

Maximum

0.922

Sum

75.126

Sum

85.368

Count

200

Count

200

measures of the centre – mean, median, mode

Mean for female is 0.4268, Standard Deviation is 01561 and the mode is 0.44. The mean for male is 0.3756, mode is 0.33 and standard deviation (SD) is 00.9886.

measure of the spread – range, IQR, standard deviation

Range of male is 0.66 and Female is 0.992.

Standard deviation of male is 0.098858373 and standard deviation for female is 0.156128617

The analysis of the data established that the mean of the cooked breakfast by female is higher than male. This means that female takes breakfast more often than male.

Graph 1: Comparison of breakfast male and female

Graph 2: Female students cooked breakfast trend

Frequency Distribution Table

Bin

Frequency

0

16

0.065857

0

0.131714

0

0.197571

3

0.263429

10

0.329286

44

0.395143

129

0.461

96

0.526857

50

0.592714

27

0.658571

13

0.724429

6

0.790286

2

0.856143

0

More

4

Conclusion

The study established that female students take breakfast more often compared to male students. It can be concluded that female students takes nutritious breakfast compared to male.

Subject: Maths

Pages: 3 Words: 900

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