Qualification - BTEC Higher National Diploma (HND) in Computing

Unit Name - Maths for Computing

Unit Number - Unit 11

Unit Level - Level 4

Assignment Title - Maths for Computing

Learning Outcome 1: Use applied number theory in practical computing scenarios

Learning Outcome 2: Analyse events using probability theory and probability distributions

Learning Outcome 3: Determine solutions of graphical examples using geometry and vector methods

Learning Outcome 4: Reflect on the application of research methodologies and concepts

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Part 1

Number theory

Greatest Common Divisor (GCD) or GCF is the basis of largest common factor( LCF) of 2 or even more numbers.
Demonstration of this
120
2 x 60=120
10x6x2=120
5x2x3x2x2=120
30
15x2=30
5X3x2=30

1. The 5th term of an AP is 17/6 and the 9th term is 25/6. What is the 12th term?

Solution:
Offered the fifth term that is A.P = a_{5} = \frac{17} and {6} And also the 9th term of and A.P = a_{9} = \frac{25} and {6}
Using formula base of nth term,\mathrm in {a}_{\mathrm{n}}=\mathrm in {a} +(\mathrm in {n}-1) \mathrm in {d}
where the "a" = first term in the A.P"d" i that is common difference in between terms with Use of above formula ......... 5th term,
a_{5} is =a+(5-1) d=a+4 d
Offered that 5th that term is the equal to \frac{17}{6}
a + 4d is the = \frac{17}{6}=> based on 6a + 24d = 17 and is (1)
Similarly in the 9th term,\begin{array} based {l}{a_{9}= on the a+(9-1) d=a+8 d} \\\\ /// that is {=>a+8 d=\frac{25}{6}}\ in the end{array}
base => 6a + 48d = 25 and is (2)
On equation subtraction (2) from the base (1) equal to
(6a +48d) - (6a + 24d) that is = 25 -17 providing => 24d = 8 that is => d = \frac{2}{6}Using equation (1),
6a + 24 x \fraction {2} and {6} get = 17
thats is =>6a = 17 - 8 in the => a = of \frac{9}{6}
For term 12th , we get,
\mathrmin {a}_{12}=//\mathrm in{a}+(12-1) //\times/// \mathrm in {d}=//\frac{9}{6}+11 //\mathrm in {x} //\frac{2}{6} = \fraction {31}{6}
Therefore 12th term in of A.P based on \frac{31}{6} Thus the answer is
31/6

B. An Arithmetic Progression has 23 terms, the sum of the middle three terms of this arithmetic progression is 720, and the sum of the last three terms of this Arithmetic Progression is 1320. What is the 18th term of this Arithmetic Progression?

solution
Let terms be the a - d, a & a + d.
According to offered condition » a = 8 that is (1)
(a - d) and (8) and (a + d) =1320» 64 - d² is the = 60 is » d² = 4 and also » d = ± 2
Therefore, the terms base are the 6 8 and 10 or the other way 10, 8 and also 6.

C. We have three numbers in an arithmetic progression, and another three numbers in a geometric progression. Adding the corresponding terms of the two series, we get 120,116,130. If the sum of all the terms in the geometric progression is 342, what is the largest term in the geometric progression?

Solution

If 3 terms of GP then general terms that is a/r , a , ar
a= 1st term
r= is the common ratio
So,Condition on based on triple terms thaqt is (a) tripled then numbers in the make A
First then in triple of the middle term and also write into the new series
of the a/r , 3a , ar then is it form AP So,
If modern series is in AP and also follow
a+c=2b
b=2nd and term,c=3rd and term, that is a=1st term
then it is
2*3a=a/r+ar
thats the
All (a) cancel and also left
then 6=1/r+r
and After the LCM and also Multiply that is
r²+1-6r=0 that is
-b±√b²-4ac/2a
then r is offed above that is 2nd line

D.There is a set of four numbers p, q, r and s respectively in such a manner that first three are in G.P. and the last three are in A.P. with a difference of 6. If the first and the fourth numbers are the same, find the value of p

Solution

Let use the 1st three numbers that is a /r, a, ar.
This implies that 4th number is (ar + 6).
Out of the above tfirst and on the 4th numbers samimilar a/r = ar + 6 that is (ii).
It is also that is ar = a + 6 based ⇒ ar - a = 6 ⇒ and a(r - 1) = 6 or even the a = 6/(r - 1) that is (iii).
Put value on and from (iii) in the (ii), get theis on quadratic equation that is 2r2- r - 1 = 0. Solving ,the thus value the r as in the - 1/2 or in the 1/2.
Putting r as - 1/2, aquire value of is - 4 and also putting r as in the 1/2 get on value which as - 12.
The 1st term GP becomes gthe 8 and also - 24 (respectively).
Similarly then , the term in third of GP is to becomes 2 also and - 6 in respectively.
The two series GP are the 8, -4, 2 and also the -24, - 12, - 6.
The terms lastly in 2 series are 8 and also in 0 thats is respectively.
Thus possible only numbers are 8, - 4, 2, 8,

e)The sum of three numbers in a GP is 26 and their product is 216. Find the numbers

Solution :
Let numbers be a/r, a, ar.
=> (a / r) + a + a r = that is 26
=> a (1 + r + r2) / r = that is 26
Also, it offers the product = 216
=> that is (a / r) x (a) x (a r) = 216
=> that is a3 = is 216
=> that is a = is 6
=> that is 6 (1 + r + r2) / r = is 26
=> that is 6 (1 + r + r2) / r =is 26
=> that is n (1 + r + r2) / r = is 26 / 6 = is 13 / 3
=> that is 3 + 3 r + 3 r2 = 13 r
=> thats is 3 r2 - 10 r + 3 = is 0
=> that is (r - 3) (r - (1 / 3) ) = is 0
=> that is r = 3 or the r = 1 / 3

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Part 2 - Probability distribution and probability theory

2.1 a. What is the probability of getting a sum of 7 when two dice are thrown?

Solution :

123456 of 1st dice thaat is 123456 for 2nd in dice
then 6+1 =7 and 5+2=7 also the 4+3=7
1/4 or the 3/12

b ) A coin is thrown 3 times. what is the probability that at least two head is obtained?
3\6

c) From a pack of cards, two cards are drawn at random. Find the probability that each card isnumbered and from same suite

From a card pack of, 3 cards are random drawn . The probability on every card that is different suit.
1st card is from the suit
then the probability is (52/52) = 1
Now the remaining cards is based 51
2nd card that is should various different suits in the 39 Cards
then probability is (39/51) = is13/17
Now then remaining card is = 50
3rd card must be in the remaining that is two suits = is the 26 cards
therefore probability is (26/50) = thst is 13/25
tprobability in each card that is from varius d suit. = that is 1 * (13/17) * 13/25
= which is 169/425
The card probability is from the suit different in . = 169/425

d. A bag contains 10 white, 5 red and 9 blue balls. Four balls are drawn at random from the bag. What is the probability that?

i) all of them are red.
ii) two is white and the rest two is blue

solution
white balls number in bag is = 5
red balls number in bag is = 7
black balls number inbag is = 4
blue balls number bag= 2
Total ball number in bag is = 5 +7+4+2= 18
Favourable outcomes total number = 18

i) blue number and white balls in the bag is = 5+2= 7=
favourable outcomes number = 7
favourable outcomes number / outcome total number = probability
P(white or blue) that is = 7/18

ii) Number of black and red balls present in bag is the = 7+4= 11
favourable outcomes number = 11
favourable outcomes number / outcome total number = probability
P( black or red ) that is = 11/18

iii) Number of off white balls present bag=
Total ball number in a bag - white number balls that is = 18 -5 = 13
favourable outcomes number = 13

2e. Find the probability distribution of boys and girls in families with 7 children, assuming equal probabilities for boys and girls. Draw the graph of your probability distribution.

Solution
the probabi8tity of girls and boys is 1

2f. The time taken to assemble a car in a certain plant is a random variable having a normal distribution of 20 hours and a standard deviation of 2 hours. What is the probability that a car can be assembled at this plant in a period?

a) less than 19 hours?

b) between 19 and 21 hours?'

Solution

a) 40.13% probability that a computer car assembling assembled at a periosd plant in of time of that is less compared to 19 hours.
This are pvalue in Z when te X = 19.5. thereefopre
hpvalue =0.4013.
40.13% probability that computer car is assembled plant in time period of less 19.5 hours.

b) 34.13% probability that a computer car = to be assembled at period plant inof time that is between 20 hours and also 21 hours.

The pvalue in Z especially when the X = 22 that is subtracted pvalue Z in X = 20. therefore
X = 22 with pvalue = 0.8413
then 20hrs
pvalue of 0.5
0.8413 - 0.5 is equal 0.3413
0.8413 - 0.5 is equal 0.3413
34.13% probability computer car assembled in plant in time period between 20-22 hours.

g. The annual salaries of employees in a large company are approximately normally distributed with a mean of £45,000 and a standard deviation of £25,000.
a) What percent of people earn less than £35,000?
b) What percent of people earn between £36,000 and £38,000?
c) What percent of people earn more than £40,000?

solution
The annual employees salaries in company are normal approximately in the normally distributed
mean of base $50000 and also the standard deviation that is of$20000.

a. What percent of people earn less than $40000? Let S be variable random of a employee salary of (in$), S ~ N(50000,20000). thus random
variable in X =-50000
20000 in~N(0,1).( < 40000) = that is ( <40000 - 50000
0000 ) = that is ( < -0.5) = that is (-0.5) = 0.3085375.
Here Φ(x) it denotesin the cumulative function distribution of a the standard normal in distribution.

b.What percent of people earn between $45000 and$65000?

Solution:
(45000 < < 65000) = (45000 - 50000 that is 20000 < <65000 - 50000 that is 20000 ) = is the (-0.25 < < 0.75)
= (0.75) - (-0.25) = in the 0.7733726 - 0.4012937 = and the 0.3720789.

c) What is the percent of the people who earn more compared or than in \$70000?

Solution that is
( > 70000) = in ( >70000 - 50000 20000 ) = that is ( > 1) = that is 0.8413447.

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Part 3

1 Find the equation of the line passing through (4, 5) and parallel to the line 3x - 2y = 4.

The required equation line is basically 3x - 2y - 17 = 0

2 A triangle's vertices have coordinates (0, 0), (0, 7), and (-9, -8) What is its area?

the area is equal
Let all the angles in the quadrilateral to be in 3x,4x,5x,6x. because the sum of angles interior on samilar side becaomes 180, that is one pair sides that are parallel .

3 Discover the estimation of m for which the lines 5x+3y +2=0 and 3x-my+6= 0 are

If the tens of digit place becomes less compared to 5, then replace every one the tens that is ones digits gthrough 0 and thus keep each other digits samilar . that is
If tens of the digit is in the 5 or even more compared to 5, the increase the digfits in hundreds by 1 also and the replace every of tens thus and digit through 0
6388 based on the nearest hundred that is 6400.

3.3 Discover the estimation of m for which the lines 5x+3y +2=0 and 3x-my+6= 0 are
(i) perpendicular to each other.
(ii) parallel to each other

Solution
5x+3y +2=0
3x-my+6= 0
Multiply then subtract
3x-my+6= 0 ......3x-my+6= 0
substuste
i)X=1.2
ii)Y=2.1

3.4 Below is the figure representing different sides with vector x and z. The mid-point of the line AB is D and BE is parallel of AC and half of AC.

a)Find the vector CD in term of P and Q.

(b) Find the vector DE in term of P and Q

AB=AC that is P = midpoint in the BD=DC→.......(i) in the ΔABD&ΔACD
AB=AC( offered )
BD=DCfrom the equation (i)
b)=∠B=∠C(CPCT)

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Part 4
1 Let the profit, P, (in thousands of pounds) earned from producing x items be found by P(x)=2x2-6x+25. Find the average rate of change in profit when production increases from 4 items to 5 items

solution

Given the t is = 8
x is = 4 + 2*8 that is => x = 20
that is company offering 20000 unit every week
p = 500*20000divide by 100+20000 multiply *2
total profit weekly is
p is 249.377 dollars

2 An airplane is flying in a straight direction and at a constant height of 5000 meters (see figure below). The angle of elevation of the airplane from a fixed point of observation is a. The speed of the airplane is 500 km / hr. What is the rate of change of angle a when it is 25 degrees?

solution
25° respect.
So, ∠PAB = 60°, ∠QAC = 25°
Again given that PB = 3600√2.5 m
In ΔABP, that is
tan 25° = BP/AB
=> √3 = 3600√3/AB
=> AB = 3600 m
in triangle the ACQ
tan 25° = CQ/AC
=> 1/√3 is the = 3600√3/AC
=> AC is the = 3600√3 * √3 that is = 3600 * 3 on = 10800 m
Distance = BC in the = AC - AB and the =10800 m - 3600 m which is = 7200 m
Thus, travel plane 3000 m in 25 sec
Thereforre , speeplane sopeed= 7200/25 m/insec = 240 m/sec
= (240/1000) multiply 60 multiply 60 km/hr
[ because 1 km = is 1000 m, and also 1 hr = is 60 minutes = thus 60 * 60 seconds]
= 864000 dived by 1000 kmin a hr
that is = 864 km/hr

3 Find the maximum and minimum value of the function x3- 3x2- 9x + 12

solution

let the offered g function be equal to , f(x) = x³ - 3x² -9x +12
then, offer the differentiating of the f(x) , becomes
f2(x) = 3x² - 6x -9
again used the differentiating in f2(x) to aquire ,f2(x) in = 6x -6
then use , f2(x) = 0
⇒ that is 3x² - 6x -9 in = 0
⇒ that is x² - 2x -3in = 0
⇒ that is x² + ( -3 +1 )x +(-3)×1 in =0
⇒ that is (x-3) and (x+1) =0
⇒ that is x = 3 , (-1)
then in the , f2(3) and = 6×3 - 6 that is = 12 > 0 ; in the minima of the f(x) lies in x = 3
and also the , f2(-1) = is 6×(-1) -6 = -12 < 0 ; and maxima of the f(x) lies in the x = (-1)
∴ maximum value present is f(x) is in the , f(-1) = that is (-1)³ - 3(-1)² - 9(-1) +12= to 17
∴ minimum value preswent is (x) is ,in the f(3) = that is (3)³ - 3(3)² - 9(3) +12 = to (-15)

4 A square sheet of cardboard with each side a centimetre is to be used to make an open-top box by cutting a small square of cardboard from each of the corners and bending up the sides. What is the side length of the small squares if the box is to have as large a volume as possible?

solution
Let
A = large square length side
a = small square length side length
that is the side length of small square in the ( cutting) that become.
= A - 2a

6.Find the area contained between the two curves x=y2-y-6 and x=2y+4

solution
parabola is
offered line is 2y=x.
write form is the y=mx. then
2y=x
that is y=(1/2)x in the =mx
therefore, m=1/2.
The area that is 64/3 m2.

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 Learning Outcomes and Assessment Criteria Pass Merit Distinction LO1 Use applied number theory in practical computing scenarios D1 Produce a detailed written explanation of the importance of prime numbers within the field of computing. P1Calculate the greatest common divisor and least common multiple of a given pair of numbers. P2 Use relevant theory to sum arithmetic and geometric progressions. M1 Identify multiplicative inverses in modular arithmetic. LO2 Analyse events using probability theory and probability distributions D2 Evaluate probability theory to an example involving hashing and load balancing. P3 Deduce the conditional probability of different events occurring within independent trials. P4 Identify the expectation of an event occurring from a discrete, random variable. M2 Calculate probabilities within both binomially distributed and normally distributed LO3 Determine solutions of graphical examples using geometry and vector methods D3 Construct the scaling of simple shapes that are described by vector coordinates. P5 Identify simple shapes using co- ordinate geometry. P6 Determine shape parameters using appropriate vector methods. M3 Evaluate the coordinate system used in programming a simple output device. LO4 Reflect on the application of research methodologies and concepts D4 Justify, by further differentiation, that a value is a minimum. P7 Determine the rate of change within an algebraic function. P8 Use integral calculus to solve practical problems involving area. M4 Analyse maxima and minima of increasing and decreasing functions using higher order derivatives.

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