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Section 1:
1. Task 1
Stage 1 of the circuit involves three gates NR2, AN2 and OR2. The input of NR2 is A and B`. The output of NR2 would be (A+B`)` which is equivalent to (A`+B). The input of AN2 is A and B` and its output is A.B`. The input of OR2 is A and B` and its output would be A+B`. In the next stage of the circuit, NR2 is present and its input is (A`+B) and A.B`. The output of this gate would be ((A`+B)+(A.B`))`. In the third stage of the system, ND2 is present and the input is x and y where x is ((A`+B)+(A.B`))` and y is (A+B`). The final output of the concerned gate would be (x.y)` which is denoted by Z. Based on the calculations performed using the properties of a boolean expression, the table below provides the view of the truth table for the concerned circuit.
A |
B |
A` |
B` |
Z = (A`+B+A.B`) |
0 |
0 |
1 |
1 |
1 |
0 |
1 |
1 |
0 |
1 |
1 |
0 |
0 |
1 |
1 |
1 |
1 |
0 |
0 |
1 |
Truth Table for the above circuit
(Source: Self-Created)
2. Task 2
The input given to the first AND gate is A` and B and its output would be A`.B. The input given to the second AND gate is A, C` and D and therefore, its output would be A.C`.D. The input given to the third AND gate is A, B` and C which in turn would generate the output as A.B`.C. All these outputs are applied to the OR gate present in the circuit and therefore, the output of the gate is (A`.B) + (A.C`.D) + (A.B`.C). This is the required sum of products that can be generated on the basis of the inputs that have been applied.
3. Task 3
2x_{1} + 6x_{2 }+ 4x_{3} = 18…….1
5x_{1} + 6x_{2} + 4x_{3} = 24……..2
-2x_{1 }+ x_{2} + 3x_{3} = 3……..3
From equation 1, the value of x3 can be calculated as
x3 = (18 - 2x1 - 6x2) /4 ……..4
Putting the value of x3 in equation 2 and 3, we get,
5x1 + 6x2 + 4*(18-2x1-6x2)/4 = 24
5x1 + 6x2 + 18-2x1-6x2 = 24
3x1 = 24-18 = 6
x1 = 6/2 = 3
Putting the value of x1 in equation 1 and 3, we get,
2*3 + 6x2 + 4x3 = 18 [Equation 1]
6x2 + 4x3 = 12………5
-2*3 + x2 +x3 = 3 [Equation 3]
x2 + x3 = 9……..6
By solving the equations 5 and 6, the values of x2 and x3 are -12 and 21 respectively. Therefore, the values of x1, x2 and x3 are
x1 = 3, x2 = -12 and x3 = 21
4. Task 4
The total price of the mobile phone and smartwatch = £ 1,250
Consider the cost of the smartwatch to be x, therefore, the cost of the mobile would be 3x.
x + 3x = 1250…….1
x = 1250/4 = 312.5
Therefore, the cost of the smartwatch is £ 312.5 and for the mobile, the cost is 312.5*3 which is equivalent to £ 937.5.
VAT = (14/100)*937.5 = £ 131.25
5. Task 5
The values of the variables given in the given equation are t1 = 2.6, t2 = 0.45, A = 15.6, a = 12.57/144 , g = 32.2, h1 = 36 and h2 = 25. Putting the values of the above variables in the equation, the value of k can be calculated as follows,
2.6 - 0.45 = (2*15.6)/ 0.0872 (√(1+k)/2*32.2)(√36 - √25)
2.15 = (31.2)/0.0872 (√(1+k)/64.4)(6-5)
2.15 = 357.79(√(1+k)/64.4)
√(1+k)/64.4 = 0.006
(1+k)/64.4 = 0.000036
1+k = 0.0023
k = 0.0023 - 1 = -0.9977
Therefore, the value of k comes out to be -0.9977.
6. Task 6
The equation given in the question is 1.2v + 0.64u = 0.85. The value of u ranges from 5 to 100. The table below provides the view of the values of v for corresponding values of u using the above equation.
u |
v = (0.85 - 0.64u)/1.2 |
5 |
-1.95 |
10 |
-6.04 |
15 |
-7.29 |
20 |
-9.95 |
30 |
-15.29 |
40 |
-20.625 |
50 |
-25.95 |
60 |
-31.29 |
70 |
-36.625 |
80 |
-41.95 |
90 |
-47.29 |
100 |
-52.625 |
Table 2: Table for showing the values of v for corresponding values of u
(Source: Self-Created)
The figure below provides the view of the graph generated in Excel showing the relation between u and v.
Relation between u and v
7. Task 7
The equation given in the question is 2u + 7v = 52. The value of v ranges from 2 to 50. The table below provides the view of the values of u for corresponding values of v using the above equation.
v |
u = (52 - 7v)/2 |
2 |
19 |
4 |
12 |
6 |
5 |
8 |
-2 |
10 |
-9 |
15 |
-26.5 |
20 |
-44 |
30 |
-79 |
35 |
-96.5 |
40 |
-114 |
45 |
-131.5 |
50 |
-149 |
Table 3: Value of u for corresponding values of v
(Source: Self-Created)
Relation between u and v
The figure above provides the view of the relationship existing between u and v as per the equation given in the question.
8. Task 8
As per the details in the questions, A indicates Returning Customer and B indicates Discount Code.
A |
B |
Y |
0 |
0 |
0 |
0 |
1 |
1 |
1 |
0 |
0 |
1 |
1 |
1 |
Table 4: Truth table
(Source: Self-Created)
The above truth table provides the idea that whenever the value of B which stands for discount codes is True, then the output of the system is also True.
Y = A`.B + A.B`
Section 2:
9. Task 9
Monthly rent for 24 students are 1500, 1350, 350, 1200, 850, 900, 1500, 1150, 1500, 900, 1400, 1100, 1250, 600,610, 960, 890, 1325, 900, 800, 2550, 495, 1200, 690.
Figure 3: Calculation of Relative Frequency for students' monthly rent
(Source: Self-Created)
Histogram
The table above provides the view of the relative frequency calculation performed in Excel for the given monthly rent of the students.
10. Task 10
Age |
Frequency |
Age x freq |
Median |
Mean |
15 |
2 |
30 |
10.5 |
19.23377 |
16 |
5 |
80 |
||
17 |
11 |
187 |
||
18 |
9 |
162 |
||
19 |
14 |
266 |
||
20 |
13 |
260 |
||
21 |
10 |
210 |
||
22 |
13 |
286 |
||
77 |
1481 |
Table 5: Calculation Table
(Source: Self-Created)
Section 3:
Task 11:
Algorithm
(Source: Self-Created)
Task 12:
The algorithm for the concerned work is
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