Comprehensive Math Assignment: Sets, Binary, Matrices, Statistics

This assignment provides solutions to a range of mathematical problems. It begins with set theory, applying the formula for the union of three sets to determine the intersection of dance, drama, and music groups. The assignment then transitions to binary numbers, identifying the largest 4-bit binary number and converting a fraction to its binary equivalent. Simplification of Boolean expressions is also covered. Further, the assignment includes coordinate geometry, finding the equation of a line and determining the relationship between two given lines. Matrix algebra is addressed through determinant calculation and inverse existence assessment. Financial mathematics appears through compound interest calculation. The assignment also includes trigonometry, calculating pole height using the tangent function and proving trigonometric identities. Polar and rectangular forms of complex numbers are covered, followed by force and friction calculations. The assignment concludes with percentage degradation problems, statistical analysis (mean, variance, standard deviation, confidence intervals), and tree structures.

Q 1. Using formula of set union for three variables is given by,
n (Dance ∪ Drama ∪ Music) = n(Dance) + n(Drama) + n(Music) – n (Dance ∩ Drama) - n
(Drama ∩ Music) - n (Music ∩ Dance) + n (Dance ∩ Drama ∩ Music)
n(Dance) = 37, n(Drama) = 13, n(Music) = 19, n (Dance ∩ Drama ∩ Music) = 5, P (Dance ∪
Drama ∪ Music) = 46
n(Dance ∩ Drama) + n(Drama ∩ Music) + n(Music ∩ Dance) = n(Dance) + n(Drama) +
n(Music) + n(Dance ∩ Drama ∩ Music) - n(Dance ∪ Drama ∪ Music)
 37 + 13 + 19 + 5 – 46
 18 Answer
Q 2.
Q 3. Biggest 4-bit Binary number = 1111, In decimal = 15, and in hexadecimal = F
Q 4. 7/8 in binary is given by 0.111
Q 5. Z = ( A+ B ) (B+ D) can be simplified to A+B.
Q 6. Z= (A+BD)(A+ B) can be simplified to A.
Q 7. The line equation can be represent by 2x –y = 0
Q 8. The line equations x-4y=2 and y+4x=3 are intersecting lines
Q 9. |A| = 1(3*3 – 8*4) – 0(3*3 – 4*2) + 2(3*8 – 3*2)
= 9 – 32 – 0 + 48 - 12 = 13
Q 10. As |A|= 3(6*25 – 2*35) – 7(4*25 – 2*15) + 5(4*35 – 6*15)
 240 – 490 + 250 = 0, therefore no inverse exist for this matrix as determinant is zero.
Q 11. P= $ 100,000, r= 1%, t= 10 years and compounded quarterly so n= 4, therefore we have,
total amount invested is given by Amount = P (1+ (r/n)) n.t. Therefore, A =
100000(1+(0.01/4))4.10= 110503.3
Q 12. Shadow =9m, Inclination angle = 55o, so height of the pole can be determined by tan θ=
P/B. So tan55=x/9 => x= tan55 * 9 => 1.428148 * 9 => 12.853332 meter
Q 13. When hypotenuse =1 is the right triangle, so, by using Pythagoras Theorem, we have
To prove: Sin2
θ + Cos2
θ =1
Proof: we know that h=1, according to Pythagoras Theorem, P2 + B2 = H2 and also sin θ = P/H so
P= Hsinθ and cosθ= B/H so B=H cos θ. So, now as H =1 therefore P2 + B2 = 1
Put values in equation, we get, (H sin θ)2 + (H cos θ)2 = 1,
So from this we get Sin2
θ + Cos2
θ =1. Hence proved.

Also tanθ=P/B = H sinθ /H cosθ => tanθ=¿ sinθ/cosθ. Hence Proved
Q 14. 6ei 3 π polar form & trigonometry form
(r=6) (ϴ=3π)
Z=[Cos ϴ+iSin ϴ]= 6[Cos3 π+iSin3 π]
Rectangular form
Z=(x+iy)
= 6 Cos3 π+6 iSin3 π = 6(-1) +0= (-6+0i)
Q 15. f= μN (Max. value of friction), but according to the question, there is no acceleration so
forces are balanced.
F= 75 Cos 20º= 75*0.4= 30.
Q 16. Laptop charges 1% in 1 cycle. & if it is degrading by 0.5% per recharge, then
Required degrade charge= 50
So required recharge cycles are= (1/0.5)*50= 100 cycles.
Q 17. A. Mean = (13 + 9 + 10 + 7 + 10) / 5 = 49/5= 9.8
Using formula, Variance = 1/N ⅀ (ai - μ)2 we get, Variance = 3.76
Standard Deviation = Square Root (Variance) = 1.9390
B. Using formula: X ± z∗s
√n , we get simplify X =510, n= 25, s = 125 and confidence =
95%, and according to z area table z value for 95% confidence is 1.96, so we get,
510 ± 1.96 * 125/√ 25 = 510 ± 49
Therefore, lower end = 510 – 49 = 461 and upper end= 510 + 49 = 559
Interval = [461, 559]
Q 18. A. Full binary tree of height =k, then internal vertices = k, and leaves = k+1
B. Full ternary tree of height =k, then internal vertices = k, and leaves = 3k – k +1 = 2k+1