Mathematics Assignment: Data Analysis, Equations, and Graphs

This mathematics assignment presents a comprehensive analysis of various mathematical concepts. The assignment begins with a statistical analysis of salary data, comparing men's and women's salaries using measures of central tendency (mean and median) and dispersion (range, interquartile range, and standard deviation). It also includes an assessment of data skewness and interpretation of box plots. The assignment then delves into solving linear and quadratic equations, including graphical representations and the determination of intercepts. Furthermore, it explores exponential equations and their applications. The document provides detailed solutions, including step-by-step calculations and interpretations, making it a valuable resource for students studying mathematics and statistics. Finally, the assignment covers geometric problems involving circumference, area, and volume calculations, including applications of the Pythagorean theorem, trigonometry, and the cosine and sine rules.

Mathematics Assignment
Student Name:
Instructor Name:
Course Number:
20th May 2019

Question 2
Part a
i)
ii) This is a continuous data.
iii) From the above table, the two measures of location that we do have are median and
mean. Using mean, then men have greater salary on average (31.7 thousands of
pounds) compared to women who have salary of 27.1 thousands of pounds. On the
other hand when median is used we still find that the same men have greater salary on
average (32 thousands of pounds) compared to women’s 26.5 thousands of pounds.
iv) From this table, the three measures of spread that we have are range, inter quartile
range and standard deviation. Considering the range of women’s salary (11 thousands
of pounds) versus men’s salary (10 thousands of pounds) we find that women’s salary
has a greater variability than those of men. The standard deviation for women salary
(3.5 thousands of pounds) also has a greater variability compared to standard
deviation of men’s salary of 3.0 thousands of pounds. Lastly the inter quartile range
for women’s salary (6 thousands of pounds) also shows a greater variability than
men’s 3 thousands of pounds. In conclusion having looked at the three measures of
spread we can say that women’s salary showed greater variability compared to men’s.
Women’s salaries
(thousands of pounds)
Men’s salaries
(thousands of pounds)
Min 22 26
median 26.5 32
max 33 36
Mean 27.1 31.7
SD 3.5 3.0
IQ range 6 3
Range 11 10

Part b
i)
ii) The data is right skewed (positively skewed). This is due to the fact that the mean
(27.1 thousands of pounds) is greater than the median (26.5 thousands of
pounds).Besides this we find that the median (26.5 thousands of pounds) is closer to
the first quartile (24 thousands of pounds) than the third quartile (30 thousands of
pound)) hence making the tail of distribution on the right hand sides to be longer than
the one on left hand side.
iii)
(1) The statement is false. From the box plot it is evident that about twenty five
percent of women are paid less than £ 24000.
(2) The statement is false. The men paid more than 32 thousands of pounds are about
40% while the Men paid less than 32 thousands of pounds are about 45%. Therefore
we can say that there are less men paid more than 32 thousands of pounds as
compared to those paid less than 32 thousands of pounds.

Part c
Histogram A represents the women’s data.
This is because the range from the histogram is 16.
Question 3
a) The graph drawn below will be used for confirming the subsequent questions answers.
3y=-5x+2
x -4 5
y 22
3 - 23
3
i) Gradient = c h ange∈ y
c h ange∈x = 4+1
−2−1 = - 5
3
ii) Equation of the line is
y−4
x+2 = - 5
3
3(y-4) = -5(x+2)
3y-12= -5x-10

3y= -5x+2
iii) 3y=-5x+2
5x+2y=2
x
2
5
+ y
1 =1
X - intercept is 2
5
b) Taking x=2 and substituting in the equation 3y= -5x+2 we have
3y=-5(2) + 2
3y= -10+2=-8
3y=-8
y= −8
3
When x=2, y=−8
3 not y=-2
Hence the point (2,-2) does not lie in the line 3y= - 5x+2
c) -4x + 15y=10
4x=15y-10
x= 3.75y-2.5
Substituting x= 3.75y-2.5 in 3x-7y=1 we have
3(3.75y-2.5)-7y=1
11.25y-7.5-7y=1
4.25y=8.5
y=2

x= 3.75y-2.5=3.75(2)-2.5=5
The two equations intersect at the point (5,2)
d)
i) y=-0.15x2+x+1.2
At y intercept, x=0 hence y=-0.15(0)2+0+1.2=1.2
y intercept =1.2metres
ii)
(1) when x=5, y=-0.15(25)+5+1.2=2.45
The line x=5 meets the parabola at (5,2.45)
(2)The kangaroo will not clear the obstacle since at point x=5 its height
will be 1.45metres above the obstacle.
iii) (1) y=-0.15x2+x+1.2
At x intercept y=0 thus -0.15x2+x+1.2=0
-0.15x2+x+1.2=0, x=−1 ± √1−4 (−0.15 ) (1.2)
2(−0.15)
x=−1± √1.72
−0.3 =−1± 1.3115
−0.3
x=−1+ 1.3115
−0.3 = -1.0 (2 s.f) OR
x=−1−1.3115
−0.3 = 7.7 (2 s.f)
x intercepts are -1.0 and 7.7 respectively.
(2) Horizontal distance from the ledge is 7.7 metres.

Question 4
a)
i) 3x+8=-4x-6
3x+4x=-6-8
7x=-14
x=-2
ii) 11- 1
3(6x+3)= 1
2x
Multiplying each term by 6 we get
66-2(6x+3)=3x
66-12x-6=3x
60=15x
4=x
iii) x
x+5 - 1
x−7 = 1
Multiplying each term by the least common multiple (x+5)(x-7) we have
x(x-7)-1(x+5)=(x+5)(x-7)
x2-7x-x-5=x2-7x+5x-35
-x-5=5x-35
-6x=-30
x=5
b) 2x2+16x-66=0
2x2+22x-6x-66=0
2x(x+11)-6(x+11)=0

( 2x-6)(x+11)=0
2x-6=0
2x=6
x=3 OR
x+11=0
x=-11
c)
i) 1
5a= 10b-c( 2b
3 -5)
Multiplying both sides by 5 we have
-a=50b-5c( 2b
3 -5)
Multiplying out bracket we have
-a=50b - 10
3 bc-25c
Multiplying both sides by 3 we have
-3a=50b-10bc-75c
Collecting like terms
50b-10bc=75c-3a
10b(5-c)=75c-3a
10b= 75 c−3 a
5−c
b= 1
10( 75 c−3 a
5−c )= 75 c−3 a
50−10 c
b= 75 c−3 a
50−10 c

ii) One mistake made by the student was that while multiplying each side by 5, he/she multiplied
each term in bracket of ( 2b
3 −5 ¿ by 5.This was not supposed to be done since 5 had already
multiplied what was outside the bracket i.e. c.
The second mistake was made while doing factorization (assuming that the working was
correct).The student was to factorize 150b-50bc as 50b (3-c) instead of b (150-50c).
Question 5
a)
i) C=Πd= 22
7 ×3.7=11.63mm where C=circumference and d=diameter
Circumference= 12mm (2 s.f)
ii) A=Πr2 where A=cross-sectional area and r=radius
Cross-sectional area of LAD= 22
7 ×1.85×1.85=10.76= 11mm2 (2s.f)
iii) 0.95(3.7mm)=3.515mm
A=Πr2 where A=cross-sectional area and r=radius
Cross-sectional area of blockage= 22
7 ×1.7575×1.7575=9.71= 9.7mm2 (2s.f)
iv)Volume of stent V= Πr2h where r=radius and h=height
V= 22
7 ×1.45×1.45×15.5=102.4mm3
Volume of stent is 100mm3 (2s.f)
b)
i) (AC)2= (AB)2 +(BC)2
(AC)2=632+10362

(AC)2=3969+1073296=1077265
AC= √ 1077265 =1037.91m
ii) Tan C= 63
1036 =0.06081
C=tan−1 0.06081=3.5°
The zip line makes an angle of 3.5 ° with the horizontal.
ci) ED=f=11, DF=e=14 and EF=d
Using cosine rule d2=e2+f2-2ef cos D
d2=142+112-2(11×14)cos 33°
d2=196+121-308cos 33°
d2=317-258.3=58.7
d= √58.7 =7.662
EF= d=7.7m (2s.f)
ii)Using sine rule we have d
sin D = f
sinF
7.7
sin 33 = 11
sinF
Sin F= 11sin 33
7.7 =0.7781
F= sin−1 0.7781=51.09 °
Angle DFE= 51° (2s.f)
iii) Area of triangle A = 1
2efsin 33°=
A = 1
2(14×11) sin 33°=83.87

Area =84 m2 (2s.f)
iv) 5.5 cm
1100 cm = x
1400 cm
x= 1400× 5.5
1100 = 7.0cm
Question 6
i) v=3000(t-6)+5000
When t=9,
v=3000(9-6)+5000
v=3000(3)+5000=14000
v=14000 words
ii) This means from 6 years to 12 years.
iii) v=3000(t-6)+5000
20000=3000(t-6)+5000
20000-5000=3000(t-6)
15000=3000(t-6)
Dividing both sides by 3000 we get
5=t-6
t=5+6=11
t= 11 years
iv) v =3000(t-6)+5000

v=3000t-18000+5000
v=3000t-13000
gradient=3000.
This represents 3000 words per year.
v) When the equation v =3000(t-6)+5000 is simplified we get v=3000t-13000.
From this equation we get that the v-intercept is -13000.It is therefore in order
to conclude that 5000 is not the v-intercept.
vi) v =3000(t-6)+5000
t 6 7 8 9 10 11 12
V=3000(t-6)[phone] 11000 1400
0
17000 2000
0
23000
b) i) v=0.27×(1.26)t

when t=18,v=0.27×(1.26)18=17.30
v=17.30 words
when t=30,v=0.27×(1.26)30=277
v=277 words
ii) scale factor= 1.26
percentage increase ( 1.262−1.26
1.26 )100=¿26%
iii) ) v=0.27×(1.26)t)
500=0.27×(1.26)t
500
0.27 =¿ 1.26t
1851.85 ¿ 1.26t
Log1851.85=t log1.26
t= log 1851.85
log 1.26 = 3.268
0.1004 =¿32.55
time=32.55 months