Mathematics Assignment: Statistical Analysis, Equations, and Graphs

This mathematics assignment presents solutions to a variety of problems, including statistical analysis of salary data, with calculations of mean, median, range, and standard deviation, alongside interpretations of data distributions and box plots. The assignment also delves into algebraic concepts, featuring the solving of linear and quadratic equations, including those involving fractions and square roots. Furthermore, it covers graphical representations, such as the plotting of linear equations, parabola analysis, and the interpretation of intercepts and gradients. The assignment includes calculations of geometric properties like circumference, area, and volume, as well as applications of trigonometric principles to solve problems related to triangles and angles. Finally, the assignment explores exponential equations and their applications, with a focus on interpreting the results and their implications.

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 =𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑦
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 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=75𝑐−3𝑎
5−𝑐
b= 1
10(75𝑐−3𝑎
5−𝑐 )=75𝑐−3𝑎
50−10𝑐
b= 𝟕𝟓𝒄−𝟑𝒂
𝟓𝟎−𝟏𝟎𝒄
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
𝑠𝑖𝑛33
= 11
𝑠𝑖𝑛𝐹
Sin F=11𝑠𝑖𝑛33
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) 5.5𝑐𝑚
1100𝑐𝑚
= 𝑥
1400𝑐𝑚
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 14000 17000 20000 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=Log1851.85
log1.26 = 3.268
0.1004=32.55
time=32.55 months