Home / 2023 May Mathematics analysis and approaches paper 2 TZ2 SL Concept Questions and Answers

Question

Topic: Method of least squares.

Given: A plant growth experiment is carried out for 42 days. The average height of the plants is recorded on seven different days. The data is shown in the table below.

Calculate: (a) The equation of the line of best fit for the data is h = ad + b, where h is the height and d is the number of days. Find the values of a and b.

(b) Find the Pearson’s product-moment correlation coefficient, r, for the data. What does it tell you about the relationship between the height and the number of days?

(c) Use the equation of the line of best fit to predict the average height of the plants on day 20.

Answer/Explanation

Answer:

(a) a =1.93258…, b = 7.21662…
a =1.93, b = 7.22

(b) r = 0.991087…
r = 0.991

(c) attempt to substitute d = 20 into their equation
height 45.8683… =
height 45.9 = (cm)

Topic :Statistics
?
Time
 
Q — 1
16:17
Eng

Question

Topic:Geometry and trigonometry

Given: Two buildings of different heights stand on a level ground. The taller building has a height of h metres. The shorter building has an unknown height. A person stands on the ground at a point that is equidistant from both buildings. The person looks up at the tops of both buildings and measures the same angle of elevation, θ. The person also measures the distances from their position to the tops of both buildings, and the horizontal distance between the tops of both buildings.

The measurements are as follows:

  • Distance from person to top of taller building: 150 m
  • Distance from person to top of shorter building: 90 m
  • Horizontal distance between tops of both buildings: 154 m

Calculate: (a) What is the angle between the lines joining the person’s position and the tops of both buildings?

(b) What is the value of h, the height of the taller building?

Answer/Explanation

Answer:

(a) attempt to substitute into cosine rule
\(154^2 = 150^2 + 90^2 – 2(150)(90)cos \hat{APB}\) OR \(cos \hat{APB}=\frac{150^2+90^2-154^2}{2(150)(90)}\)
\(\hat{APB} = 75.2286…^0\) OR 1.31298… radians
\(\hat{APB}=75.2^0\) OR 1.31 radians

(b) valid approach to find θ
\(θ= \frac{180^0 – \hat{APB}}{2}\) OR \(θ=\frac{180^0-75.2286…^0}{2}\) (=52.3856…) OR
\(θ=\frac{\pi – 1.31298…}{2}\) (=0.914302…)
valid approach to express h in terms of θ
\(sin θ = \frac{h}{150}\) OR h = 150sin52.3856…\(^0\)
h = 118.820…
h = 119 (m)

Topic:Height and Distance
?
Time
 
Q – 2
8:27
Eng
Scroll to Top