Question 1
Given,
- Below table shows the number of days, d, that the experiment has been running for studying growth of plant and the average height, hcm, of the plants on each of those days. Experiment was for 7 days.
Find:
(a) If The regression line of h on d for this data can be written in the form h = ad + b. find value of a and the value of b
(b) What is value of the Pearson’s product-moment correlation coefficient, r.
(c) Use regression line to estimate the average height of the plants when the experiment has been running for 20 days.
▶️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)
Question 2
Given
- There are two buildings A and B . Point P on the ground is between the two buildings.
- The angle of elevation to the top of each building from point P on the ground , is θ
3. The distance from point P to point A at the top of the taller building is 150 metres.
4. The distance from point P to point B at the top of the shorter building is 90 metres.
5. The distance between A and B is 154 metres.
Find
(a) what is value \(\hat{APB}\).
(b) What is height of taller building, h.
▶️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)
Question 3
Given
- A Model for amount of drug (in mg) in patent as function of time is developed as A(t) = 500\(e^{-kt}\), where k is a positive constant and t is the time in hours after the initial dose is given.
Find
(a) What will be amount of the drug in the patient’s body when t = 0.
(b) If After three hours, the amount of the drug in the patient’s body has decreased to 280mg , What will be value of k .
(c) If second dose is given T hours after the initial dose, when the amount of the drug in the patient’s body is 140mg. What is value of T.
▶️Answer/Explanation
Answer:
(a) A(0) = 500 (mg)
(b) 280 = 500\(e^{-3k}\)
k = 0.193272…
k = 0.193(=-\(\frac{1}{3}In(\frac{380}{500}\)))
(c) 500\(e^{-0.193272…T}\) = 140
T = 6.58636…
T = 6.59 (h)
Question 4
Given
- A normal distribution is modelled for The weights, W grams, of bags of rice packaged in a factory ,with mean 204 grams and standard deviation 5 grams.
Find:
(a) A bag of rice is selected at random , what will be probability that it weighs more than 210 grams?
(b) If According to this model, 80% of the bags of rice weigh between w grams and 210 grams. What will be probability that a randomly selected bag of rice weighs less than w grams.
(c) What will be value of w.
(d) if Ten bags of rice are selected at random, What will be probability that exactly one of the bags weighs less than w grams.
▶️Answer/Explanation
Answer:
(a) evidence of attempting to find correct area under normal curve
P(W>210) OR sketch
P(W>210) = 0.115069…
P(W>210) = 0.115
(b) recognizing P(W<w) = 1 – P(w<W<210)-P(W>210)
P(W<w) = 1 – 0.8 – 0.115069…
P(W<w) = 0.084930…
P(W<w) = 0.0849
(c) evidence of attempting to use inverse normal function
w =197.136…
w =197 (grams)
(d) recognition of binomial distribution
X ∼ B(10,0.0849302…)
P(X=1) = 0.382076…
P(X=1) = 0.382
Question 5
Given
- expansion of \((x + h)^8\) , where h > 0 is expanded as \(x^8+ax^7 + bx^6 + cx^5 + dx^4 + … + h^8\), where a, c, b, c, d, … \(\epsilon \mathbb{R}\)
(a) Find an expression, in terms of h, for
(i) a;
(ii) b;
(iii) d .
(b) If t a, b, and d are the first three terms of a geometric sequence, What will be value of h.
▶️Answer/Explanation
Answer:
(a) attempt to use the binomial expansion of \((x+h)^8\)
(i) a = 8h (accept \(^8C_1\)h )
(ii) b = 28\(h^2\) (accept \(^8C_2h^2\))
(iii) d = 70\(h^4\) (accept \(^8C_4 h^4\))
(b) recognition that there is a common ratio between their terms
\(8h \times r = 28 h^2\) OR \(28h^2 \times r= 70 h^4\) OR \(8h \times r^2 = 70h^4\)
correct equation in terms of h
\(\frac{28h^2}{8h} = \frac{70h^4}{28h^2}\)(or equivalent)
h = 1.4
Question 6
Given
- Velocity of a particle moving in straight line is given by v(t) = 4\(e^{-\frac{t}{3}}\) cos(\(\frac{t}{2}-\frac{\pi}{4}\)), for 0 ≤ t ≤ 4\(\pi \) where \(vms^{-1}\) , at time t seconds. Graph for same is
(a) What will be value of \(t_1\), time when first time when the particle’s acceleration is zero.
(b) Find the value of \(t_2\), second time when the particle is instantaneously at rest.
(c) What will be distance travelled by the particle between t = \(t_1\) and t = \(t_2\).
▶️Answer/Explanation
Answer:
(a) recognize that acceleration is zero when v′(t) = 0 OR at a local maximum on the graph of v
\(t_1\) = 0.394791…
\(t_1\) = 0.395(=arctan(\(\frac{5}{12}\)))(seconds)
(b) recognition that v = 0
sketch OR t = 4.71238… OR t =10.9955…
\(t_2\) = 10.9955…
\(t_2\) = 11.0(=\(\frac{7\pi}{2}\))
distance = 7.83118…
= 7.83 (m)
Question 7
Given:
- For the function h(x) = \(\sqrt{4x-2}\), for x ≥ \(\frac{1}{2}\).
Find
(a) (i) What is \(h^{-1}\)(x), the inverse of h(x), what is its domain.
(ii) What is the range of \(h^{-1}\)(x).
(b) The graph of h intersects the graph of \(h^{-1}\) at two points. What is x-coordinates of these two points?
(c) What is area enclosed by the graph of h and the graph of \(h^{-1}\).
(d) What is h'(x)
(e) What is value of x for which the graph of h and the graph of \(h^{-1}\) have the same gradient.
▶️Answer/Explanation
Answer:
(a) (i) swapping x and y , or \(h(h^{-1}(x))\) = x
\(h^{-1}(x) = \frac{x^2+2}{4}\)
recognizing range of h is domain of \(h^{-1}\)
Domain: x ≥ 0
(ii) range of \(h^{-1}\) is y ≥ \(\frac{1}{2}\)
(b) \(\sqrt{4x-2} = \frac{x^2+2}{4}\) OR \(\sqrt{4x-2} = x\) OR \(\frac{x^2 + 2}{4} = x\)
x = 0.585786…, x = 3.414213…(=2+\(\sqrt{2}\))
x = 0.586, x = 3.41
(c) attempt to form integral of the difference between h (x) and their \(h^{−1}\) , using their limits from part (b)
6.5996632… – 4.7140452…
1.88561…
area = 1.89
(d) attempt to use chain rule or power rule
\(h'(x) = 4.\frac{1}{2}(4x-2)^{\frac{1}{2}}\)
\(h'(x) = \frac{2}{\sqrt{4x-2}}\)
(e) EITHER
\((h^{-1})'(x)=\frac{x}{2}\)
equating their h’(x) to the derivative of their \(h^{-1}(x)\) and attempting to solve for x
\(\frac{2}{\sqrt{4x-2}} = \frac{x}{2}\)
OR
finding intersection of graphs of their derivatives
THEN
1.772776…
x = 1.77
Question 7
Given:
spring with a weight is pulled down and released so that it moves up and down
The Height H is modelled by the function H(t) = acos(7.8t) + b, for a, b ∈ \(\mathbb{R}\) and 0 ≤ t ≤ 10, where t is the time in seconds after the weight is released.
Find
(a) What is the period of the function.
The weight is released when its base is at a minimum height of 1 metre above the ground, and it reaches a maximum height of 1.8 metres above the ground. The graph of H is shown in the following diagram.
(b) What is the value of ?
(i) a;
(ii) b.
(c) What is the number of times that the weight reaches its maximum height in the first five seconds of its motion?
(d) first time that the base of the weight reaches a height of 1.5 metres?
(e) If A camera is set to take a picture of the weight at a random time during the first five seconds
of its motion, what is the probability that the height of the base of the weight is greater than 1.5 metres at the time the picture is taken?
▶️Answer/Explanation
Answer:
(a) 7.8 = \(\frac{2 \pi}{period}\)
\(\frac{7 \pi}{7.8} = 0.805536…\)
period = 0.806(=\(\frac{10 \pi}{39}\))
(b) METHOD 1
(i) amplitude = \(\frac{max-min}{2}\)
\(\frac{1.8 – 1}{2}\)
a = -0.4
(ii) a = -0.4
METHOD 2
attempt to form two simultaneous equations in a and b
H(0) = 1 \(\Rightarrow \) a + b = 1, H(\(\frac{\pi }{7.8}\)) = 1.8 \(\Rightarrow \) – a + b = 1.8
a = -0.4, b = 1.4
(c) EITHER
\(\frac{5}{period}\) = 6.207… < 6 \(\frac{1}{2}\)
OR
consideration of number of maximums on graph in first 5 seconds
OR
maximums when t = 0.403, 1.21, 2.01, 2.82, 3.62, 4.43
THEN
6 times
(d) recognizing that H (t) = 1.5
-0.4 cos(7.8t) + 1.4 = 1.5
0.233779…
t = 0.234 (seconds)
(e) finding second time height is 1.5 metres
t = 0.571757…
in each period, height is greater than 1.5 metres for 0.337978… seconds
multiplying their value by 6 and dividing by 5
\(\frac{0.337978… \times 6}{5}\) OR \(\frac{2.02787…}{5}\)
= 0.405574…
P(height is greater than 1.5m) = 0.406
Question 8
A bag contains n balls. It is known that ten of the balls are green, and the rest of the balls are red. Balls are drawn from the bag, one after the other, without replacement.
(a) Find, in terms of n, the probability that
(i) the first ball drawn is green;
(ii) the first two balls are green.
For the following parts of this question, let n = 25.
(b) Show that the probability that the first two balls are red is 0.35 .
(c) Find the probability that the first three balls are all red.
(d) Find the probability that at least one of the first three balls is green.
A game is played where four balls are drawn, one after the other, from the bag of 25 balls, without replacement. A player earns points based on when the first green ball is drawn. At the end of each game, the four balls are put back in the bag.
A player earns zero points if no green ball is picked, or if the first green ball is picked on the first or second draw.
A player earns 10 points if the first green ball is picked on the third draw and earns 50 points if the first green ball is picked on the fourth draw.
Millie plays this game k times. She finds her score by adding together her points from each game.
(e) Find the least value of k such that Millie’s expected score is greater than 100.
Answer/Explanation
Answer:
(a) (i) \(\frac{10}{n}\)
(ii) multiplying probabilities for GG
P(GG) = \(\frac{10}{n} \times \frac{9}{n-1}\)
P(GG) = \(\frac{90}{n^2-n}\) (accept \(\frac{90}{n(n-1)}\))
(b) P(First red) = \(\frac{15}{25}\) and P(Second red) = \(\frac{14}{24}\) (seen anywhere)
P(RR) = \(\frac{15}{25} \times \frac{14}{24}\) (or equivalent)
= 0.35
(c) \(\frac{15}{25} \times \frac{14}{24} \times \frac{13}{23}\) OR \(0.35 \times \frac{13}{23}\)
0.197826…
P(three red) = 0.198 (exact answer is \(\frac{91}{460}\))
(d) P(at least one green) = 1 – P(three red) OR
P(at least one G) = P(one G) + P(two G) + P(three G)
0.802173…
P(at least one green) = 0.802 (exact answer is \(\frac{369}{460}\))
(e) P(first green on third draw) = \(\frac{15}{25} \times \frac{14}{24} \times \frac{10}{23} \times \frac{22}{22}\) (=\(\frac{7}{46} = 0.152173…\))
P(first green on fourth draw) = \(\frac{15}{25} \times \frac{14}{24} \times \frac{13}{23} \times \frac{10}{22}\) (=\(\frac{91}{1012} = 0.0899209…\))
attempt to substitute their probabilities into expected value formula
expected points per game = \(10 \times \frac{7}{46} + 50 \times \frac{3045}{506} = 6.01778…\))
setting up inequality or equation in k
\(\frac{3045}{506}\)k>100
k>16.6174… (=\(\frac{10120}{609}\))
Millie must play at least 17 times.