2023 IB DP Mathematics - Applications and Interpretation 模擬試題連答案詳解

(a) Using the compound interest formula:
\(A = P\left(1 + \frac{r}{100k}\right)^{kt}\)
\(A = 8000\left(1 + \frac{4.5}{1200}\right)^{12 \times 5}\)
\(A = 8000(1.00375)^{60} \approx 10014.37\)

To the nearest dollar, the amount is $10014.

(b) To find when the values are equal, set:
\(22000(0.85)^t = 8000\left(1 + \frac{4.5}{1200}\right)^{12t}\)

Using a graphic display calculator (GDC) to find the intersection of \(y_1 = 22000(0.85)^x\) and \(y_2 = 8000(1.00375)^{12x}\):
\(t \approx 4.88\) years.

(a)
[M1] for substituting correctly into the compound interest formula.
[A1] for 10014.

(b)
[M1] for setting up the equation representing the equality of the two models.
[M1] for using GDC to solve the equation.
[A1] for 4.88.

(a) Let \(B_n\) be the number of books at the end of year \(n\).
At the end of Year 1:
\(B_1 = 3500 \times (1 - 0.06) + 400 = 3500 \times 0.94 + 400 = 3690\)

At the end of Year 2:
\(B_2 = 3690 \times 0.94 + 400 = 3468.6 + 400 = 3868.6 \approx 3869\).

(b) This recurrence relation can be represented as:
\(B_n = 3500(0.94)^n + 400 \sum_{i=0}^{n-1} 0.94^i\)

Or by using the sequence feature on a GDC:
For \(n = 10\), \(B_{10} \approx 4961.05\)

To the nearest integer, the number of books is 4961.

(a)
[M1] for calculating the books after 1 year: 3690.
[M1] for multiplying by 0.94 and adding 400 for the second year.
[A1] for showing 3868.6 which rounds to 3869.

(b)
[M1] for establishing a correct method to find the 10th term (recurrence list on GDC or formula).
[A2] for 4961 (accept 4960 from different rounding methods, but 4961 is the accurate rounded value).

(a) The height of the platform is when \(d = 0\):
\(h(0) = 5\) meters.

(b) The maximum height occurs at the vertex of the quadratic function.
\(d = -\frac{b}{2a} = -\frac{1.2}{2(-0.05)} = 12\) meters.
\(h(12) = -0.05(12)^2 + 1.2(12) + 5 = -7.2 + 14.4 + 5 = 12.2\) meters.
(Alternatively, using GDC to find maximum point).

(c) The projectile hits the ground when \(h(d) = 0\):
\(-0.05d^2 + 1.2d + 5 = 0\)

Using the quadratic formula or GDC:
\(d \approx 27.6\) meters (since \(d > 0\)).

(a)
[A1] for 5 (or 5 m).

(b)
[M1] for finding the x-coordinate of the vertex (12) or attempting to find maximum via GDC.
[A1] for 12.2 (or 12.2 m).

(c)
[M1] for setting \(h(d) = 0\).
[A2] for 27.6 (or 27.6 m).

(a) Midpoint of \(AB\):
\(M_{AB} = \left(\frac{2+8}{2}, \frac{8+6}{2}\right) = (5, 7)\)

Gradient of \(AB\):
\(m_{AB} = \frac{6-8}{8-2} = -\frac{2}{6} = -\frac{1}{3}\)

Gradient of the perpendicular bisector:
\(m_{\perp} = -\frac{1}{m_{AB}} = 3\)

Equation of the perpendicular bisector:
\(y - 7 = 3(x - 5) \implies y = 3x - 8\)

(b) To find the circumcenter, we find the perpendicular bisector of another side, e.g., \(BC\).

Midpoint of \(BC\):
\(M_{BC} = \left(\frac{8+4}{2}, \frac{6+2}{2}\right) = (6, 4)\)

Gradient of \(BC\):
\(m_{BC} = \frac{2-6}{4-8} = 1\)

Gradient of the perpendicular bisector of \(BC\):
\(m'_{\perp} = -1\)

Equation of this perpendicular bisector:
\(y - 4 = -1(x - 6) \implies y = -x + 10\)

To find the circumcenter, solve the system of equations:
\(3x - 8 = -x + 10 \implies 4x = 18 \implies x = 4.5\)

Substitute \(x = 4.5\):
\(y = -4.5 + 10 = 5.5\)

So the coordinates are \((4.5, 5.5)\).

(a)
[M1] for finding the midpoint of \(AB\).
[M1] for finding the gradient of \(AB\) and then the perpendicular gradient.
[A1] for \(y = 3x - 8\).

(b)
[M1] for finding the equation of a second perpendicular bisector (e.g., \(y = -x + 10\)).
[M1] for solving the system of two linear equations.
[A1] for \((4.5, 5.5)\).

(a) Draw a North line at \(Q\).
The bearing from \(Q\) to \(P\) is \(040^\circ + 180^\circ = 220^\circ\).
The bearing from \(Q\) to \(R\) is \(110^\circ\).
The angle \(\angle PQR\) is the difference between these bearings:
\(\angle PQR = 220^\circ - 110^\circ = 110^\circ\).

(b) Using the cosine rule on \(\triangle PQR\):
\(PR^2 = PQ^2 + QR^2 - 2(PQ)(QR)\cos(\angle PQR)\)
\(PR^2 = 12^2 + 18^2 - 2(12)(18)\cos(110^\circ)\)
\(PR^2 = 144 + 324 - 432\cos(110^\circ) \approx 615.75\)
\(PR = \sqrt{615.75} \approx 24.8\text{ km}\) (to 3 s.f.).

(c) Using the sine rule to find \(\angle QPR\):
\(\frac{\sin(\angle QPR)}{QR} = \frac{\sin(\angle PQR)}{PR}\)
\(\frac{\sin(\angle QPR)}{18} = \frac{\sin(110^\circ)}{24.814}\)
\(\sin(\angle QPR) = \frac{18 \sin(110^\circ)}{24.814} \approx 0.68164\)
\(\angle QPR \approx 42.97^\circ\)

The bearing of \(R\) from \(P\) is:
\(\text{Bearing} = 040^\circ + 42.97^\circ = 82.97^\circ \approx 083^\circ\).

(a)
[M1] for a diagram or calculation showing clear understanding of alternate interior angles or bearings.
[A1] for \(110^\circ\).

(b)
[M1] for substituting correctly into the cosine rule.
[A1] for \(24.8\text{ km}\) (or 24.8).

(c)
[M1] for using the sine rule (or cosine rule) to find \(\angle QPR\).
[A1] for finding \(\angle QPR \approx 43.0^\circ\).
[A1] for adding \(40^\circ\) to find the bearing \(083^\circ\) (or \(83^\circ\)).

(a) Let \(X\) be the weight of an apple. \(X \sim N(150, 12^2)\).
Using GDC: \(P(X > 170) \approx 0.04779 \approx 0.0478\).

(b) We are given \(P(X < w) = 0.15\).
Using inverse normal on the GDC:
\(w \approx 137.56 \approx 138\).

(c) Let \(Y\) be the number of small apples in a sample of 5.
\(Y \sim B(5, 0.15)\).
Using binomial probability on the GDC:
\(P(Y = 2) = \binom{5}{2} (0.15)^2 (0.85)^3 \approx 0.138\).

(a)
[M1] for a correct probability statement \(P(X > 170)\) or normal distribution sketch.
[A1] for 0.0478.

(b)
[M1] for setting up the equation \(P(X < w) = 0.15\) or equivalent.
[A1] for 138 (accept 137.6).

(c)
[M1] for identifying the binomial distribution with \(n=5\) and \(p=0.15\).
[A1] for 0.138.

(a) \(H_0\): Preferred music genre and age group are independent (or there is no association between preferred music genre and age group).

(b) Degrees of freedom:
\(\text{df} = (\text{rows} - 1)(\text{columns} - 1) = (2 - 1)(\text{3} - 1) = 2\).

(c) Total "Under 30" = \(45 + 32 + 13 = 90\).
Total "Rock" = \(32 + 38 = 70\).
Grand Total = \(90 + 90 = 180\).
Expected Frequency = \(\frac{90 \times 70}{180} = 35\).

(d) Using GDC for \(\chi^2\) two-way test on the matrix:
\(\chi^2 \approx 11.129\)
\(p\text{-value} \approx 0.00383\) (or 0.0038)

Since \(0.00383 < 0.05\) (the significance level), we reject the null hypothesis. There is significant evidence to suggest that preferred music genre and age group are associated.

(a)
[A1] for stating the correct null hypothesis (must mention independence or no association).

(b)
[A1] for 2.

(c)
[M1] for using row total \(\times\) column total / grand total.
[A1] for 35.

(d)
[A1] for \(p\text{-value} \approx 0.00383\).
[R1] for comparing \(p\text{-value}\) to 0.05 and making the correct decision (reject \(H_0\)).

(a) The volume of the box is \(V = x^2 h = 500\), which gives \(h = \frac{500}{x^2}\).
The surface area of an open box with a square base of side \(x\) and height \(h\) is:
\(A = x^2 + 4xh\)

Substitute \(h = \frac{500}{x^2}\) into the surface area equation:
\(A = x^2 + 4x \left(\frac{500}{x^2}\right) = x^2 + \frac{2000}{x\}}.

(b) Differentiating \)A\) with respect to \(x\):
\(\frac{\text{d}A}{\text{d}x} = 2x - 2000x^{-2} = 2x - \frac{2000}{x^2}\).

(c) Set \(\frac{\text{d}A}{\text{d}x} = 0\) to find the minimum:
\(2x - \frac{2000}{x^2} = 0 \implies 2x^3 = 2000 \implies x^3 = 1000 \implies x = 10\text{ cm}\).

(a)
[M1] for writing down correct expressions for volume and surface area.
[A1] for expressing \(h\) in terms of \(x\) and substituting to obtain the given formula.

(b)
[M1] for attempting to differentiate.
[A1] for \(2x - \frac{2000}{x^2}\).

(c)
[M1] for setting derivative equal to 0.
[A1] for \(x = 10\).

(a) The depreciation of the food truck can be modeled by the geometric sequence formula: \( V = 45000 \times (1 - 0.12)^t = 45000 \times 0.88^t \).\
\
For \( t = 5 \):\
\( V = 45000 \times 0.88^5 \approx 23747.98 \).\
\
Rounding to 3 significant figures gives \$23,700.\
\
(b) To find when the value falls below \$15,000, we set up the inequality:\
\( 45000 \times 0.88^t < 15000 \)\
\( 0.88^t < \frac{1}{3} \)\
\( t \ln(0.88) < \ln\left(\frac{1}{3}\right) \)\
\( t > \frac{\ln(1/3)}{\ln(0.88)} \approx 8.59 \) years.\
\
Since we need the number of complete years, it will take 9 years for the value to fall below \$15,000.

(M1) for writing a correct depreciation expression: \( 45000 \times 0.88^5 \).\
(A1) for \( 23700 \) (or \( 23747.98 \)).\
(M1) for setting up the inequality or equation: \( 45000 \times 0.88^t < 15000 \).\
(A1) for solving to find \( t \approx 8.59 \).\
(A1) for rounding up to 9 complete years.

(a) When \( t = 0 \), \( T(0) = 85 \).\
\( 22 + a b^{0} = 85 \Rightarrow 22 + a = 85 \Rightarrow a = 63 \).\
\
(b) When \( t = 10 \), \( T(10) = 45 \).\
\( 22 + 63 b^{-10} = 45 \)\
\( 63 b^{-10} = 23 \)\
\( b^{-10} = \frac{23}{63} \)\
\( b = \left(\frac{63}{23}\right)^{0.1} \approx 1.10597 \).\
\
To 3 significant figures, \( b = 1.11 \).\
\
(c) For \( t = 20 \):\
\( T(20) = 22 + 63 \times (1.10597)^{-20} = 22 + 63 \times \left(\frac{23}{63}\right)^2 \approx 30.4^\circ\text{C} \).

(A1) for \( a = 63 \).\
(M1) for substituting \( a = 63 \) and \( t = 10 \) into the formula.\
(A1) for \( b \approx 1.11 \).\
(M1) for substituting \( t = 20 \) into the model.\
(A1) for \( 30.4^\circ\text{C} \).

(a) Find the midpoint of \( AB \):\
\( M = \left(\frac{2+8}{2}, \frac{8+6}{2}\right) = (5, 7) \).\
\
Find the gradient of \( AB \):\
\( m_{AB} = \frac{6-8}{8-2} = -\frac{1}{3} \).\
\
The perpendicular gradient is \( m = 3 \).\
\
Using the point-slope formula with \( M(5, 7) \):\
\( y - 7 = 3(x - 5) \Rightarrow y = 3x - 8 \).\
\
(b) Calculate the distance from \( P(5, 5.5) \) to each hub:\
\( PA = \sqrt{(5-2)^2 + (5.5-8)^2} = \sqrt{9 + 6.25} = \sqrt{15.25} \approx 3.91 \).\
\( PB = \sqrt{(5-8)^2 + (5.5-6)^2} = \sqrt{9 + 0.25} = \sqrt{9.25} \approx 3.04 \).\
\( PC = \sqrt{(5-4)^2 + (5.5-2)^2} = \sqrt{1 + 12.25} = \sqrt{13.25} \approx 3.64 \).\
\
Since \( PB \) is the shortest distance, Hub B is the closest delivery hub to the customer.

(M1) for finding the midpoint \( (5, 7) \).\
(M1) for finding the gradient of \( AB \) and determining the perpendicular gradient \( 3 \).\
(A1) for the correct equation \( y = 3x - 8 \).\
(M1) for calculating the distances from \( P \) to the hubs.\
(A1) for identifying Hub B as the closest.

(a) \( H_0 \): Preferred music genre and age group are independent.\
\
(b) Total number of "Under 30" participants \( = 55 + 40 + 15 = 110 \).\
Total number of "Classical" preference participants \( = 15 + 35 = 50 \).\
Grand total \( = 200 \).\
\
Expected Value \( = \frac{110 \times 50}{200} = 27.5 \).\
\
(c) Using a GDC for a \( 2 \times 3 \) contingency table, the test statistics are:\
\( \chi^2 \approx 18.87 \)\
\( p \)-value \\approx 7.99 \\times 10^{-5} \) (or \( 0.0000799 \)).\
\
Since the \( p \)-value \( < 0.05 \), we reject the null hypothesis. Thus, preferred music genre is not independent of age group.

(A1) for a correct null hypothesis stating independence.\
(M1) for expected value calculation setup.\
(A1) for \( 27.5 \).\
(G2) for \( p \)-value \\approx 7.99 \\times 10^{-5} \) (accept \( 0.00008 \)).\
(R1) for comparing the \( p \)-value with \( 0.05 \) and rejecting \( H_0 \).

(a) The three fenced sides consist of two widths of length \( x \) and one length parallel to the wall. The parallel side has length \( 120 - 2x \).\
\
Area \( A = x(120 - 2x) = 120x - 2x^2 \).\
\
(b) Differentiating \( A \) with respect to \( x \):\
\( \frac{\text{d}A}{\text{d}x} = 120 - 4x \).\
\
(c) For maximum area, set the derivative to zero:\
\( 120 - 4x = 0 \Rightarrow x = 30 \).\
\
Substituting \( x = 30 \) back into the area equation:\
\( A = 30(120 - 2(30)) = 30 \times 60 = 1800\text{ m}^2 \).

(M1) for expressing the length as \( 120 - 2x \).\
(A1) for \( A = 120x - 2x^2 \).\
(A1) for \( 120 - 4x \).\
(M1) for setting their derivative to 0.\
(A1) for \( 1800\text{ m}^2 \).

(a) Volume of a cone: \( V = \frac{1}{3}\pi r^2 h \).\
\
\( V = \frac{1}{3}\pi (1.5)^2 (4.0) = 3\pi \approx 9.42\text{ m}^3 \).\
\
(b) Using Pythagoras' theorem for slant height \( l \):\
\( l = \sqrt{r^2 + h^2} = \sqrt{1.5^2 + 4.0^2} = \sqrt{18.25} \approx 4.27\text{ m} \).\
\
(c) Curved surface area: \( A = \pi r l \).\
\
\( A = \pi (1.5)(\sqrt{18.25}) \approx 20.137\text{ m}^2 \).\
\
Correct to three significant figures, the area is \( 20.1\text{ m}^2 \).

(M1) for using the cone volume formula.\
(A1) for \( 9.42 \).\
(M1) for Pythagoras' theorem setup.\
(A1) for \( 4.27 \).\
(M1) for using the curved surface area formula.\
(A1) for \( 20.1 \).

(a) Let \( X \) represent the mass of an apple, where \( X \sim N(150, 12^2) \).\
\
Using a GDC to find \( P(140 < X < 165) \):\
\( P(140 < X < 165) \approx 0.6897 \).\
\
Correct to three significant figures, the probability is \( 0.690 \).\
\
(b) We are given \( P(X < w) = 0.08 \).\
\
Using the inverse normal cumulative function on a GDC:\
\( w = \text{invNorm}(0.08, 150, 12) \approx 133.14 \) grams.\
\
To the nearest gram, \( w = 133 \).

(M1) for correct normal probability notation/setup: \( P(140 < X < 165) \).\
(A1) for \( 0.690 \).\
(M1) for setting up equation \( P(X < w) = 0.08 \).\
(A1) for \( 133 \).

(a) The width of each interval is \( h = 2 \).\
\
Using the trapezoidal rule:\
\( \text{Distance} \approx \frac{2}{2} [ v(0) + 2(v(2) + v(4) + v(6)) + v(8) ] \)\
\( = 1 \times [ 0 + 2(3.5 + 5.2 + 4.8) + 2.1 ] \)\
\( = 2(13.5) + 2.1 = 29.1\text{ meters} \).\
\
(b) The exact distance is found by integrating the velocity function:\
\( \int_{0}^{8} (-0.15t^2 + 1.5t) \text{d}t = \left[ -0.05t^3 + 0.75t^2 \right]_{0}^{8} \)\
\( = [-0.05(512) + 0.75(64)] - [0] \)\
\( = -25.6 + 48 = 22.4\text{ meters} \).

(M1) for establishing \( h = 2 \).\
(M1) for correct substitution into the trapezoidal formula.\
(A1) for \( 29.1 \).\
(M1) for setting up the integration of \( v(t) \).\
(A1) for \( 22.4 \).

(a) Let \(X\) be the lifespan of a bulb in hours, so \(X \sim N(8000, 600^2)\). To find \(P(7500 \le X \le 8500)\), we use a graphic display calculator (GDC) with a normal cumulative distribution function (lower limit = 7500, upper limit = 8500, \(\mu = 8000\), \(\sigma = 600\)). This yields \(P(7500 \le X \le 8500) \approx 0.59534...\), which rounds to 0.595 to 3 significant figures. (b) We want to find the value of \(t\) such that \(P(X < t) = 0.015\). Using the inverse normal function on a GDC with area = 0.015, \(\mu = 8000\), and \(\sigma = 600\), we obtain \(t \approx 6697.95...\) hours. Rounding to the nearest hour, we get \(t = 6698\).

(a) [2.47 marks total]: M1.47 for a setup indicating normal cumulative distribution with correct limits and parameters. A1 for 0.595 (accept 0.59534...). (b) [4.00 marks total]: M1.50 for setting up the inverse normal equation \(P(X < t) = 0.015\). M1.50 for finding the unrounded value \(t \approx 6697.95...\). A1 for rounding to the nearest hour to get 6698.

(a) The total volume of the tank is the sum of the volume of the cylinder and the volume of the hemisphere: \( V = \pi r^2 h + \frac{2}{3}\pi r^3 \). Setting \( V = 50 \): \( \pi r^2 h + \frac{2}{3}\pi r^3 = 50 \Rightarrow \pi r^2 h = 50 - \frac{2}{3}\pi r^3 \Rightarrow h = \frac{50}{\pi r^2} - \frac{2}{3}r \). (b) The inner surface area of the tank is: \( A = 2\pi r h + 2\pi r^2 \). Substituting the expression for \( h \) from part (a): \( A = 2\pi r \left( \frac{50}{\pi r^2} - \frac{2}{3}r \right) + 2\pi r^2 = \frac{100}{r} - \frac{4}{3}\pi r^2 + 2\pi r^2 = \frac{100}{r} + \frac{2}{3}\pi r^2 \). (c) Differentiating \( A \) with respect to \( r \): \( \frac{dA}{dr} = -\frac{100}{r^2} + \frac{4}{3}\pi r \). (d) To minimize the surface area, set \( \frac{dA}{dr} = 0 \): \( -\frac{100}{r^2} + \frac{4}{3}\pi r = 0 \Rightarrow \frac{4}{3}\pi r = \frac{100}{r^2} \Rightarrow r^3 = \frac{75}{\pi} \Rightarrow r = \left(\frac{75}{\pi}\right)^{1/3} \approx 2.88 \text{ m} \). (e) Substituting \( r \approx 2.8794 \) into the expression for \( A \): \( A \approx \frac{100}{2.8794} + \frac{2}{3}\pi (2.8794)^2 \approx 34.729 + 17.365 = 52.1 \text{ m}^2 \). (f) Find the second derivative: \( \frac{d^2A}{dr^2} = \frac{200}{r^3} + \frac{4}{3}\pi \). Since \( r > 0 \), both terms are positive, meaning \( \frac{d^2A}{dr^2} > 0 \) for all positive \( r \). Thus, the critical point represents a minimum.

(a) M1 for writing the correct total volume equation. A1 for correct algebraic manipulation leading to the given expression. (b) M1 for identifying the correct components of surface area. A1 for substituting the expression for h. M1 A1 for simplifying algebraically to reach the exact target expression. (c) M1 A1 for differentiating the 100/r term. M1 A1 for differentiating the (2/3)pi r^2 term. (d) M1 for setting their derivative equal to 0. A1 for isolating r^3. A1 for finding r = 2.88. (e) M1 for substituting their value of r into the area formula. A1 for area = 52.1. (f) M1 for finding the second derivative expression. R1 for evaluating or stating it is positive, R1 for concluding a minimum.

(a) \( H_0 \): The type of book preferred is independent of the reader's age group. \( H_1 \): The type of book preferred is not independent of the reader's age group. (b) \( \text{Degrees of Freedom} = (\text{rows} - 1)(\text{columns} - 1) = (3 - 1)(3 - 1) = 4 \). (c) Expected frequency \( = \frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}} = \frac{90 \times 70}{300} = 21 \). (d) Using a GDC on the 3x3 table of observed values: \( \chi^2 \approx 25.386 \approx 25.4 \) (3 s.f.), \( p \text{-value} \approx 4.21 \times 10^{-5} \approx 0.0000421 \). (e) Since the \( p \)-value \( (0.0000421) < 0.05 \) (or the critical value), we reject the null hypothesis \( H_0 \). There is sufficient evidence to conclude that book preference is associated with age group. (f)(i) \( P(25\text{-}50 \cup \text{Non-fiction}) = \frac{120 + 95 - 45}{300} = \frac{170}{300} = \frac{17}{30} \approx 0.567 \). (ii) \( P(\text{Fiction} \mid \text{Over 50}) = \frac{45}{90} = 0.5 \). (iii) \( P(\text{Both Under 25}) = \frac{90}{300} \times \frac{89}{299} = \frac{3}{10} \times \frac{89}{299} = \frac{267}{2990} \approx 0.0893 \).

(a) A1 for correct H0, A1 for correct H1. (b) A1 for df = 4. (c) M1 for showing the formula structure, A1 for concluding 21. (d) M2 for correct GDC inputs, A1 for Chi-sq = 25.4, A1 for p-value = 4.21 x 10^-5. (e) R1 for comparing p-value with 0.05, A1 for rejecting H0 and writing a contextual conclusion. (f)(i) M1 for summing probabilities, A1 for 17/30 (0.567). (f)(ii) M1 for conditional probability setup, A1 for 0.5. (f)(iii) M1 for product of fractions without replacement, A1 for 267/2990 (0.0893).

(a) In triangle \( APB \): The bearing from \( A \) to \( B \) is \( 065^\circ \) and from \( A \) to \( P \) is \( 340^\circ \). The internal angle \( \widehat{PAB} = 65^\circ + (360^\circ - 340^\circ) = 85^\circ \). The bearing from \( B \) to \( A \) is \( 065^\circ + 180^\circ = 245^\circ \), and the bearing from \( B \) to \( P \) is \( 305^\circ \). The internal angle \( \widehat{ABP} = 305^\circ - 245^\circ = 60^\circ \). Therefore, \( \widehat{APB} = 180^\circ - (85^\circ + 60^\circ) = 35^\circ \). (b) Using the Sine Rule in \( \triangle APB \): \( \frac{AP}{\sin(60^\circ)} = \frac{150}{\sin(35^\circ)} \Rightarrow AP = \frac{150 \sin(60^\circ)}{\sin(35^\circ)} \approx 226.445 \approx 226 \text{ m} \). (c) In right-angled triangle \( \triangle TPA \): \( \tan(23^\circ) = \frac{TP}{AP} \Rightarrow TP = 226.445 \times \tan(23^\circ) \approx 96.115 \approx 96.1 \text{ m} \). (d) Using the Sine Rule to find \( BP \): \( \frac{BP}{\sin(85^\circ)} = \frac{150}{\sin(35^\circ)} \Rightarrow BP = \frac{150 \sin(85^\circ)}{\sin(35^\circ)} \approx 260.525 \approx 261 \text{ m} \). (e) In right-angled triangle \( \triangle TPB \), let the angle of elevation be \( \theta \): \( \tan(\theta) = \frac{TP}{BP} = \frac{96.115}{260.525} \approx 0.3689 \Rightarrow \theta = \arctan(0.3689) \approx 20.3^\circ \).

(a) M1 for finding angle PAB = 85, M1 for finding angle ABP = 60, A1 for angle APB = 35. (b) M1 for correct substitution into Sine Rule, A1 for rearranging, A1 for 226 m. (c) M1 for setting up tan(23) = TP/AP, A1 for multiplying, A1 for 96.1 m. (d) M1 for correct Sine Rule substitution, A1 for 261 m. (e) M1 for identifying the correct right-angled triangle, M1 for tan(theta) = TP/BP, A1 for 20.3 degrees.

(a) Initial velocity is when \( t = 0 \): \( v(0) = 3(0)^2 - 14(0) + 8 = 8 \text{ m s}^{-1} \). (b) Momentarily at rest when \( v(t) = 0 \): \( 3t^2 - 14t + 8 = 0 \Rightarrow (3t - 2)(t - 4) = 0 \Rightarrow t = \frac{2}{3} \text{ s} \) and \( t = 4 \text{ s} \). (c) Acceleration \( a(t) = v'(t) = 6t - 14 \). When \( t = 3 \): \( a(3) = 6(3) - 14 = 4 \text{ m s}^{-2} \). (d) Total distance traveled is the integral of the absolute value of velocity: \( D = \int_0^4 |3t^2 - 14t + 8| dt = \int_0^{2/3} v(t) dt + \int_{2/3}^4 -v(t) dt \). Integrating gives \( \left[ t^3 - 7t^2 + 8t \right]_0^{2/3} + \left[ -t^3 + 7t^2 - 8t \right]_{2/3}^4 = \frac{68}{27} + \left( 16 + \frac{68}{27} \right) = \frac{568}{27} \approx 21.0 \text{ m} \). (e) Displacement is \( s(t) = \int (3t^2 - 14t + 8) dt = t^3 - 7t^2 + 8t + C \). Given \( s(1) = 10 \): \( 1^3 - 7(1)^2 + 8(1) + C = 10 \Rightarrow 2 + C = 10 \Rightarrow C = 8 \). Thus, \( s(t) = t^3 - 7t^2 + 8t + 8 \).

(a) A1 for 8 m s^-1. (b) M1 for setting velocity to 0, A1 for t = 2/3, A1 for t = 4. (c) M1 for differentiating, A1 for 4 m s^-2. (d) M1 for integrating absolute velocity (or split integral), M1 for evaluating at limits, A1 for each part or GDC usage, A1 for final answer of 21.0 m. (e) M1 for integrating, A1 for correct algebraic terms, M1 for substituting s(1)=10 to find C, A1 for correct final function.

(a) Let \( X \sim N(150, 15^2) \). (i) \( P(X > 170) \approx 0.0912 \) (from GDC). (ii) \( P(130 < X < 165) \approx 0.750 \) (from GDC). (b) We need \( P(X < w) = 0.08 \). Using inverse normal: \( w \approx 128.92 \approx 129 \text{ grams} \). (c) Let \( Y \sim B(20, p) \) where \( p = P(X > 170) = 0.091211 \). (i) \( P(Y = 3) = \binom{20}{3} p^3 (1-p)^{17} \approx 0.171 \) (from GDC binomial PDF). (ii) \( P(Y \ge 2) = 1 - P(Y \le 1) \approx 1 - 0.4497 = 0.550 \) (from GDC binomial CDF). (d) We are testing for a decrease in the population mean: \( H_0: \mu = 150 \) and \( H_1: \mu < 150 \).

(a)(i) M1 for normal CDF setup, A1 for 0.0912. (a)(ii) M1 for correct limits, A1 for 0.750. (b) M1 for inverse normal setup, A1 for 129. (c)(i) M1 for identifying binomial, A1 for p, A1 for 0.171. (c)(ii) M1 for writing P(Y>=2) = 1 - P(Y<=1), A1 for 0.550. (d) A1 for H0, A1 for H1.

(a)(i) Loan amount is \( 90\% \text{ of } 25000 = 22500 \text{ USD} \). (ii) Using a GDC Finance Solver: \( N = 48 \), \( I\% = 6 \), \( PV = 22500 \), \( FV = 0 \), \( P/Y = 12 \), \( C/Y = 12 \). Solving for monthly payment: \( PMT \approx 528.41 \text{ USD} \). (iii) Total payments \( = 528.41 \times 48 = 25363.68 \). Interest paid \( = 25363.68 - 22500 = 2863.68 \text{ USD} \). (b)(i) Option B: \( PV = 18000 \), \( r = 0.045 \), \( k = 4 \), \( n = 5 \). \( FV = 18000 \times \left(1 + \frac{0.045}{4}\right)^{20} \approx 22513.52 \text{ USD} \). (ii) Interest earned \( = 22513.52 - 18000 = 4513.52 \text{ USD} \). (c) This is a geometric series (annuity due) with 10 terms. The first deposit compound is \( 500 \times 1.035^{10} \) and the last is \( 500 \times 1.035^1 \). \( u_1 = 500 \times 1.035 = 517.5 \), \( r = 1.035 \). \( S_{10} = \frac{517.5(1.035^{10} - 1)}{1.035 - 1} \approx 6071.00 \text{ USD} \).

(a)(i) A1 for $22,500. (a)(ii) M2 for correct finance solver parameters, A1 for $528.41. (a)(iii) M1 for total payments calculation, A1 for $2,863.68. (b)(i) M1 for compound interest formula setup, A1 for correct substitution, A1 for $22,513.52. (b)(ii) A1 for $4,513.52. (c) M1 for recognizing geometric series with first term 517.5, M1 for common ratio 1.035, M1 for formula substitution, A1 for $6,071.00.

(a)(i) \( a = \frac{14 - 6}{2} = 4 \). (ii) \( d_0 = \frac{14 + 6}{2} = 10 \). (b) The time from maximum to minimum is \( 10.25 - 4 = 6.25 \) hours. This is half the period, so Period \( = 2 \times 6.25 = 12.5 \) hours. Therefore, \( b = \frac{2\pi}{12.5} \approx 0.50265 \approx 0.503 \). (c) Since the maximum occurs at \( t = 4 \), we can write \( c = 4 \). (d) At 14:30, \( t = 14.5 \). \( d(14.5) = 4 \cos(0.50265(14.5 - 4)) + 10 = 4 \cos(5.2778) + 10 \approx 12.119 \approx 12.1 \text{ m} \) (using radians). (e) We need \( 4 \cos(0.50265(t - 4)) + 10 \ge 11.5 \Rightarrow \cos(0.50265(t - 4)) \ge 0.375 \). The boundaries in \( [12, 24] \) are when \( 0.50265(t - 4) = 2\pi \pm 1.1864 \Rightarrow t - 4 \approx 10.14 \text{ and } 14.86 \Rightarrow t \approx 14.14 \text{ and } 18.86 \). Thus, the interval is \( 14.1 \le t \le 18.9 \) (or between 14:08 and 18:52).

(a)(i) A1 for amplitude 4, A1 for working. (a)(ii) A1 for midline 10, A1 for working. (b) M1 for finding time diff = 6.25 hours, A1 for showing period = 12.5, A1 for b = 0.503. (c) A1 for c = 4. (d) M1 for substituting t = 14.5, A1 for using radians, A1 for 12.1 m. (e) M1 for inequality setup, M1 for finding critical values on GDC, A1 for t1 = 14.1, A1 for t2 = 18.9.

**(a)**
For Lake Alpha:
- Inflow of pollutant: from Lake Beta, which carries concentration \(\frac{y}{200}\text{ kg/m}^3\) at a rate of \(15\text{ m}^3/\text{day}\). Thus, inflow rate is \(15 \times \frac{y}{200} = 0.075y\text{ kg/day}\).
- Outflow of pollutant: to Lake Beta, which carries concentration \(\frac{x}{100}\text{ kg/m}^3\) at a rate of \(40\text{ m}^3/\text{day}\). Thus, outflow rate is \(40 \times \frac{x}{100} = 0.4x\text{ kg/day}\).
- Therefore, \(\frac{dx}{dt} = -0.4x + 0.075y\).

For Lake Beta:
- Inflow of pollutant: from Lake Alpha at rate \(40 \times \frac{x}{100} = 0.4x\text{ kg/day}\).
- Outflow of pollutant: return flow to Lake Alpha at rate \(15 \times \frac{y}{200} = 0.075y\text{ kg/day}\), and outflow to river at rate \(25 \times \frac{y}{200} = 0.125y\text{ kg/day}\).
- Total outflow rate of pollutant: \(0.075y + 0.125y = 0.2y\text{ kg/day}\).
- Therefore, \(\frac{dy}{dt} = 0.4x - 0.2y\).

**(b)**
In matrix form, the system is:
\[\frac{d\mathbf{x}}{dt} = \begin{pmatrix} -0.4 & 0.075 \\ 0.4 & -0.2 \end{pmatrix} \mathbf{x}\]
To find the eigenvalues, solve \(\det(\mathbf{A} - \lambda\mathbf{I}) = 0\):
\[\det \begin{pmatrix} -0.4 - \lambda & 0.075 \\ 0.4 & -0.2 - \lambda \end{pmatrix} = 0\]
\[(-0.4 - \lambda)(-0.2 - \lambda) - 0.03 = 0\]
\[\lambda^2 + 0.6\lambda + 0.08 - 0.03 = 0 \implies \lambda^2 + 0.6\lambda + 0.05 = 0\]
\[(\lambda + 0.1)(\lambda + 0.5) = 0\]
So the eigenvalues are \(\lambda_1 = -0.1\) and \(\lambda_2 = -0.5\).

To find the eigenvector for \(\lambda_1 = -0.1\):
\[(\mathbf{A} - (-0.1)\mathbf{I})\mathbf{v}_1 = \mathbf{0} \implies \begin{pmatrix} -0.3 & 0.075 \\ 0.4 & -0.1 \end{pmatrix} \begin{pmatrix} v_{11} \\ v_{12} \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\]
From the first row, \(-0.3v_{11} + 0.075v_{12} = 0 \implies v_{12} = 4v_{11}\).
An eigenvector is \(\mathbf{v}_1 = \begin{pmatrix} 1 \\ 4 \end{pmatrix}\).

To find the eigenvector for \(\lambda_2 = -0.5\):
\[(\mathbf{A} - (-0.5)\mathbf{I})\mathbf{v}_2 = \mathbf{0} \implies \begin{pmatrix} 0.1 & 0.075 \\ 0.4 & 0.3 \end{pmatrix} \begin{pmatrix} v_{21} \\ v_{22} \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\]
From the first row, \(0.1v_{21} + 0.075v_{22} = 0 \implies v_{22} = -\frac{4}{3}v_{21}\).
An eigenvector is \(\mathbf{v}_2 = \begin{pmatrix} 3 \\ -4 \end{pmatrix}\).

**(c)**
The general solution is:
\[\begin{pmatrix} x(t) \\ y(t) \end{pmatrix} = C_1 e^{-0.1t} \begin{pmatrix} 1 \\ 4 \end{pmatrix} + C_2 e^{-0.5t} \begin{pmatrix} 3 \\ -4 \end{pmatrix}\]
So, \(x(t) = C_1 e^{-0.1t} + 3C_2 e^{-0.5t}\) and \(y(t) = 4C_1 e^{-0.1t} - 4C_2 e^{-0.5t}\).

**(d)**
Using initial conditions \(x(0) = 80\) and \(y(0) = 0\):
\[C_1 + 3C_2 = 80\]
\[4C_1 - 4C_2 = 0 \implies C_1 = C_2\]
Substitute \(C_1 = C_2\) into the first equation:
\[4C_1 = 80 \implies C_1 = 20 \text{ and } C_2 = 20\]
Therefore, the particular solution is:
\[x(t) = 20e^{-0.1t} + 60e^{-0.5t}\]
\[y(t) = 80e^{-0.1t} - 80e^{-0.5t}\]

**(e)**
(i) To find \(t_{\text{max}}\) where \(y(t)\) is maximized, set \(\frac{dy}{dt} = 0\):
\[\frac{dy}{dt} = -8e^{-0.1t} + 40e^{-0.5t} = 0\]
\[40e^{-0.5t} = 8e^{-0.1t} \implies e^{0.4t} = 5\]
\[0.4t = \ln(5) \implies t_{\text{max}} = 2.5 \ln(5) \approx 4.02 \text{ days}\]

(ii) Calculate the maximum mass:
\[y(4.02) = 80 e^{-0.1(2.5 \ln(5))} - 80 e^{-0.5(2.5 \ln(5))} = 80(5^{-0.25}) - 80(5^{-1.25}) = 80 \cdot 5^{-0.25} \left(1 - \frac{1}{5}\right) = 64 \cdot 5^{-0.25} \approx 42.8 \text{ kg}\]

**(f)**
Since the maximum level of pollutant in Lake Beta is \(42.8\text{ kg}\), which is greater than \(30\text{ kg}\), the protocol must be activated.
To find the interval, solve \(y(t) = 30\):
\[80 e^{-0.1t} - 80 e^{-0.5t} = 30\]
Using a GDC to solve \(8e^{-0.1t} - 8e^{-0.5t} - 3 = 0\):
Let \(u = e^{-0.1t}\), then \(8u - 8u^5 - 3 = 0\).
The two positive real roots for \(u\) are \(u_1 \approx 0.3833\) and \(u_2 \approx 0.8683\).
- For \(e^{-0.1t} = 0.8683 \implies t_1 \approx 1.412 \approx 1.41\text{ days}\).
- For \(e^{-0.1t} = 0.3833 \implies t_2 \approx 9.590 \approx 9.59\text{ days}\).

The total duration is \(t_2 - t_1 = 9.590 - 1.412 \approx 8.18\text{ days}\) (accept range \(8.17\) to \(8.19\)).

**(a)**
M1: Setting up rates for Lake Alpha: Inflow rate = \(15 \times \frac{y}{200}\) and Outflow rate = \(40 \times \frac{x}{100}\).
A1: Combining to yield \(\frac{dx}{dt} = -0.4x + 0.075y\).
M1: Setting up rates for Lake Beta: Inflow rate = \(40 \times \frac{x}{100}\) and Outflow rates = \(15 \times \frac{y}{200}\) and \(25 \times \frac{y}{200}\).
A1: Combining to yield \(\frac{dy}{dt} = 0.4x - 0.2y\).

**(b)**
M1: Writing down matrix form correctly.
M1: Attempting to solve \(\det(\mathbf{A} - \lambda\mathbf{I}) = 0\).
A1: Obtaining quadratic equation \(\lambda^2 + 0.6\lambda + 0.05 = 0\).
A1: Correct eigenvalues \(\lambda = -0.1, -0.5\).
M1: Attempt to solve \((\mathbf{A} - \lambda\mathbf{I})\mathbf{v} = \mathbf{0}\) for either eigenvalue.
A1: Correct eigenvector for \(\lambda = -0.1\) (or scalar multiple).
A1: Correct eigenvector for \(\lambda = -0.5\) (or scalar multiple).

**(c)**
M1: Using the formula \(\mathbf{x}(t) = C_1 e^{\lambda_1 t}\mathbf{v}_1 + C_2 e^{\lambda_2 t}\mathbf{v}_2\).
A2: Writing correct system general equations (award A1 if only one of \(x(t)\) or \(y(t)\) is correct).

**(d)**
M1: Setting up the linear system using initial conditions \(x(0)=80\), \(y(0)=0\).
A1: Finding \(C_1 = 20\).
A1: Finding \(C_2 = 20\).
A1: Final particular equations written explicitly.

**(e)**
M1: Taking the derivative \(\frac{dy}{dt}\) and setting it to zero.
A1: Simplifying to \(e^{0.4t} = 5\) or equivalent.
A1: \(t_{\text{max}} = 2.5\ln(5)\) (exact) or \(4.02\text{ days}\) (at least 3sf).
M1: Substituting \(t_{\text{max}}\) back into \(y(t)\).
A1: \(42.8\text{ kg}\) (accept \(42.7 - 42.9\)).

**(f)**
R1: Comparing maximum value to 30 and confirming activation.
M1: Writing equation \(80e^{-0.1t} - 80e^{-0.5t} = 30\) and attempting GDC root finder.
A1: Obtaining critical times \(t_1 \approx 1.41\) and \(t_2 \approx 9.59\).
A1: Finding duration as \(t_2 - t_1 \approx 8.18\text{ days}\).

**(a)**
(i) Let the initial distribution of bicycles be represented by the row vector:
\[\mathbf{s}_0 = \begin{pmatrix} 200 & 200 & 200 \end{pmatrix}\]
After 1 day, the distribution \(\mathbf{s}_1\) is:
\[\mathbf{s}_1 = \mathbf{s}_0 \mathbf{P} = \begin{pmatrix} 200 & 200 & 200 \end{pmatrix} \begin{pmatrix} 0.5 & 0.3 & 0.2 \\ 0.2 & 0.6 & 0.2 \\ 0.3 & 0.1 & 0.6 \end{pmatrix}\]
- Expected at C: \(200(0.5) + 200(0.2) + 200(0.3) = 100 + 40 + 60 = 200\)
- Expected at R: \(200(0.3) + 200(0.6) + 200(0.1) = 60 + 120 + 20 = 200\)
- Expected at U: \(200(0.2) + 200(0.2) + 200(0.6) = 40 + 40 + 120 = 200\)

So after 1 day, there are \(200\) bicycles at Town Centre, \(200\) at Railway Station, and \(200\) at University Campus.

(ii) Since the state remains unchanged, \(\mathbf{s}_2 = \mathbf{s}_1\mathbf{P} = \mathbf{s}_1\). Therefore, after 2 days, the expected number of bicycles at each station is still \(200\) at C, \(200\) at R, and \(200\) at U.

**(b)**
Let \(\mathbf{s} = \begin{pmatrix} c & r & u \end{pmatrix}\) be the steady-state distribution, where \(c + r + u = 1\).
Solve the equation \(\mathbf{s} \mathbf{P} = \mathbf{s}\):
\[\begin{pmatrix} c & r & u \end{pmatrix} \begin{pmatrix} 0.5 & 0.3 & 0.2 \\ 0.2 & 0.6 & 0.2 \\ 0.3 & 0.1 & 0.6 \end{pmatrix} = \begin{pmatrix} c & r & u \end{pmatrix}\]
This yields the system of equations:
1) \(0.5c + 0.2r + 0.3u = c \implies -0.5c + 0.2r + 0.3u = 0 \implies -5c + 2r + 3u = 0\)
2) \(0.3c + 0.6r + 0.1u = r \implies 0.3c - 0.4r + 0.1u = 0 \implies 3c - 4r + u = 0\)
3) \(0.2c + 0.2r + 0.6u = u \implies c + r - 2u = 0 \implies r = 2u - c\)

Substitute \(r = 2u - c\) into (2):
\[3c - 4(2u - c) + u = 0 \implies 7c - 7u = 0 \implies c = u\]
Since \(c = u\), we have \(r = 2u - u = u\).
Using \(c + r + u = 1 \implies u + u + u = 1 \implies 3u = 1 \implies u = \frac{1}{3}\).
Thus, \(c = \frac{1}{3}\), \(r = \frac{1}{3}\), \(u = \frac{1}{3}\).
The steady-state distribution vector is \(\begin{pmatrix} \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \end{pmatrix}\).

**(c)**
(i) Since \(10\%\) of bicycles at \(U\) are moved to \(C\), while \(C\) and \(R\) remain unchanged, the rebalancing matrix \(\mathbf{M}\) is:
\[\mathbf{M} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0.1 & 0 & 0.9 \end{pmatrix}\]

(ii) Show \(\mathbf{Q} = \mathbf{P}\mathbf{M}\):
\[\mathbf{Q} = \begin{pmatrix} 0.5 & 0.3 & 0.2 \\ 0.2 & 0.6 & 0.2 \\ 0.3 & 0.1 & 0.6 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0.1 & 0 & 0.9 \end{pmatrix}\]
\[= \begin{pmatrix} 0.5(1) + 0.2(0.1) & 0.3 & 0.2(0.9) \\ 0.2(1) + 0.2(0.1) & 0.6 & 0.2(0.9) \\ 0.3(1) + 0.6(0.1) & 0.1 & 0.6(0.9) \end{pmatrix} = \begin{pmatrix} 0.52 & 0.30 & 0.18 \\ 0.22 & 0.60 & 0.18 \\ 0.36 & 0.10 & 0.54 \end{pmatrix}\]

(iii) Solve \(\mathbf{w} \mathbf{Q} = \mathbf{w}\) where \(\mathbf{w} = \begin{pmatrix} c & r & u \end{pmatrix}\) and \(c + r + u = 1\):
1) \(0.52c + 0.22r + 0.36u = c \implies -0.48c + 0.22r + 0.36u = 0\)
2) \(0.30c + 0.60r + 0.10u = r \implies 0.3c - 0.4r + 0.1u = 0 \implies u = 4r - 3c\)
3) \(0.18c + 0.18r + 0.54u = u \implies 9c + 9r - 23u = 0\)

Substitute \(u = 4r - 3c\) into equation (3):
\[9c + 9r - 23(4r - 3c) = 0 \implies 78c - 83r = 0 \implies r = \frac{78}{83}c\]
Then, \(u = 4\left(\frac{78}{83}c\right) - 3c = \frac{63}{83}c\).
Using \(c + r + u = 1\):
\[c + \frac{78}{83}c + \frac{63}{83}c = 1 \implies \frac{224}{83}c = 1 \implies c = \frac{83}{224} \approx 0.3705\]
\[r = \frac{78}{224} \approx 0.3482\text{, and } u = \frac{63}{224} = \frac{9}{32} \approx 0.2813\]

The original steady-state proportion at University was \(\frac{1}{3} \approx 0.3333\).
The new steady-state proportion is \(\frac{9}{32} \approx 0.2813\).

Reduction: \(\frac{1}{3} - \frac{9}{32} = \frac{5}{96} \approx 0.0521\) (or a reduction of \(5.21\%\)).

**(d)**
- **Hypotheses:**
\(H_0\): The observed trip transitions are consistent with the transition probabilities predicted by the original model \(\mathbf{P}\).
\(H_1\): The observed trip transitions are not consistent with the transition probabilities predicted by the original model \(\mathbf{P}\).

- **Expected Frequencies:**
Multiply the total trips from each starting station by the corresponding transition probabilities:
- Starting from C (Total 300):
Expected to C: \(300 \times 0.5 = 150\)
Expected to R: \(300 \times 0.3 = 90\)
Expected to U: \(300 \times 0.2 = 60\)
- Starting from R (Total 400):
Expected to C: \(400 \times 0.2 = 80\)
Expected to R: \(400 \times 0.6 = 240\)
Expected to U: \(400 \times 0.2 = 80\)
- Starting from U (Total 300):
Expected to C: \(300 \times 0.3 = 90\)
Expected to R: \(300 \times 0.1 = 30\)
Expected to U: \(300 \times 0.6 = 180\)

- **Degrees of Freedom:**
Since there are 3 independent starting stations (representing 3 distinct multinomial distributions), and each has 3 outcome categories:
\(df = 3 \times (3 - 1) = 6\).

- **Calculation of \(\chi^2\) Statistic:**
\[\chi^2 = \sum \frac{(O - E)^2}{E}\]
\[= \frac{(160 - 150)^2}{150} + \frac{(80 - 90)^2}{90} + \frac{(60 - 60)^2}{60} + \frac{(90 - 80)^2}{80} + \frac{(250 - 240)^2}{240} + \frac{(60 - 80)^2}{80} + \frac{(110 - 90)^2}{90} + \frac{(40 - 30)^2}{30} + \frac{(150 - 180)^2}{180}\]
\[= \frac{100}{150} + \frac{100}{90} + 0 + \frac{100}{80} + \frac{100}{240} + \frac{400}{80} + \frac{400}{90} + \frac{100}{30} + \frac{900}{180}\]
\[\approx 0.6667 + 1.1111 + 0 + 1.25 + 0.4167 + 5 + 4.4444 + 3.3333 + 5 = 21.2222 \approx 21.2\]

- **Critical Value / \(p\)-value:**
For \(df = 6\) at the \(5\%\) significance level:
- Critical value \(\chi^2_{0.05} \approx 12.592\).
- The \(p\)-value corresponding to \(\chi^2 = 21.22\) is \(0.00167\).

- **Conclusion:**
Since \(\chi^2_{\text{calc}} = 21.22 > 12.592\) (or \(p\text{-value} \approx 0.00167 < 0.05\)), we reject the null hypothesis \(H_0\).
There is sufficient evidence at the \(5\%\) level of significance to conclude that the observed trip data is not consistent with the original model's transition probabilities.

**(a)**
M1: Setting up calculation \(\mathbf{s}_0\mathbf{P}\).
A1: Correct values for day 1 (C: 200, R: 200, U: 200).
M1: Recognizing that because \(\mathbf{s}_1 = \mathbf{s}_0\), \(\mathbf{s}_2 = \mathbf{s}_1\mathbf{P}\).
A2: Correct values for day 2 (C: 200, R: 200, U: 200) (Award A1 for any two correct).

**(b)**
M1: Writing down system of equations or using GDC matrix steady-state finder.
A2: Correct equations in terms of proportions \(c, r, u\).
M2: Systematic algebraic steps to solve the linear system.
A1: Final correct fractions: \(c = 1/3, r = 1/3, u = 1/3\).

**(c)**
(i) A1: Correct matrix \(\mathbf{M}\).
(ii) M1: Multiplying \(\mathbf{P}\mathbf{M}\) with clear working shown.
A1: Verifying all entries of the matrix \(\mathbf{Q}\).
(iii) M1: Setting up equations for new steady state \(\mathbf{w}\mathbf{Q} = \mathbf{w}\).
A1: Solving to get new proportions: \(c = 83/224, r = 78/224, u = 9/32\) (accept decimal equivalents to 3sf: \(0.371, 0.348, 0.281\)).
M1: Subtracting new proportion of University from original (\(1/3 - 9/32\) or \(0.3333 - 0.2813\)).
A1: Finding the reduction as \(5/96 \approx 0.0521\) (or \(5.21\%\)).

**(d)**
**Hypotheses:**
A1: Correct null and alternative hypotheses stated in context.
**Expected Frequencies:**
M1: Showing method to find expected frequencies.
A2: All expected frequencies calculated correctly (Award A1 if only 1 or 2 mistakes).
**Degrees of Freedom:**
A1: Stating \(df = 6\) with justification or calculation shown.
**Chi-squared Statistic:**
M1: Setting up the formula \(\sum \frac{(O-E)^2}{E}\).
A2: Correctly calculated value \(\chi^2 \approx 21.2\) (accept \(21.22\)). (Award A1 for minor arithmetic error).
**Decision and Conclusion:**
A1: Stating critical value (\(12.59\)) or \(p\)-value (\(0.00167\)) and comparing with \(0.05\).
R1: Rejecting \(H_0\) and writing a correct, context-based conclusion.