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(a) Using the compound depreciation formula:
\( V = P(1 - r)^t \)
where \( P = 24000 \), \( r = 0.085 \), and \( t = 6 \).

\( V = 24000(1 - 0.085)^6 = 24000(0.915)^6 \approx 14084.05 \)

Correct to the nearest dollar, the value is $14,084.

(b) We want to find the smallest integer \( t \) such that:
\( 24000(0.915)^t < 10000 \)

\( (0.915)^t < \frac{10000}{24000} \approx 0.416667 \)

Taking the natural logarithm of both sides:
\( t \ln(0.915) < \ln(0.416667) \)

Since \( \ln(0.915) \) is negative, we reverse the inequality when dividing:
\( t > \frac{\ln(0.416667)}{\ln(0.915)} \approx 9.85 \)

Therefore, the number of complete years is 10.

M1 for setting up the correct equation/expression for part a: \( 24000(1-0.085)^6 \).
A1 for $14,084 (accept 14084).
M1 for setting up the inequality or equation for part b: \( 24000(0.915)^t < 10000 \) (or equivalent).
M1 for an attempt to solve using logarithms or GDC table.
A1 for 9.85.
A1 for 10 (years) as the final integer answer.

(a) The profit function is a quadratic function of the form \( P(x) = ax^2 + bx + c \) where \( a = -0.4 \) and \( b = 6.8 \). The maximum occurs at the vertex:
\( x = -\frac{b}{2a} = -\frac{6.8}{2(-0.4)} = 8.5 \)

Since \( x \) is in hundreds of items, the number of items is:
\( 8.5 \times 100 = 850 \) items.

(b) Substitute \( x = 8.5 \) into the profit function to find the maximum profit:
\( P(8.5) = -0.4(8.5)^2 + 6.8(8.5) - 12 \)
\( P(8.5) = -0.4(72.25) + 57.8 - 12 = -28.9 + 57.8 - 12 = 16.9 \)

Since \( P \) is in thousands of dollars, the maximum weekly profit is:
\( 16.9 \times 1000 = \$16,900 \).

M1 for an attempt to find the vertex of the quadratic, e.g., using \( x = -\frac{b}{2a} \) or setting the derivative equal to 0: \( P'(x) = -0.8x + 6.8 = 0 \).
A1 for \( x = 8.5 \).
A1 for converting to the number of items: 850.
M1 for substituting their value of \( x \) back into the profit function.
A1 for 16.9 (thousand dollars).
A1 for translating this to the final answer in dollars: $16,900.

(a) First find the midpoint of \( AB \):
\( M = \left(\frac{2+8}{2}, \frac{8+6}{2}\right) = (5, 7) \)

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

The gradient of the perpendicular bisector is the negative reciprocal:
\( m_{\perp} = 3 \)

The equation of the perpendicular bisector of \( AB \) is:
\( y - 7 = 3(x - 5) \implies y = 3x - 8 \)

(b) The point \( P \) is the circumcentre, which lies at the intersection of the perpendicular bisectors.
Let's find the perpendicular bisector of \( BC \).
Midpoint of \( BC \):
\( N = \left(\frac{8+4}{2}, \frac{6+2}{2}\right) = (6, 4) \)

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

Gradient of perpendicular bisector of \( BC \):
\( m = -1 \)

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

Equating the two perpendicular bisectors:
\( 3x - 8 = -x + 10 \implies 4x = 18 \implies x = 4.5 \)

Substituting back to find \( y \):
\( y = -4.5 + 10 = 5.5 \)

Thus, the coordinates of \( P \) are \( (4.5, 5.5) \).

M1 for finding the midpoint of \( AB \): \( (5, 7) \).
M1 for finding the gradient of \( AB \): \( -\frac{1}{3} \) and its perpendicular gradient: \( 3 \).
A1 for the correct equation of the perpendicular bisector of \( AB \): \( y = 3x - 8 \) (or equivalent).
M1 for finding the equation of a second perpendicular bisector (e.g., \( BC \): \( y = -x + 10 \) or \( AC \): \( y = -0.5x + 7.75 \)).
M1 for solving the system of equations of the two perpendicular bisectors.
A1 for the correct coordinates of \( P \): \( (4.5, 5.5) \).

(a) Let \( X \) be the weight of an apple, so \( X \sim N(150, 12^2) \).
We want to find \( P(140 < X < 165) \).
Using a graphic display calculator with the lower limit as 140, upper limit as 165, \( \mu = 150 \), and \( \sigma = 12 \):
\( P(140 < X < 165) \approx 0.692019... \)
Correct to 3 significant figures, this is 0.692.

(b) We are given that \( P(X < w) = 0.15 \).
Using the inverse normal function on a graphic display calculator with area = 0.15, \( \mu = 150 \), and \( \sigma = 12 \):
\( w \approx 137.562... \)
Correct to 1 decimal place, \( w = 137.6 \) grams.

M1 for expressing the probability statement \( P(140 < X < 165) \) (or showing standardizing steps).
A2 for 0.692 (accept 0.692019...).
M1 for stating the inverse normal probability equation \( P(X < w) = 0.15 \) or equivalent.
A1 for 137.562...
A1 for 137.6 (grams) (correct to 1 decimal place as requested).

(a) The null hypothesis \( H_0 \): A student's favorite sport is independent of their gender.

(b) The expected value is calculated as:
\( E = \frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}} \)
For Female and Soccer:
\( E = \frac{100 \times 75}{200} = 37.5 \).

(c) Using the GDC's \( \chi^2 \) two-way test with the observed matrix:
\( \chi^2 \approx 5.333 \)
with degrees of freedom \( df = (2-1) \times (3-1) = 2 \).
The corresponding \( p \)-value is:
\( p \approx 0.06948... \approx 0.0695 \).
Since the \( p \)-value \( (0.0695) \) is greater than the significance level \( 0.05 \), we do not reject (or we fail to reject) the null hypothesis. There is insufficient evidence to suggest that favorite sport and gender are dependent.

A1 for stating the correct null hypothesis (must mention independence).
M1 for the expected value formula: \( \frac{100 \times 75}{200} \).
A1 for 37.5.
A1 for the correct \( p \)-value of 0.0695 (accept 0.06948...).
R1 for comparing the \( p \)-value to 0.05 (e.g., \( 0.0695 > 0.05 \)).
A1 for the correct conclusion in context (do not reject the null hypothesis).

(a) To find the points of intersection, we set the two equations equal:
\( 6x - x^2 = 2x \)
\( 4x - x^2 = 0 \)
\( x(4 - x) = 0 \)
Thus, the \( x \)-coordinates are \( x = 0 \) and \( x = 4 \).

(b) The area \( A \) is given by the definite integral of the upper function minus the lower function from \( x = 0 \) to \( x = 4 \):
\( A = \int_{0}^{4} ((6x - x^2) - 2x) \, dx = \int_{0}^{4} (4x - x^2) \, dx \)
Integrating each term:
\( A = \left[ 2x^2 - \frac{x^3}{3} \right]_0^4 \)
\( A = \left( 2(4)^2 - \frac{4^3}{3} \right) - 0 \)
\( A = 32 - \frac{64}{3} = \frac{96 - 64}{3} = \frac{32}{3} \approx 10.6667... \)
Correct to 3 significant figures, the area is 10.7.

M1 for setting up the equation \( 6x - x^2 = 2x \).
A1 for finding the correct \( x \)-values: 0 and 4.
M1 for setting up the correct definite integral: \( \int_{0}^{4} (4x - x^2) \, dx \).
M1 for finding the antiderivative \( 2x^2 - \frac{x^3}{3} \) (or for directly using GDC numerical integration).
A1 for the exact value \( \frac{32}{3} \) or decimal value 10.7 (accept 10.6667...).

(a) The trapezoidal rule formula is:
\( A \approx \frac{h}{2} \left( y_0 + 2(y_1 + y_2 + y_3 + \dots + y_{n-1}) + y_n \right) \)
Here, the interval width \( h = 5 \) and \( n = 4 \).
\( A \approx \frac{5}{2} \left( 12 + 2(18 + 22 + 17) + 10 \right) \)
\( A \approx 2.5 \left( 12 + 2(57) + 10 \right) \)
\( A \approx 2.5 \left( 12 + 114 + 10 \right) \)
\( A \approx 2.5 \times 136 = 340 \text{ m}^2 \).

(b) The accuracy of this estimate can be improved by taking more measurements at smaller, more frequent intervals (decreasing \( h \)).

M1 for identifying \( h = 5 \).
M1 for substituting the values correctly into the trapezoidal rule formula: \( \frac{5}{2} \left[ 12 + 2(18+22+17) + 10 \right] \).
A2 for the correct estimate of 340 (\(\text{m}^2\)).
R1 for stating that taking more measurements / smaller intervals / smaller width would improve accuracy.

(a) Based on the given information, we write down the equations:
Monday: \( 2x + 3y + z = 23.50 \)
Tuesday: \( 3x + y + 2z = 21.50 \)
Wednesday: \( x + 2y + 3z = 24.00 \)

(b) We can write this system in matrix form:
\( \begin{pmatrix} 2 & 3 & 1 \\ 3 & 1 & 2 \\ 1 & 2 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 23.50 \\ 21.50 \\ 24.00 \end{pmatrix} \)

Using a GDC to find the reduced row-echelon form (RREF) or the inverse matrix:
\( \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 2 & 3 & 1 \\ 3 & 1 & 2 \\ 1 & 2 & 3 \end{pmatrix}^{-1} \begin{pmatrix} 23.50 \\ 21.50 \\ 24.00 \end{pmatrix} = \begin{pmatrix} 3.00 \\ 4.50 \\ 4.00 \end{pmatrix} \)

Thus, the price of an Espresso is $3.00, the price of a Latte is $4.50, and the price of a Cappuccino is $4.00.

A2 for writing down all three correct equations (award A1 if only two are correct).
M2 for setting up the matrix equation \( AX = B \) or indicating the use of GDC to solve the system.
A2 for the correct solutions: \( x = 3 \) (or Espresso: $3.00), \( y = 4.5 \) (or Latte: $4.50), \( z = 4 \) (or Cappuccino: $4.00).

We can model this using a financial calculator or TVM solver with the following parameters:
- Number of periods, \(N = 60\)
- Nominal annual interest rate, \(I\% = 3.6\)
- Present Value, \(PV = 0\)
- Monthly payment, \(PMT = -500\)
- Payments per year, \(P/Y = 12\)
- Compounding periods per year, \(C/Y = 12\)

Using the TVM solver on a graphic display calculator:
\(FV = 32824.60\)

Rounding to the nearest dollar, the total value of her savings is $32,825.

M1 for identifying the correct parameters for the TVM solver: \(N = 60, I\% = 3.6, PMT = -500\).
A1 for setup/inputs used correctly on the GDC.
A1 for the unrounded value \(32824.60\).
A1 for rounding correctly to the nearest dollar to obtain 32825.

(a) Substitute \(t = 0\) and \(T(0) = 85\) into the equation:
\(85 = 22 + a e^0\)
\(85 = 22 + a \implies a = 63\)

(b) Substitute \(t = 10\), \(a = 63\), and \(T(10) = 48\) into the equation:
\(48 = 22 + 63 e^{-10k}\)
\(26 = 63 e^{-10k}\)
\(e^{-10k} = \frac{26}{63}\)
\(-10k = \ln\left(\frac{26}{63}\right) \approx -0.88503\)
\(k \approx 0.0885\) (to 3 significant figures)

(c) Substitute \(t = 20\) into the model:
\(T(20) = 22 + 63 e^{-20 \times 0.088503}\)
\(T(20) = 22 + 63 \left(\frac{26}{63}\right)^2 \approx 32.73\)
To 3 significant figures, the temperature is \(32.7^\circ\text{C}\).

(a) M1 for substituting \(t = 0\), A1 for \(a = 63\).
(b) M1 for substituting their \(a\) and \(t = 10\) into the equation, M1 for taking the natural logarithm, A1 for \(k \approx 0.0885\).
(c) M1 for substituting \(t = 20\) into their equation, A1 for \(32.7^\circ\text{C}\).

(a) First, find the midpoint of the line segment \(AB\):
\(M = \left(\frac{2+8}{2}, \frac{8+6}{2}\right) = (5, 7)\)

Next, find the gradient of the line segment \(AB\):
\(m_{AB} = \frac{6 - 8}{8 - 2} = \frac{-2}{6} = -\frac{1}{3}\)

The gradient of the perpendicular bisector is the negative reciprocal of \(m_{AB}\):
\(m_{\perp} = -\frac{1}{-1/3} = 3\)

Using the point-slope form with midpoint \((5, 7)\):
\(y - 7 = 3(x - 5)\)
\(y = 3x - 15 + 7 \implies y = 3x - 8\)

(b) The vertex of the Voronoi cells occurs at the intersection of the perpendicular bisectors. Solve the system of equations:
\(y = 3x - 8\)
\(y = x - 2\)

Set the equations equal to each other:
\(3x - 8 = x - 2\)
\(2x = 6 \implies x = 3\)

Substitute \(x = 3\) back into one of the equations to find \(y\):
\(y = 3 - 2 = 1\)

The coordinates of the vertex are \((3, 1)\).

(a) M1 for finding the midpoint \((5, 7)\), M1 for finding the gradient of \(AB\) as \(-\frac{1}{3}\), M1 for using the perpendicular gradient \(3\), A1 for \(y = 3x - 8\).
(b) M1 for setting up the system of linear equations, A1 for finding the correct intersection point \((3, 1)\).

(a) The null hypothesis \(H_0\): Preferred art form and grade level are independent.

(b) The row total for Grade 10 is \(35 + 15 + 20 = 70\).
The column total for Drama is \(15 + 25 = 40\).
The grand total is 150.
Expected frequency = \(\frac{70 \times 40}{150} \approx 18.67\) (or 18.7 to 3 s.f.).

(c) Enter the matrix of observed values into the GDC to conduct a \(\chi^2\) two-way test.
The calculator yields:
\(\chi^2 \approx 5.524\), df = 2, and \(p\)-value \(\approx 0.0632\).

(d) Since the \(p\)-value \((0.0632)\) is greater than the significance level \((0.05)\), we do not reject the null hypothesis.

(a) A1 for stating the null hypothesis clearly.
(b) M1 for finding row and column totals, A1 for \(18.7\) (accept \(18.67\)).
(c) A2 for \(0.0632\) (accept \(0.0631\) to \(0.0633\)).
(d) R1 for comparing their \(p\)-value to \(0.05\) and drawing the correct conclusion (Do not reject).

Let \(X\) be the weight of a cereal box, where \(X \sim N(505, 4^2)\).

(a) Using normal cumulative distribution on a GDC:
\(P(X < 500) \approx 0.10565 \approx 0.106\).

(b) We require \(P(X < w) = 0.025\).
Using the inverse normal distribution function on a GDC:
\(w \approx 497.16\) grams.
To the nearest tenth of a gram, the minimum acceptable weight is \(497.2\) g.

(c) First, find the probability that a box weighs between 502 and 510 grams:
\(P(502 \le X \le 510) \approx 0.66775\).

The expected number of boxes is:
\(1000 \times 0.66775 \approx 667.75\).
To the nearest integer, the expected number is 668 boxes.

(a) M1 for setting up normal cdf with correct parameters, A1 for \(0.106\).
(b) M1 for using inverse normal with area 0.025, A1 for \(497.2\) g.
(c) M1 for calculating \(P(502 \le X \le 510) \approx 0.668\), A1 for multiplying by 1000 to get \(668\) (accept \(667\)).

(a) Using the formula for the \(n\)-th term of an arithmetic sequence, \(u_n = u_1 + (n-1)d\):
\(u_3 = u_1 + 2d = 14\)
\(u_{12} = u_1 + 11d = 50\)

Subtracting the first equation from the second equation:
\(9d = 36 \implies d = 4\)

Substitute \(d = 4\) back into the first equation:
\(u_1 + 2(4) = 14 \implies u_1 = 6\)

(b) The sum of the first \(n\) terms is \(S_n = \frac{n}{2}[2u_1 + (n-1)d]\).
For \(n = 25\):
\(S_{25} = \frac{25}{2}[2(6) + 24(4)]\)
\(S_{25} = 12.5 [12 + 96] = 12.5 \times 108 = 1350\).

(a) M1 for setting up two simultaneous equations, A1 for \(d = 4\), A1 for \(u_1 = 6\).
(b) M1 for substituting their values into the arithmetic sum formula, A1 for \(1350\).

(a) Let \(y\) be the side parallel to the brick wall.
The total perimeter of the three fenced sides is:
\(2x + y = 80 \implies y = 80 - 2x\)

The area \(A\) of the rectangle is:
\(A(x) = x \cdot y = x(80 - 2x) = 80x - 2x^2\).

(b) Differentiate \(A(x)\) with respect to \(x\):
\(\frac{\mathrm{d}A}{\mathrm{d}x} = 80 - 4x\).

(c) To find the maximum area, set the derivative equal to 0:
\(80 - 4x = 0 \implies x = 20\)

Substitute \(x = 20\) back into the area equation:
\(A(20) = 80(20) - 2(20)^2 = 1600 - 800 = 800\text{ m}^2\).

(a) M1 for expressing the perimeter constraint \(2x + y = 80\), A1 for showing the substitution to arrive at the given equation.
(b) A1 for \(\frac{\mathrm{d}A}{\mathrm{d}x} = 80 - 4x\).
(c) M1 for setting their derivative to 0, A1 for finding \(x = 20\), A1 for the final area of \(800\text{ m}^2\).

(a) Acceleration is the derivative of velocity:
\(a(t) = v'(t) = 6t - 12\)

At \(t = 3\):
\(a(3) = 6(3) - 12 = 6\text{ m s}^{-2}\)

(b) The particle is at rest when \(v(t) = 0\):
\(3t^2 - 12t + 9 = 0\)
\(3(t^2 - 4t + 3) = 0\)
\(3(t - 1)(t - 3) = 0\)

So, the particle is at rest at \(t = 1\) second and \(t = 3\) seconds.

(c) The total distance travelled is given by \(\int_0^3 |v(t)| \mathrm{d}t\).
Since the particle changes direction at \(t = 1\), we split the integral:
\(\text{Distance} = \int_0^1 (3t^2 - 12t + 9) \mathrm{d}t - \int_1^3 (3t^2 - 12t + 9) \mathrm{d}t\)

The antiderivative is \(s(t) = t^3 - 6t^2 + 9t\).

- \(s(0) = 0\)
- \(s(1) = 1 - 6 + 9 = 4\)
- \(s(3) = 27 - 54 + 27 = 0\)

Distance in \([0,1]\) is \(|s(1) - s(0)| = 4\).
Distance in \([1,3]\) is \(|s(3) - s(1)| = |0 - 4| = 4\).

Total distance travelled is \(4 + 4 = 8\) meters.

(a) M1 for differentiating \(v(t)\) to get \(6t - 12\), A1 for \(6\).
(b) M1 for setting \(v(t) = 0\), A1 for finding both \(t = 1\) and \(t = 3\).
(c) M1 for recognizing that integration of the absolute value is required, M1 for dividing into correct sub-intervals (or using GDC directly), A1 for \(8\) m.

(a) Let \(X\) be the weight of an apple in grams. \(X \sim N(150, 12^2)\).
Using a graphic display calculator:
\(P(X > 165) \approx 0.105649... \approx 0.106\) (to 3 significant figures).

(b) We require \(P(X < w) = 0.15\).
Using the inverse normal function on a graphic display calculator:
\(w \approx 137.562... \approx 138\) (to 3 significant figures).

(c) Let \(Y\) be the number of apples weighing more than \(165\text{ g}\) in a box of \(8\).
\(Y\) follows a binomial distribution with \(n = 8\) and probability of success \(p = 0.105649...\).
\(Y \sim B(8, 0.105649...)\).
We want to find \(P(Y = 2)\).
Using a binomial probability distribution function on a graphic display calculator:
\(P(Y = 2) = \binom{8}{2} (0.105649...)^2 (1 - 0.105649...)^6 \approx 0.160408... \approx 0.160\) (to 3 significant figures).

(a)
(M1) for setting up the normal distribution calculation, e.g. \(P(X > 165)\).
(A1) for \(0.106\) (accept \(0.105649...\)).

(b)
(M1) for setting up the inverse normal calculation, e.g. \(P(X < w) = 0.15\).
(A1) for \(138\) (accept \(137.562...\)).

(c)
(M1) for identifying the binomial distribution, e.g. \(Y \sim B(8, \text{their } p)\).
(M1) for setting up the binomial probability \(P(Y=2)\).
(A0.47) for \(0.160\) (accept \(0.160408...\); award (A0.47)(ft) if using their rounded \(0.106\) from (a) which yields \(0.161\)).

(a) The volume of the box is given by \(V = x^2 h = 4000\), which implies \(h = \frac{4000}{x^2}\).
The cost of constructing the base is \(5 \times x^2 = 5x^2\).
The cost of the four sides is \(4 \times (x \times h) \times 3 = 12 x h\).
Substituting \(h = \frac{4000}{x^2}\) into the cost equation:
\(C(x) = 5x^2 + 12x\left(\frac{4000}{x^2}\right) = 5x^2 + \frac{48000}{x}\).

(b) Differentiating \(C(x)\) with respect to \(x\):
\(C'(x) = 10x - \frac{48000}{x^2}\).

(c) To minimize the cost, set \(C'(x) = 0\):
\(10x - \frac{48000}{x^2} = 0 \implies 10x^3 = 48000 \implies x^3 = 4800\).
\(x = \sqrt[3]{4800} \approx 16.9\text{ m}\) (16.8687... m).

(d) Substituting the value of \(x\) back into the cost function:
\(C(16.8687) = 5(16.8687)^2 + \frac{48000}{16.8687} \approx 1422.62 + 2845.50 = \$4268.12\).
To 3 significant figures, the minimum cost is $4270 (or $4268 exact).

(e) We require \(5x^2 + \frac{48000}{x} \le 6500\).
Using the GDC to find the intersection points of \(y = 5x^2 + \frac{48000}{x}\) and \(y = 6500\):
\(x_1 \approx 7.74\text{ m}\) and \(x_2 \approx 31.6\text{ m}\).
Therefore, the range of values is \(7.74 \le x \le 31.6\).

(a) M1: For expressing volume \(V = x^2 h = 4000\) and isolating \(h\).
M1: For writing the correct cost components (base and sides).
M1: For substituting \(h\) in terms of \(x\).
A1: For arriving at the correct given cost equation.

(b) M1: For power rule application.
A1: For the correct derivative \(10x - \frac{48000}{x^2}\).

(c) M1: For setting \(C'(x) = 0\).
M1: For algebraic steps to get \(x^3 = 4800\).
A1: For \(x \approx 16.9\text{ m}\).

(d) M1: For substituting their value of \(x\) into \(C(x)\).
A1: For the correct minimum cost (accept $4270 or $4268).

(e) M1: For setting up the inequality or equation \(5x^2 + \frac{48000}{x} = 6500\).
A1: For locating both boundary points \(7.74\) and \(31.6\).
A1: For writing the final answer as an inequality or interval \([7.74, 31.6]\).

(a) \(H_0\): Employee department and preferred work style are independent.

(b) Column total for Hybrid is \(45 + 35 + 40 = 120\). Row total for Engineering is \(50 + 35 + 15 = 100\). Total sample size is 300.
\(E = \frac{100 \times 120}{300} = 40\).

(c) Degrees of freedom \(= (r-1)(c-1) = (3-1)(3-1) = 4\).

(d) Using a GDC to conduct a \(\chi^2\) test for independence on the contingency table:
\(\chi^2 \approx 27.1\) (27.124...)
\(p\text{-value} \approx 1.87 \times 10^{-5}\) (0.0000187).

(e) Since the \(p\)-value \(\approx 1.87 \times 10^{-5} < 0.05\), we reject \(H_0\). There is significant evidence to suggest that department and preferred work style are not independent.

(f) \(H_0: \mu_1 = \mu_2\) and \(H_1: \mu_1 > \mu_2\), where \(\mu_1\) is the population mean for Sales and \(\mu_2\) is the population mean for Engineering.

(g) Using a GDC to conduct a two-sample t-test (one-tailed, unequal variances):
\(t \approx 6.09\), and \(p\text{-value} \approx 2.22 \times 10^{-6}\).

(h) Since \(p\text{-value} \approx 2.22 \times 10^{-6} < 0.01\), we reject the null hypothesis. There is strong evidence that the mean hours spent in meetings by Sales employees is greater than that of Engineering employees.

(a) A1: For correctly stating that the two variables are independent.

(b) M1: For calculating the expected frequency formula.
A1: For the correct expected value \(40\).

(c) A1: For the correct degrees of freedom \(4\).

(d) M1: For attempting to use the GDC for a Chi-squared test.
A1: For \(\chi^2 \approx 27.1\).
A1: For \(p\text{-value} \approx 1.87 \times 10^{-5}\).

(e) R1: For comparing the \(p\)-value to \(0.05\).
A1: For the correct conclusion of rejecting \(H_0\).

(f) A1: For both hypotheses stated correctly using appropriate notation.

(g) M1: For inputting data into the GDC and choosing a one-tailed unpooled t-test.
A1: For \(p\text{-value} \approx 2.22 \times 10^{-6}\).

(h) R1: For comparing \(p\)-value with \(0.01\).
A1: For rejecting \(H_0\) and stating the conclusion in context.

(a) \(W \sim N(150, 15^2)\)
(i) Using GDC normal CDF with lower = 130, upper = 160, \(\mu = 150\), \(\sigma = 15\):
\(P(130 < W < 160) \approx 0.656\).
(ii) Using GDC normal CDF with lower = 175, upper = infinity:
\(P(W > 175) \approx 0.0478\).

(b) \(P(W < 120) = P\left(Z < \frac{120 - 150}{15}\right) = P(Z < -2) \approx 0.02275\).
Since \(0.02275 \approx 0.0228\), the result is shown.

(c) Let \(X\) be the number of small apples in a sample of 50. Then \(X \sim B(50, 0.0228)\).
(i) \(E(X) = n p = 50 \times 0.0228 = 1.14\).
(ii) Using binomial CDF with \(n = 50\), \(p = 0.0228\), and \(x = 2\):
\(P(X \le 2) \approx 0.889\).

(d) Let \(C\) be the packaging cost of a single apple. The probability of an apple being small is \(0.0228\) and other is \(1 - 0.0228 = 0.9772\).
The expected packaging cost of one apple is:
\(E(C) = 0.05 \times 0.0228 + 0.12 \times 0.9772 = 0.00114 + 0.117264 = \$0.118404\).
For a sample of 100 apples, the expected total packaging cost is:
\(100 \times E(C) = 100 \times 0.118404 = \$11.84\).

(a) M1: For identifying the normal distribution model.
A1: For the correct probability \(0.656\) in (i).
M1: For set-up of the upper-tailed probability.
A1: For the correct probability \(0.0478\) in (ii).

(b) M1: For standardizing the value \(120\) or setting up the GDC calculation.
A1: For getting \(0.02275\) and concluding it is approximately \(0.0228\).

(c) M1: For applying \(E(X) = np\).
A1: For \(1.14\).
M1: For identifying binomial distribution model \(B(50, 0.0228)\).
A1: For \(0.889\) (accept \(0.891\) if using exact value \(0.02275\)).

(d) M1: For setting up the expected cost equation for 1 apple.
A1: For the correct probabilities used in the cost calculation.
M1: For multiplying the single-apple expected cost by 100.
A1: For the final cost of $11.84 (accept $11.83 if using exact probabilities).

(a) The vertical increase in height from \(P\) to \(Q\) is given by:
\(\Delta z = 10 \sin(45^\circ) = 10 \times \frac{\sqrt{2}}{2} = 5\sqrt{2} \approx 7.07\text{ m}\).
Since \(P\) has height 15, the total vertical height of \(Q\) is:
\(z_Q = 15 + 5\sqrt{2} \approx 22.07\text{ m}\).
To 3 significant figures, this is \(22.1\text{ m}\).

(b) The horizontal projection of \(PQ\) onto the ground has length:
\(d = 10 \cos(45^\circ) = 5\sqrt{2} \approx 7.071\text{ m}\).
This projection lies along the line \(y = x\) in the first quadrant, so:
\(x^2 + y^2 = d^2 \implies 2x^2 = (5\sqrt{2})^2 = 50 \implies x^2 = 25 \implies x = 5\).
Since \(x = y\) and \(x > 0\), the coordinates are \(x = 5, y = 5\).
Combining with the vertical height from part (a), \(Q\) has coordinates \((5, 5, 22.1)\).

(c) Point \(R\) is at \((12, 5, 0)\). The exact coordinates of \(Q\) are \((5, 5, 15 + 5\sqrt{2}) \approx (5, 5, 22.071)\).
Using the 3D distance formula:
\(QR = \sqrt{(12-5)^2 + (5-5)^2 + (0-22.071)^2} = \sqrt{7^2 + 0^2 + 487.13} = \sqrt{49 + 487.13} = \sqrt{536.13} \approx 23.2\text{ m}\).

(d) The projection of \(Q\) onto the ground is \(Q'(5, 5, 0)\). \(QRQ'\) forms a right-angled triangle at \(Q'\).
The horizontal distance is \(Q'R = 12 - 5 = 7\text{ m}\).
The height of \(Q\) is \(QQ' \approx 22.071\text{ m}\).
Let \(\theta\) be the angle between \(QR\) and the horizontal ground:
\(\tan(\theta) = \frac{22.071}{7} \approx 3.153 \implies \theta \approx 72.4^\circ\) (or \(1.26\text{ radians}\)).

(e) The coordinates of the vertices in the 2D ground plane are \(O(0, 0)\), \(R(12, 5)\), and \(Q'(5, 5)\).
Using the formula for the area of a triangle given coordinates:
\(\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|\)
\(\text{Area} = \frac{1}{2} |0(5 - 5) + 12(5 - 0) + 5(0 - 5)| = \frac{1}{2} |60 - 25| = 17.5\text{ m}^2\).

(a) M1: For calculating the vertical displacement \(10 \sin(45^\circ)\).
A1: For the correct vertical height \(22.1\text{ m}\) (or \(22.07\text{ m}\)).

(b) M1: For calculating the horizontal projection length \(10 \cos(45^\circ)\).
M1: For setting up the equation \(x^2 + y^2 = d^2\) with \(y = x\).
A1: For showing that \(x = 5, y = 5\).
A1: For stating final coordinates with correctly rounded height of \(22.1\).

(c) M1: For attempting to use the 3D distance formula.
M1: For substituting the values correctly.
A1: For the final answer of \(23.2\text{ m}\) (accept \(23.1\) to \(23.3\)).

(d) M1: For identifying the right triangle with horizontal side 7 and vertical side \(22.1\).
M1: For using \(\tan\) or \(\sin\) to set up the angle equation.
A1: For \(72.4^\circ\) (or \(1.26\text{ rad}\)).

(e) M1: For applying a valid 2D area method (vector cross product, determinant, or geometric formula).
A1: For the correct area of \(17.5\text{ m}^2\).

(a) Using GDC TVM Solver for Bank A:
\(N = 25 \times 12 = 300\)
\(I\% = 4.2\)
\(PV = 350000\)
\(FV = 0\)
\(P/Y = 12\)
\(C/Y = 12\)
Solving for \(PMT\):
\(PMT \approx -1888.37\).
So the monthly payment is $1888.37.

(b) Total amount paid to Bank A over 25 years:
\(\text{Total} = 1888.37 \times 300 = \$566,511\).

(c) Total interest paid:
\(\text{Interest} = 566511 - 350000 = \$216,511\).

(d) Using GDC TVM Solver for Bank B:
\(N = 20 \times 12 = 240\)
\(I\% = 3.8\)
\(PV = 350000\)
\(FV = 0\)
\(P/Y = 12\)
\(C/Y = 12\)
Solving for \(PMT\):
\(PMT \approx -2084.79\).
So the monthly payment is $2084.79.

(e) Total paid for Bank B including fee:
\(\text{Total} = 2084.79 \times 240 + 5000 = 500349.60 + 5000 = \$505,349.60\) (or $505,350 to 3sf).

(f) Comparing the two total payments:
Bank A: $566,511
Bank B: $505,349.60
Bank B is cheaper.
Difference \(= 566511 - 505349.60 = \$61,161.40\).
Therefore, Bank B is cheaper by $61,200 (to 3sf) or $61,161.40 exact.

(a) M1: For correctly identifying parameters for the TVM solver.
M1: For attempting to calculate monthly payment with PMT formula or GDC.
A1: For monthly payment of $1888.37.

(b) M1: For multiplying monthly payment by 300.
A1: For the total payment $566,511 (or $566,500).

(c) A1: For correct interest calculation $216,511.

(d) M1: For correct inputs for Bank B (N = 240, I% = 3.8).
M1: For attempting to solve on GDC.
A1: For monthly payment of $2084.79.

(e) M1: For multiplying PMT by 240 and adding $5,000.
A1: For the correct total of $505,350 (or $505,349.60).

(f) R1: For comparison between Bank A and Bank B total costs.
A1: For identifying Bank B as cheaper.
A1: For the correct difference of $61,161 (accept $61,200).

(a) Initial velocity is at \(t = 0\):
\(v(0) = 3(0)^2 - 16(0) + 16 = 16\text{ m s}^{-1}\).

(b) Set \(v(t) = 0\):
\(3t^2 - 16t + 16 = 0 \implies (3t - 4)(t - 4) = 0\).
So, \(t = \frac{4}{3}\text{ s}\) (1.33 s) and \(t = 4\text{ s}\).

(c) Acceleration is the derivative of velocity:
\(a(t) = v'(t) = 6t - 16\).
At \(t = 3\):
\(a(3) = 6(3) - 16 = 2\text{ m s}^{-2}\).

(d) The total distance traveled in the first 4 seconds is:
\(\text{Distance} = \int_0^4 |v(t)| dt = \int_0^4 |3t^2 - 16t + 16| dt\).
Using GDC or analytical integration:
\(\int_0^{4/3} (3t^2 - 16t + 16) dt - \int_{4/3}^4 (3t^2 - 16t + 16) dt\)
Let \(s(t) = t^3 - 8t^2 + 16t\).
\(s(0) = 0\)
\(s(4/3) = \frac{256}{27} \approx 9.48\)
\(s(4) = 0\)
\(\text{Total Distance} = \frac{256}{27} + \left|0 - \frac{256}{27}\right| = \frac{512}{27} \approx 19.0\text{ m}\).

(e) Displacement is given by the definite integral without absolute value:
\(\text{Displacement} = \int_0^6 (3t^2 - 16t + 16) dt\)
\(= [t^3 - 8t^2 + 16t]_0^6 = (6^3 - 8(6)^2 + 16(6)) - 0\)
\(= 216 - 288 + 96 = 24\text{ m}\).

(a) A1: For correct initial velocity \(16\text{ m s}^{-1}\).

(b) M1: For setting \(v(t) = 0\).
A1: For factoring or finding roots correctly.
A1: For both correct times \(t = 1.33\text{ s}\) and \(t = 4\text{ s}\).

(c) M1: For differentiating \(v(t)\).
A1: For substituting \(t = 3\) to get \(2\text{ m s}^{-2}\).

(d) M1: For setting up the absolute value integral \(\int_0^4 |v(t)| dt\).
M1: For splitting the integral or using the GDC absolute value function.
A1: For calculating the distance in the segments (approx. 9.48 each).
A1: For the total distance of \(19.0\text{ m}\) (accept \(\frac{512}{27}\)).

(e) M1: For setting up the definite integral for displacement \(\int_0^6 v(t) dt\).
M1: For correct antiderivative expression \(t^3 - 8t^2 + 16t\).
A1: For applying the limits 0 and 6.
A1: For the final answer of \(24\text{ m}\).

(a) The midline value \(d\) is the average of the maximum and minimum depths:
\(d = \frac{12 + 4}{2} = 8\).

(b) The amplitude \(a\) is half the difference between the maximum and minimum depths:
\(a = \frac{12 - 4}{2} = 4\).
Since the function starts at high tide, we have shown \(a = 4\).

(c) The time from high tide (03:00, or \(t = 3\)) to the subsequent low tide (09:15, or \(t = 9.25\)) represents half the tidal period.
\(\text{Half Period} = 9.25 - 3 = 6.25\text{ hours}\).
\(\text{Period} = 2 \times 6.25 = 12.5\text{ hours}\).
Then:
\(b = \frac{2\pi}{\text{Period}} = \frac{2\pi}{12.5} \approx 0.50265\text{ rad/hour}\).
Rounding to three significant figures, we get \(b \approx 0.503\).

(d) Since high tide (maximum) occurs at \(t = 3\) and cosine has its maximum when its argument is 0, the horizontal shift is \(c = 3\).

(e) At 14:00, \(t = 14\). Substituting \(a=4\), \(b=0.50265\), \(c=3\), \(d=8\) into the model (ensure GDC is in radians):
\(H(14) = 4 \cos(0.50265(14 - 3)) + 8 = 4 \cos(5.5292) + 8 \approx 4(0.72896) + 8 \approx 10.9\text{ m}\).

(f) We want to solve \(H(t) \ge 10\) for \(t \ge 12\).
\(4 \cos(0.50265(t - 3)) + 8 = 10 \implies \cos(0.50265(t - 3)) = 0.5\).
\(0.50265(t - 3) = \frac{\pi}{3} + 2k\pi\) or \(-\frac{\pi}{3} + 2k\pi\).
For the rising tide after the low tide at 09:15, the next high tide is at \(t = 3 + 12.5 = 15.5\).
Solving the equation for the rising phase:
\(0.50265(t - 3) = -\frac{\pi}{3} + 2\pi \implies 0.50265(t-3) = \frac{5\pi}{3} \approx 5.236\)
\(t - 3 = \frac{5.236}{0.50265} \approx 10.417 \implies t \approx 13.417\) hours after midnight.
Converting to hours and minutes:
\(13.417\text{ hours} = 13\text{ hours and } 25\text{ minutes}\).
So the vessel can enter at 13:25 (or 1:25 PM).

(a) M1: For average formula of midline.
A1: For \(d = 8\).

(b) M1: For amplitude formula.
A1: For showing \(a = 4\).

(c) M1: For using \(9:15 = 9.25\) hours.
M1: For determining the period of 12.5 hours.
M1: For setting up \(b = 2\pi / T\).
A1: For showing \(b \approx 0.503\).

(d) M1: For associating the maximum with the cosine peak at \(t = 3\).
A1: For \(c = 3\).

(e) M1: For substituting \(t = 14\) in the model in radians.
A1: For depth of \(10.9\text{ m}\) (accept \(10.9 - 11.0\)).

(f) M1: For setting up the equation \(H(t) = 10\) or using GDC intersection.
A1: For \(t \approx 13.4\) or 13:25.

(a) Each year, 8% of the trees are lost, meaning \(100\% - 8\% = 92\%\) or \(0.92 u_n\) remain at the end of the year.
Then, 500 new trees are planted.
Therefore, the number of trees at the start of Year \(n+1\) is:
\(u_{n+1} = 0.92 u_n + 500\).

(b) We start with \(u_1 = 8000\).
\(u_2 = 0.92(8000) + 500 = 7360 + 500 = 7860\).
\(u_3 = 0.92(7860) + 500 = 7231.2 + 500 = 7731.2\).
\(u_4 = 0.92(7731.2) + 500 = 7112.704 + 500 = 7612.704 \approx 7613\) trees.

(c) To find the constants in the solution \(u_n = A(0.92)^{n-1} + B\):
For a steady-state \(u_n = B\) as a constant function:
\(B = 0.92B + 500 \implies 0.08B = 500 \implies B = \frac{500}{0.08} = 6250\).
Now, using the initial condition \(u_1 = 8000\) at \(n = 1\):
\(u_1 = A(0.92)^0 + 6250 = 8000 \implies A + 6250 = 8000 \implies A = 1750\).
Thus, \(u_n = 1750(0.92)^{n-1} + 6250\).

(d) As \(n \to \infty\), the term \((0.92)^{n-1} \to 0\).
Therefore, \(u_n \to B = 6250\).
The long-term steady-state population is 6250 trees.

(e) We want to find when \(u_n < 6500\):
\(1750(0.92)^{n-1} + 6250 < 6500 \implies 1750(0.92)^{n-1} < 250\)
\((0.92)^{n-1} < \frac{250}{1750} = \frac{1}{7}\)
Taking the natural logarithm on both sides:
\((n-1) \ln(0.92) < \ln\left(\frac{1}{7}\right)\)
Since \\ln(0.92) < 0\), we reverse the inequality:
\(n-1 > \frac{\ln(1/7)}{\ln(0.92)} \approx \frac{-1.9459}{-0.08338} \approx 23.34\)
\(n > 24.34\).
Since \(n\) must be an integer, \(n = 25\).
The population first drops below 6500 at the start of Year 25.

(a) M1: For identifying that 92% of trees remain.
A1: For constructing the complete recurrence relation.

(b) M1: For calculating \(u_2\).
M1: For calculating \(u_3\).
A1: For the correct population of \(7613\) (or \(7612.7\)).

(c) M1: For setting up the steady-state equation to find \(B\).
A1: For \(B = 6250\).
M1: For substituting \(n=1\) and \(u_1 = 8000\) into the formula.
A1: For finding \(A = 1750\).
A1: For writing the final formula \(u_n = 1750(0.92)^{n-1} + 6250\).

(d) M1: For taking the limit of the sequence as \(n \to \infty\).
A1: For stating the steady-state is 6250 trees.

(e) M1: For setting up the inequality \(1750(0.92)^{n-1} + 6250 < 6500\) or equivalent GDC table search.
A1: For concluding Year 25.

**Part A: Equilibrium and Phase Portrait**

1. To find the equilibrium level, set \(\frac{dx}{dt} = 0\) and \(\frac{dy}{dt} = 0\):
\(-3x + y + 12 = 0 \implies y = 3x - 12\)
Substitute into the second equation:
\(2x - 2(3x - 12) + 8 = 0 \implies 2x - 6x + 24 + 8 = 0\)
\(-4x = -32 \implies x_0 = 8\)
Substitute back to find \(y\):
\(y_0 = 3(8) - 12 = 12\)
So the equilibrium point is \((8, 12)\).

2. Since \(X = x - 8\) and \(Y = y - 12\), we have \(\frac{dX}{dt} = \frac{dx}{dt}\) and \(\frac{dY}{dt} = \frac{dy}{dt}\).
Substitute \(x = X + 8\) and \(y = Y + 12\) into the differential equations:
\(\frac{dX}{dt} = -3(X + 8) + (Y + 12) + 12 = -3X + Y\)
\(\frac{dY}{dt} = 2(X + 8) - 2(Y + 12) + 8 = 2X - 2Y\)
Thus, the system in matrix form is:
\(\frac{d\mathbf{X}}{dt} = \begin{pmatrix} -3 & 1 \\ 2 & -2 \end{pmatrix} \mathbf{X}\)
So \\mathbf{M} = \begin{pmatrix} -3 & 1 \\ 2 & -2 \end{pmatrix}\).

3. To find the eigenvalues, solve \(\det(\mathbf{M} - \lambda \mathbf{I}) = 0\):
\(\det\begin{pmatrix} -3 - \lambda & 1 \\ 2 & -2 - \lambda \end{pmatrix} = (-3 - \lambda)(-2 - \lambda) - 2 = 0\)
\(\lambda^2 + 5\lambda + 6 - 2 = 0 \implies \lambda^2 + 5\lambda + 4 = 0\)
\((\lambda + 1)(\lambda + 4) = 0\)
Thus, the eigenvalues are \(\lambda_1 = -1\) and \(\lambda_2 = -4\).

**Part B: Exact Solutions**

4. For \(\lambda_1 = -1\):
\(\mathbf{M} - (-1)\mathbf{I} = \begin{pmatrix} -2 & 1 \\ 2 & -1 \end{pmatrix}\)
Solving \(\begin{pmatrix} -2 & 1 \\ 2 & -1 \end{pmatrix} \begin{pmatrix} u \\ v \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \implies -2u + v = 0 \implies v = 2u\).
An eigenvector is \(\mathbf{v}_1 = \begin{pmatrix} 1 \\ 2 \end{pmatrix}\).

For \(\lambda_2 = -4\):
\(\mathbf{M} - (-4)\mathbf{I} = \begin{pmatrix} 1 & 1 \\ 2 & 2 \end{pmatrix}\)
Solving \(\begin{pmatrix} 1 & 1 \\ 2 & 2 \end{pmatrix} \begin{pmatrix} u \\ v \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \implies u + v = 0 \implies v = -u\).
An eigenvector is \(\mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}\).

5. The general solution for the deviation variables is:
\(\mathbf{X}(t) = c_1 e^{-t} \begin{pmatrix} 1 \\ 2 \end{pmatrix} + c_2 e^{-4t} \begin{pmatrix} 1 \\ -1 \end{pmatrix}\)
Therefore, the general solution for the concentrations is:
\(x(t) = 8 + c_1 e^{-t} + c_2 e^{-4t}\)
\(y(t) = 12 + 2c_1 e^{-t} - c_2 e^{-4t}\).

6. Apply initial conditions \(x(0) = 14\) and \(y(0) = 18\):
\(14 = 8 + c_1 + c_2 \implies c_1 + c_2 = 6\)
\(18 = 12 + 2c_1 - c_2 \implies 2c_1 - c_2 = 6\)
Adding these equations gives:
\(3c_1 = 12 \implies c_1 = 4\)
Substituting back gives \(c_2 = 2\).
Thus, the particular solution is:
\(x(t) = 8 + 4e^{-t} + 2e^{-4t}\)
\(y(t) = 12 + 8e^{-t} - 2e^{-4t}\).

**Part C: Numerical Approximation**

7. Let \(f(x, y) = -3x + y + 12\) and \(g(x, y) = 2x - 2y + 8\).
At \(t = 0\), \(x_0 = 14\), \(y_0 = 18\), and \(h = 0.5\).
\(\frac{dx}{dt} = -3(14) + 18 + 12 = -12\)
\(\frac{dy}{dt} = 2(14) - 2(18) + 8 = 0\)

First Euler step:
\(x(0.5) \approx x_0 + h \left(\frac{dx}{dt}\right)_0 = 14 + 0.5(-12) = 8\)
\(y(0.5) \approx y_0 + h \left(\frac{dy}{dt}\right)_0 = 18 + 0.5(0) = 18\).

8. At \(t = 0.5\), \(x_1 = 8\), \(y_1 = 18\).
\(\frac{dx}{dt} = -3(8) + 18 + 12 = 6\)
\(\frac{dy}{dt} = 2(8) - 2(18) + 8 = -12\)

Second Euler step:
\(x(1.0) \approx x_1 + h \left(\frac{dx}{dt}\right)_1 = 8 + 0.5(6) = 11\)
\(y(1.0) \approx y_1 + h \left(\frac{dy}{dt}\right)_1 = 18 + 0.5(-12) = 12\).

9. The exact value of \(x(1.0)\) is:
\(x(1) = 8 + 4e^{-1} + 2e^{-4} \approx 8 + 4(0.367879) + 2(0.018316) \approx 9.50815\)
Percentage error:
\(\text{Percentage Error} = \frac{|11 - 9.50815|}{9.50815} \times 100\% \approx 15.6899\% \approx 15.7\%\).

**Part A: Equilibrium and Phase Portrait**
- 1. M1 for setting both derivatives to 0. A1 for correct value of \(x_0 = 8\), A1 for correct value of \(y_0 = 12\).
- 2. M1 for differentiating deviation variables and substituting. A1 for writing the matrix \(\mathbf{M} = \begin{pmatrix} -3 & 1 \\ 2 & -2 \end{pmatrix}\) correctly.
- 3. M1 for setting up the characteristic equation \(\det(\mathbf{M} - \lambda\mathbf{I}) = 0\). A1 for correct characteristic polynomial \(\lambda^2 + 5\lambda + 4 = 0\). A1 for correct eigenvalues \(\lambda = -1\) and \(\lambda = -4\).

**Part B: Exact Solutions**
- 4. M1 for setting up matrix equation for \(\lambda = -1\), A1 for \(\begin{pmatrix} 1 \\ 2 \end{pmatrix}\). M1 for setting up matrix equation for \(\lambda = -4\), A1 for \(\begin{pmatrix} 1 \\ -1 \end{pmatrix}\).
- 5. M1 for using the structure of the general solution for deviation variables. A1 for adding the equilibrium values. A1 for the correct explicit formulas.
- 6. M1 for substituting \(t = 0\) to set up a system for \(c_1, c_2\). A1 for finding \(c_1 = 4\) and \(c_2 = 2\). A1 for final correct particular solution.

**Part C: Numerical Approximation**
- 7. M1 for calculating derivatives at initial state. A1 for derivative values \(-12\) and \(0\). A1 for \(x(0.5) \approx 8\), A1 for \(y(0.5) \approx 18\).
- 8. M1 for calculating derivatives at state 1. A1 for derivative values \(6\) and \(-12\). A1 for \(x(1.0) \approx 11\) and \(y(1.0) \approx 12\).
- 9. M1 for finding the exact value of \(x(1.0)\). M1 for using the percentage error formula. A1 for obtaining \(15.7\%\).

**Part A: Short-term Transitions**

1. We are looking for the transition probability from Green Space (State 3) to Commercial (State 2) in two steps. This is the \((3, 2)\) element of \(\mathbf{P}^2\).
\(\mathbf{P}^2 = \begin{pmatrix} 0.7 & 0.2 & 0.1 \\ 0.1 & 0.8 & 0.1 \\ 0.2 & 0.2 & 0.6 \end{pmatrix} \begin{pmatrix} 0.7 & 0.2 & 0.1 \\ 0.1 & 0.8 & 0.1 \\ 0.2 & 0.2 & 0.6 \end{pmatrix}\)
For row 3, column 2 of the product:
\(P^2_{32} = 0.2(0.2) + 0.2(0.8) + 0.6(0.2) = 0.04 + 0.16 + 0.12 = 0.32\).

2. The initial vector in 2020 is \(\mathbf{s}_0 = \begin{pmatrix} 0.5 & 0.3 & 0.2 \end{pmatrix}\).
After one decade (2030):
\(\mathbf{s}_1 = \mathbf{s}_0 \mathbf{P} = \begin{pmatrix} 0.5 & 0.3 & 0.2 \end{pmatrix} \begin{pmatrix} 0.7 & 0.2 & 0.1 \\ 0.1 & 0.8 & 0.1 \\ 0.2 & 0.2 & 0.6
\end{pmatrix} = \begin{pmatrix} 0.42 & 0.38 & 0.20 \end{pmatrix}\)
After two decades (2040):
\(\mathbf{s}_2 = \mathbf{s}_1 \mathbf{P} = \begin{pmatrix} 0.42 & 0.38 & 0.20 \end{pmatrix} \begin{pmatrix} 0.7 & 0.2 & 0.1 \\ 0.1 & 0.8 & 0.1 \\ 0.2 & 0.2 & 0.6 \end{pmatrix} = \begin{pmatrix} 0.372 & 0.428 & 0.200 \end{pmatrix}\).
So the expected distribution is 37.2% Residential, 42.8% Commercial, and 20% Green Space.

**Part B: Steady State**

3. The steady state satisfies \(\mathbf{\pi} \mathbf{P} = \mathbf{\pi}\) and \(\pi_1 + \pi_2 + \pi_3 = 1\).
From \(\mathbf{\pi} \mathbf{P} = \mathbf{\pi}\):
\(0.7\pi_1 + 0.1\pi_2 + 0.2\pi_3 = \pi_1 \implies -0.3\pi_1 + 0.1\pi_2 + 0.2\pi_3 = 0\)
\(0.2\pi_1 + 0.8\pi_2 + 0.2\pi_3 = \pi_2 \implies 0.2\pi_1 - 0.2\pi_2 + 0.2\pi_3 = 0\)
Also, \(\pi_1 + \pi_2 + \pi_3 = 1\).

4. Solving the system:
From the second equation: \(\pi_1 - \pi_2 + \pi_3 = 0 \implies \pi_2 = \pi_1 + \pi_3\).
Substitute into the first equation:
\(-0.3\pi_1 + 0.1(\pi_1 + \pi_3) + 0.2\pi_3 = 0 \implies -0.2\pi_1 + 0.3\pi_3 = 0 \implies \pi_1 = 1.5\pi_3\).
Then \(\pi_2 = 1.5\pi_3 + \pi_3 = 2.5\pi_3\).
Using the normalization equation:
\(1.5\pi_3 + 2.5\pi_3 + \pi_3 = 1 \implies 5\pi_3 = 1 \implies \pi_3 = 0.2\).
Thus, \(\pi_1 = 0.3\) and \(\pi_2 = 0.5\).
So the steady state is \(\begin{pmatrix} 0.3 & 0.5 & 0.2 \end{pmatrix}\).

5. The steady-state percentage of Green Space is 20%. This implies that despite the continuous development and transformation of land, the total proportion of green areas will stabilize and be preserved at 20% in the long term, avoiding total depletion.

**Part C: Diagonalisation and General Powers**

6. Solve \(\det(\mathbf{P} - \lambda \mathbf{I}) = 0\):
Since \(\mathbf{P}\) is a transition matrix, \(\lambda_1 = 1\) is an eigenvalue.
The trace of \(\mathbf{P}\) is \(0.7 + 0.8 + 0.6 = 2.1\).
Thus, \(1 + \lambda_2 + \lambda_3 = 2.1 \implies \lambda_2 + \lambda_3 = 1.1\).
The determinant is:
\(\det(\mathbf{P}) = 0.7(0.48 - 0.02) - 0.2(0.06 - 0.02) + 0.1(0.02 - 0.16) = 0.322 - 0.008 - 0.014 = 0.3\).
Thus, \(1 \cdot \lambda_2 \cdot \lambda_3 = 0.3 \implies \lambda_2 \lambda_3 = 0.3\).
Solving \(\lambda^2 - 1.1\lambda + 0.3 = 0 \implies (\lambda - 0.6)(\lambda - 0.5) = 0\).
So the eigenvalues are indeed \(1\), \(0.6\), and \(0.5\).

7. Compute \(\mathbf{v}_2 \mathbf{P}\):
\(\begin{pmatrix} 1 & 0 & -1 \end{pmatrix} \begin{pmatrix} 0.7 & 0.2 & 0.1 \\ 0.1 & 0.8 & 0.1 \\ 0.2 & 0.2 & 0.6 \end{pmatrix} = \begin{pmatrix} 1(0.7) - 1(0.2) & 1(0.2) - 1(0.2) & 1(0.1) - 1(0.6) \end{pmatrix} = \begin{pmatrix} 0.5 & 0 & -0.5 \end{pmatrix} = 0.5 \begin{pmatrix} 1 & 0 & -1 \end{pmatrix}\).
Since \(\mathbf{v}_2 \mathbf{P} = 0.5 \mathbf{v}_2\), \(\mathbf{v}_2\) is indeed a left eigenvector with eigenvalue \(\lambda = 0.5\).

8. Set up the linear combination:
\(\begin{pmatrix} 0.5 & 0.3 & 0.2 \end{pmatrix} = \alpha \begin{pmatrix} 3 & 5 & 2 \end{pmatrix} + \beta \begin{pmatrix} 1 & 0 & -1 \end{pmatrix} + \gamma \begin{pmatrix} 1 & -1 & 0 \end{pmatrix}\).
This gives the system of equations:
\(3\alpha + \beta + \gamma = 0.5\)
\(5\alpha - \gamma = 0.3 \implies \gamma = 5\alpha - 0.3\)
\(2\alpha - \beta = 0.2 \implies \beta = 2\alpha - 0.2\).
Substitute into the first equation:
\(3\alpha + (2\alpha - 0.2) + (5\alpha - 0.3) = 0.5 \implies 10\alpha - 0.5 = 0.5 \implies \alpha = 0.1\).
Then, \(\beta = 2(0.1) - 0.2 = 0\).
And \(\gamma = 5(0.1) - 0.3 = 0.2\).
Thus, \(\alpha = 0.1\), \(\beta = 0\), and \(\gamma = 0.2\).

**Part A: Short-term Transitions**
- 1. M1 for identifying the required probability is from the transition matrix squared. M1 for executing the correct matrix multiplication step for element \((3, 2)\). A1 for \(0.32\).
- 2. M1 for multiplying the initial state vector by \(\mathbf{P}\). A1 for correct \(\mathbf{s}_1 = \begin{pmatrix} 0.42 & 0.38 & 0.2 \end{pmatrix}\). M1 for multiplying \(\mathbf{s}_1\) by \(\mathbf{P}\). A1 for correct final vector / percentages.

**Part B: Steady State**
- 3. M1 for setting up \(\mathbf{\pi}\mathbf{P} = \mathbf{\pi}\). A1 for at least two correct equations, A1 for including the normalization condition \(\pi_1 + \pi_2 + \pi_3 = 1\).
- 4. M1 for an attempt to substitute and reduce the equations. A1 for expressing variables in terms of one of them (e.g., \(\pi_1 = 1.5\pi_3\)). M1 for using the sum-to-1 constraint. A1 for finding \(\pi_3 = 0.2\). A1 for the correct steady-state probabilities: Residential = 30%, Commercial = 50%, Green Space = 20%.
- 5. R1 for identifying 20% stability. R1 for concluding this means that despite active transitions, Green Space will not vanish and stabilizes at its initial share.

**Part C: Diagonalisation and General Powers**
- 6. M1 for recognizing \(\lambda = 1\) as a root. M1 for using trace relationship \(\text{Tr}(\mathbf{P}) = 2.1\). M1 for determinant relationship \(\det(\mathbf{P}) = 0.3\). A1 for solving the quadratic equation to get \(\lambda = 0.6\) and \(\lambda = 0.5\).
- 7. M1 for multiplying \(\mathbf{v}_2\) by \(\mathbf{P}\). A1 for showing the result is indeed \(0.5\mathbf{v}_2\).
- 8. M1 for setting up the vector equation system. M1 for expressing \(\beta\) and \(\gamma\) in terms of \(\alpha\). A1 for finding \(\alpha = 0.1\). A1 for obtaining \(\beta = 0\) and \(\gamma = 0.2\).