MetadataPastPaper.sampleTitle

(a) The null hypothesis is that music preference and age group are independent.

(b) Row total for '30 and over' = \(25 + 35 + 35 = 95\).
Column total for 'Classical' = \(15 + 35 = 50\).
Grand total = \(205\).
Expected frequency = \(\frac{95 \times 50}{205} = \frac{4750}{205} \approx 23.2\) (to 3 s.f.).

(c) Using a GDC for a \(\chi^2\) test for independence with a \(2 \times 3\) contingency table:
Degrees of freedom: \(df = (2-1)(3-1) = 2\).
\(\chi^2\) statistic \(\approx 15.4\).
\(p\)-value \(\approx 0.000465\) (or \(4.65 \times 10^{-4}\)).
Since the \(p\)-value is less than the significance level of \(0.05\), we reject the null hypothesis.

(a) **(M1)** for stating independence (e.g., 'Music preference is independent of age group'). **(A1)**
(b) **(M1)** for calculating the row, column, and grand totals, and using the formula \(\frac{\text{row total} \times \text{col total}}{\text{grand total}}\). **(A1)** for \(23.2\) (accept \(23.17\)).
(c) **(A1)** for \(p\)-value \(\approx 0.000465\) (accept range \(0.000460\) to \(0.000470\)). **(R1)** for comparing the \(p\)-value with the significance level (e.g., \(0.000465 < 0.05\)). **(A1)** for concluding that \(H_0\) is rejected.

(a) Value after 5 years under Option A:
\(V_A(5) = 8500 - 950 \times 5 = 8500 - 4750 = \$3750\).

(b) Value after 5 years under Option B:
\(V_B(5) = 8500 \times (1 - 0.15)^5 = 8500 \times 0.85^5 \approx \$3771.50\) (or \(\$3770\) to 3 s.f.).

(c) We compare the values at each year:
At \(t = 4\):
\(V_A(4) = 8500 - 950 \times 4 = 4700\)
\(V_B(4) = 8500 \times 0.85^4 \approx 4437.05\) (Option B is less than Option A)
At \(t = 5\):
\(V_A(5) = 3750\)
\(V_B(5) \approx 3771.50\) (Option B is greater than Option A)
Thus, it takes 5 years for the value under Option B to become greater than the value under Option A for the first time.

(a) **(M1)** for using the linear depreciation formula. **(A1)** for \(3750\).
(b) **(M1)** for using the reducing balance depreciation formula. **(A1)** for \(3770\) (or \(3771.50\)).
(c) **(M1)** for setting up an inequality or comparing values for different years. **(A1)** for finding that at \(t=4\), \(V_B < V_A\) and at \(t=5\), \(V_B > V_A\). **(A1)** for concluding 5 years.

(a) Gradient of \(AB = \frac{10 - 8}{5 - 1} = \frac{2}{4} = 0.5\).

(b) The midpoint of \(AB\) is \(\left(\frac{1+5}{2}, \frac{8+10}{2}\right) = (3, 9)\).
The gradient of the perpendicular bisector is the negative reciprocal of \(0.5\), which is \(-2\).
Equation: \(y - 9 = -2(x - 3) \implies y = -2x + 15\).

(c) The Voronoi vertex is the intersection point of the perpendicular bisectors of \(AB\) and \(BC\).
We solve the system of equations:
\(y = -2x + 15\)
\(x - 3y + 15 = 0\)
Substituting the first equation into the second:
\(x - 3(-2x + 15) + 15 = 0\)
\(x + 6x - 45 + 15 = 0 \implies 7x = 30 \implies x = \frac{30}{7} \approx 4.29\) km.
Substituting \(x\) back to find \(y\):
\(y = -2\left(\frac{30}{7}\right) + 15 = -\frac{60}{7} + \frac{105}{7} = \frac{45}{7} \approx 6.43\) km.
The coordinates of the vertex are \((\frac{30}{7}, \frac{45}{7}) \approx (4.29, 6.43)\).

(a) **(A1)** for \(0.5\) (or \(\frac{1}{2}\)).
(b) **(M1)** for finding the midpoint \((3, 9)\). **(M1)** for identifying the perpendicular gradient as \(-2\). **(A1)** for \(y = -2x + 15\).
(c) **(M1)** for attempting to solve the system of simultaneous equations. **(A1)** for \(x = 4.29\) (or \(\frac{30}{7}\)), and **(A1)** for \(y = 6.43\) (or \(\frac{45}{7}\)).

(a) Let the total height of the poster board be \(y\).
The total area is \(xy = 1500 \implies y = \frac{1500}{x}\).
The printable area has width \((x - 6)\) and height \((y - 10)\).
Printable area \(A(x) = (x - 6)(y - 10) = (x - 6)\left(\frac{1500}{x} - 10\right)\)
\(A(x) = 1500 - 10x - \frac{9000}{x} + 60 = 1560 - 10x - \frac{9000}{x}\).

(b) To maximize the printable area, find the derivative \(A'(x)\):
\(A'(x) = -10 + \frac{9000}{x^2}\).
Set \(A'(x) = 0\):
\(-10 + \frac{9000}{x^2} = 0 \implies 10x^2 = 9000 \implies x^2 = 900 \implies x = 30\text{ cm}\) (since \(x > 0\)).

(c) Substitute \(x = 30\) into the equation for \(A\):
\(A(30) = 1560 - 10(30) - \frac{9000}{30} = 1560 - 300 - 300 = 960\text{ cm}^2\).

(a) **(M1)** for expressing height \(y = \frac{1500}{x}\). **(M1)** for writing the area expression \((x - 6)(y - 10)\). **(A1)** for demonstrating the algebraic expansion to reach the given formula.
(b) **(M1)** for finding \(A'(x) = -10 + \frac{9000}{x^2}\). **(M1)** for setting the derivative to 0 and solving. **(A1)** for \(x = 30\).
(c) **(A1)** for \(960\).

(a)(i) \(c\) is the average of the maximum and minimum temperatures:
\(c = \frac{28 + 14}{2} = 21\).
(a)(ii) \(a\) is the amplitude of the temperature wave:
\(a = \frac{28 - 14}{2} = 7\).

(b) The period of the temperature fluctuation is 24 hours.
\(\text{Period} = \frac{2\pi}{b} = 24 \implies b = \frac{2\pi}{24} = \frac{\pi}{12} \approx 0.262\) (to 3 s.f.).

(c) At 14:00, \(t = 14\).
\(T(14) = 7 \sin\left(\frac{\pi}{12}(14 - 12)\right) + 21 = 7 \sin\left(\frac{2\pi}{12}\right) + 21\)
\(T(14) = 7 \sin\left(\frac{\pi}{6}\right) + 21 = 7(0.5) + 21 = 24.5^\circ\text{C}\).

(a)(i) **(A1)** for \(c = 21\).
(a)(ii) **(A1)** for \(a = 7\).
(b) **(M1)** for equating the period formula to 24. **(A1)** for \(b = \frac{\pi}{12}\) (or \(0.262\)).
(c) **(M1)** for substituting \(t = 14\) into the model. **(A1)** for \(24.5\).

(a) Let \(X\) be the weight of an apple. \(X \sim N(150, \sigma^2)\).
\(\text{P}(X > 180) = 0.15 \implies \text{P}\left(Z > \frac{180-150}{\sigma}\right) = 0.15\).
Using inverse normal on GDC:
\(\frac{30}{\sigma} = 1.03643 \implies \sigma = \frac{30}{1.03643} \approx 28.9\text{ g}\).

(b) Using \(\sigma = 28.945...\):
\(\text{P}(130 \le X \le 160) = \text{P}(-0.691 \le Z \le 0.346) \approx 0.390\).
(Using \(\sigma = 28.9\), \(\text{P}(130 \le X \le 160) \approx 0.391\)).

(c) Let \(Y\) be the number of apples weighing more than \(180\text{ g}\). \(Y \sim B(8, 0.15)\).
\(\text{P}(Y \ge 2) = 1 - \text{P}(Y \le 1)\).
Using binomial CDF:
\(\text{P}(Y \le 1) \approx 0.6572\).
\(\text{P}(Y \ge 2) = 1 - 0.6572 = 0.343\) (to 3 s.f.).

(a) **(M1)** for setting up standard normal equation \(\frac{180-150}{\sigma} = 1.03643\). **(A1)** for \(\sigma = 28.9\).
(b) **(M1)** for setting up the normal probability \(\text{P}(130 \le X \le 160)\). **(A1)** for \(0.390\) (or \(0.391\)).
(c) **(M1)** for recognizing binomial distribution \(B(8, 0.15)\). **(M1)** for calculating \(1 - \text{P}(Y \le 1)\) or equivalent sum. **(A1)** for \(0.343\).

(a)(i) \(\bar{x} = \frac{68}{10} = 6.8\) hours.
(a)(ii) \(\bar{y} = \frac{310}{10} = 31\) cones.

(b) The slope \(a\) is:
\(a = \frac{S_{xy}}{S_{xx}} = \frac{\sum xy - \frac{(\sum x)(\sum y)}{n}}{\sum x^2 - \frac{(\sum x)^2}{n}} = \frac{2280 - \frac{68 \times 310}{10}}{512 - \frac{68^2}{10}}\)
\(a = \frac{2280 - 2108}{512 - 462.4} = \frac{172}{49.6} \approx 3.47\) (to 3 s.f.).

The intercept \(b\) is:
\(b = \bar{y} - a\bar{x} = 31 - 3.46774 \times 6.8 = 31 - 23.5806 \approx 7.42\) (to 3 s.f.).

Equation of the regression line is: \(y = 3.47x + 7.42\) (or using exact values \(y = 3.4677x + 7.419\)).

(c) Substitute \(x = 8.5\):
\(y = 3.4677 \times 8.5 + 7.419 \approx 36.9\).

Rounding to the nearest integer gives 37 ice cream cones.

(a)(i) **(A1)** for \(6.8\).
(a)(ii) **(A1)** for \(31\).
(b) **(M1)** for calculating \(S_{xx}\) and \(S_{xy}\) (or equivalent step on GDC). **(M1)** for slope \(a = 3.47\) and intercept \(b = 7.42\). **(A1)** for \(y = 3.47x + 7.42\).
(c) **(M1)** for substituting \(x = 8.5\) into their equation. **(A1)** for \(37\).

(a) Let \(O\) be the center of the square base \(ABCD\).
Then \(OM\) is parallel to \(AD\), and its length is \(\frac{12}{2} = 6\text{ cm}\).
In the right-angled triangle \(VOM\), by Pythagoras' theorem:
\(VM = \sqrt{VO^2 + OM^2} = \sqrt{8^2 + 6^2} = \sqrt{100} = 10\text{ cm}\).

(b) Total surface area comprises the base and 4 triangular faces:
\(\text{Base Area} = 12 \times 12 = 144\text{ cm}^2\).
\(\text{Area of one triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 12 \times 10 = 60\text{ cm}^2\).
\(\text{Total Area} = 144 + 4(60) = 144 + 240 = 384\text{ cm}^2\).

(c) The distance from the center \(O\) to a vertex \(A\) is half the diagonal of the base square:
\(OA = \frac{1}{2} \sqrt{12^2 + 12^2} = 6\sqrt{2} \approx 8.485\text{ cm}\).
In the right-angled triangle \(VOA\), let \(\theta\) be the angle \(VAO\):
\(\tan(\theta) = \frac{VO}{OA} = \frac{8}{6\sqrt{2}} \approx 0.9428\).
\(\theta = \arctan(0.9428) \approx 43.3^\circ\) (or \(0.756\text{ radians}\)).

(a) **(M1)** for identifying the right-angled triangle \(VOM\) with sides \(8\) and \(6\). **(A1)** for \(10\).
(b) **(M1)** for finding the area of one triangular face. **(M1)** for adding the area of 4 faces to the base area. **(A1)** for \(384\).
(c) **(M1)** for finding \(OA = 6\sqrt{2}\). **(M1)** for using trigonometry to express the angle (e.g. \(\tan\theta = \frac{8}{8.485}\)). **(A1)** for \(43.3^\circ\) (or \(0.756\text{ radians}\)).

(a) Using the compound interest formula: \(A = P \left(1 + \frac{r}{100k}\right)^{kt}\) where \(P = 12000\), \(r = 4.5\), \(k = 12\), and \(t = 5\). This gives \(A = 12000 \left(1 + \frac{4.5}{1200}\right)^{60} \approx 15021.55\) USD. (b) To find the real value, we discount the future value by the inflation rate over the 5-year period: \(V = \frac{15021.55}{(1 + 0.021)^5} \approx 13539.03\) USD.

(a) M1 for substituting into the compound interest formula or setting up TVM solver. A1 for correct values. A1 for 15 021.55 (accept 15 000). (b) M1 for dividing by inflation factor. A1 for correct setup. A1 for 13 539.03 (accept 13 500).

(a) The minimum value of \(\cos\left(\frac{\pi}{12} t\right)\) is \(-1\), which occurs when the cosine term is \(1\) since the coefficient is negative. This first occurs at \(t = 0\). The minimum temperature is \(T(0) = -6(1) + 20 = 14^\circ\text{C}\). (b) At 09:00, \(t = 9\). Substituting \(t = 9\) into the model: \(T(9) = -6 \cos\left(\frac{9\pi}{12}\right) + 20 \approx 24.2^\circ\text{C}\). (c) We solve \(T(t) = 22 \implies -6 \cos\left(\frac{\pi}{12} t\right) + 20 = 22 \implies \cos\left(\frac{\pi}{12} t\right) = -\frac{1}{3}\). This gives the boundary times \(t_1 \approx 7.2979\) and \(t_2 \approx 16.702\). The temperature is above \(22^\circ\text{C}\) between these times, so the duration is \(16.702 - 7.2979 \approx 9.40\) hours.

(a) A1 for 14 °C, A1 for t = 0 (or midnight). (b) M1 for substituting t = 9. A1 for 24.2. (c) M1 for setting the equation equal to 22. A1 for finding the two intersection times (7.30 and 16.7). A1 for 9.40 hours.

(a) \(H_0\): Preferred exercise type is independent of age group (or there is no association between preferred exercise type and age group). (b) Expected frequency \(= \frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}} = \frac{70 \times 41}{150} \approx 19.1\). (c) Running a \(\chi^2\) two-way test on a GDC with the given matrix yields a \(p\)-value of \(0.0459\) (to 3 s.f.). (d) Since the \(p\)-value (\(0.0459\)) is less than the significance level (\(0.05\)), we reject the null hypothesis.

(a) A1 for stating independence. (b) M1 for correct fraction setup. A1 for 19.1 (or 19.13). (c) A2 for 0.0459. (d) R1 for comparing p-value with 0.05 and making the correct decision to reject.

(a) Midpoint of \(AB\) is \(\left(\frac{2+8}{2}, \frac{8+10}{2}\right) = (5, 9)\). Gradient of \(AB\) is \(\frac{10-8}{8-2} = \frac{1}{3}\). The perpendicular gradient is \(-3\). The equation is \(y - 9 = -3(x - 5)\), which simplifies to \(y = -3x + 24\). (b) Midpoint of \(AC\) is \(\left(\frac{2+6}{2}, \frac{8+2}{2}\right) = (4, 5)\). Gradient of \(AC\) is \(\frac{2-8}{6-2} = -\frac{3}{2}\). The perpendicular gradient is \(\frac{2}{3}\). The equation of the perpendicular bisector of \(AC\) is \(y - 5 = \frac{2}{3}(x - 4) \implies y = \frac{2}{3}x + \frac{7}{3}\). Solving the system of equations: \(-3x + 24 = \frac{2}{3}x + \frac{7}{3} \implies -9x + 72 = 2x + 7 \implies 11x = 65 \implies x = \frac{65}{11} \approx 5.91\). Substituting back gives \(y = -3\left(\frac{65}{11}\right) + 24 = \frac{69}{11} \approx 6.27\). The coordinates of the water fountain are \((5.91, 6.27)\).

(a) M1 for finding midpoint of AB. M1 for perpendicular gradient. A1 for correct equation. (b) M1 for finding equation of another perpendicular bisector (e.g., AC: \(y = \frac{2}{3}x + \frac{7}{3}\) or BC: \(y = -0.25x + 7.75\)). M1 for setting up the system of equations. A1 for correct coordinates \((5.91, 6.27)\).

(a) (i) \(\bar{x} = \frac{4+6+8+10+12+14+16+18}{8} = 11\)
(ii) \(\bar{y} = \frac{52+60+64+73+72+85+81+95}{8} = 72.75 \approx 72.8\)

(b) Using a GDC, the Pearson's product-moment correlation coefficient is:
\(r \approx 0.971\) (0.970737...)

(c) There is a strong, positive correlation between training hours and performance scores.

(d) Using a GDC, the regression line of \(y\) on \(x\) is:
\(y = 2.79x + 42.1\) (or \(y = 2.7857...x + 42.107...\))

(e) Substitute \(x = 15\) into the regression equation:
\(y = 2.7857...(15) + 42.107... \approx 83.9\)

(f) The estimate is reliable because:
1. \(x = 15\) is within the experimental range of the data (interpolation between \(4\) and \(18\)).
2. The Pearson correlation coefficient \(r \approx 0.971\) indicates a very strong linear relationship.

(g) To find Spearman's rank correlation coefficient, we first rank the variables:
$$\begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline
\text{Employee} & \text{A} & \text{B} & \text{C} & \text{D} & \text{E} & \text{F} & \text{G} & \text{H} \\
\hline
\text{Rank of } x & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline
\text{Rank of } y & 1 & 2 & 3 & 5 & 4 & 7 & 6 & 8 \\
\hline
\text{Difference } d & 0 & 0 & 0 & -1 & 1 & -1 & 1 & 0 \\
\hline
\text{Difference } d^2 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 0 \\
\hline
\end{array}$$
\(\sum d^2 = 4\)
Using the formula:
\(r_s = 1 - \frac{6 \sum d^2}{n(n^2 - 1)} = 1 - \frac{6(4)}{8(64 - 1)} = 1 - \frac{24}{504} \approx 0.952\) (0.95238...)

(h) A manager might prefer Spearman's rank correlation coefficient because:
- It is more robust against potential outliers that could skew a Pearson linear analysis.
- It assesses monotonic relationships, which is helpful if performance improvements taper off (non-linear) rather than continuing in a strictly straight-line fashion.

(a) (i) M1 for finding the sum divided by 8, A1 for \(\bar{x} = 11\). (ii) A1 for \(\bar{y} = 72.8\) (72.75).
(b) G1 for finding \(r\), A1 for \(r = 0.971\) (or exact 0.970737...).
(c) A1 for stating both "strong" and "positive" in the context.
(d) G1 for gradient \(m \approx 2.79\), G1 for intercept \(c \approx 42.1\).
(e) M1 for substituting 15 into their regression equation, A1 for \(83.9\) (accept 84).
(f) R1 for mentioning interpolation / within the data range, R1 for mentioning a strong correlation.
(g) M1 for correct ranking of at least one variable, M1 for calculating \(\sum d^2 = 4\), A1 for \(r_s \approx 0.952\).
(h) R1 for mentioning resistance to outliers, R1 for mentioning non-linear/monotonic relationships.

(a) Distance \(AB = \sqrt{(9 - 1)^2 + (12 - 8)^2} = \sqrt{8^2 + 4^2} = \sqrt{80} \approx 8.94\) km.

(b) Midpoint of \(AB\) is \(M_{AB} = \left(\frac{1+9}{2}, \frac{8+12}{2}\right) = (5, 10)\).
Gradient of \(AB\) is \(m_{AB} = \frac{12 - 8}{9 - 1} = \frac{4}{8} = 0.5\).
Gradient of the perpendicular bisector is \(m_1 = -\frac{1}{0.5} = -2\).
Equation: \(y - 10 = -2(x - 5) \Rightarrow y = -2x + 20\).

(c) Midpoint of \(BC\) is \(M_{BC} = \left(\frac{9+5}{2}, \frac{12+2}{2}\right) = (7, 7)\).
Gradient of \(BC\) is \(m_{BC} = \frac{2 - 12}{5 - 9} = \frac{-10}{-4} = 2.5\).
Gradient of the perpendicular bisector is \(m_2 = -\frac{1}{2.5} = -0.4\).
Equation: \(y - 7 = -0.4(x - 7) \Rightarrow y - 7 = -0.4x + 2.8 \Rightarrow 0.4x + y = 9.8\).
Multiplying by 5 to get integer coefficients:
\(2x + 5y = 49\) (as shown).

(d) The circumcenter \(P\) is the intersection of the two perpendicular bisectors:
\(y = -2x + 20\)
\(2x + 5y = 49\)
Substitute the first into the second:
\(2x + 5(-2x + 20) = 49\)
\(2x - 10x + 100 = 49 \Rightarrow -8x = -51 \Rightarrow x = 6.375 \approx 6.38\)
Substitute \(x\) back:
\(y = -2(6.375) + 20 = 7.25\)
Therefore, \(P(6.38, 7.25)\) or exactly \((6.375, 7.25)\).

(e) Find the distance from \(P(6.375, 7.25)\) to \(A(1, 8)\):
\(AP = \sqrt{(6.375 - 1)^2 + (7.25 - 8)^2} = \sqrt{5.375^2 + (-0.75)^2} = \sqrt{28.890625 + 0.5625} = \sqrt{29.453125} \approx 5.43\) km.

(f) The boundary line separating the regions closest to \(B\) and \(D\) is the perpendicular bisector of \(BD\).
Midpoint of \(BD\) is \(M_{BD} = \left(\frac{9+11}{2}, \frac{12+4}{2}\right) = (10, 8)\).
Gradient of \(BD\) is \(m_{BD} = \frac{4 - 12}{11 - 9} = \frac{-8}{2} = -4\).
Gradient of the perpendicular bisector is \(m_3 = -\frac{1}{-4} = 0.25\).
Equation: \(y - 8 = 0.25(x - 10) \Rightarrow y - 8 = 0.25x - 2.5 \Rightarrow y = 0.25x + 5.5\).

(a) M1 for correct substitution into distance formula, A1 for \(8.94\) (accept \(\sqrt{80}\) or \(4\sqrt{5}\)).
(b) M1 for finding the midpoint \((5, 10)\), M1 for finding gradient \(-2\), A1 for \(y = -2x + 20\).
(c) M1 for finding the midpoint \((7, 7)\), M1 for finding gradient \(-0.4\), A1 for showing the steps leading to \(2x + 5y = 49\).
(d) M1 for setting up a system of equations, A1 for \(x = 6.38\) (6.375), A1 for \(y = 7.25\).
(e) M1 for using distance formula from their \(P\) to \(A\), \(B\), or \(C\), A1 for \(5.43\) km.
(f) M1 for finding midpoint of \(BD\) as \((10, 8)\), M1 for perpendicular gradient \(0.25\), A1 for \(y = 0.25x + 5.5\).

(a) The volume of a cylinder is \(V = \pi r^2 h = 150\).
Therefore, \(h = \frac{150}{\pi r^2}\).

(b) The cost formula is the sum of the costs of the top, bottom, and curved surface:
\(C(r) = 12 \times (\text{area of top}) + 18 \times (\text{area of bottom}) + 8 \times (\text{area of curved surface})\)
\(C(r) = 12(\pi r^2) + 18(\pi r^2) + 8(2\pi r h)\)
\(C(r) = 30\pi r^2 + 16\pi r h\)
Substitute \(h = \frac{150}{\pi r^2}\):
\(C(r) = 30\pi r^2 + 16\pi r \left(\frac{150}{\pi r^2}\right)\)
\(C(r) = 30\pi r^2 + \frac{2400}{r}\) (as shown).

(c) Differentiating \(C(r)\) with respect to \(r\):
\(C'(r) = \frac{\text{d}}{\text{d}r}\left(30\pi r^2 + 2400r^{-1}\right) = 60\pi r - 2400r^{-2}\)
\(C'(r) = 60\pi r - \frac{2400}{r^2}\).

(d) To find the minimum cost, set \(C'(r) = 0\):
\(60\pi r - \frac{2400}{r^2} = 0 \Rightarrow 60\pi r = \frac{2400}{r^2}\)
\(r^3 = \frac{2400}{60\pi} = \frac{40}{\pi}\)
\(r = \sqrt[3]{\frac{40}{\pi}} \approx 2.33\) meters (2.3326...).

(e) Substitute the optimal radius \(r \approx 2.3326\) back into the cost function:
\(C(2.3326) = 30\pi (2.3326)^2 + \frac{2400}{2.3326} \approx 513.75 + 1028.89 = 1542.64 \approx \$1540\).

(f) Evaluate \(C'(3)\):
\(C'(3) = 60\pi(3) - \frac{2400}{3^2} = 180\pi - \frac{2400}{9} \approx 565.49 - 266.67 = 298.82 > 0\).
Since the derivative \(C'(3)\) is positive, the cost function is increasing at \(r = 3\). Thus, if the radius is slightly increased, the cost will increase.

(a) M1 for setting up \(\pi r^2 h = 150\), A1 for \(h = \frac{150}{\pi r^2}\).
(b) M1 for setting up cost sum with coefficients, A1 for \(30\pi r^2\), M1 for side surface cost term, A1 for substituting \(h\) and completing the show-that step.
(c) M1 for power rule on both terms, A1 for \(60\pi r\), A1 for \(-\frac{2400}{r^2}\).
(d) M1 for setting \(C'(r) = 0\), M1 for algebraic manipulation to \(r^3 = \frac{40}{\pi}\), A1 for \(2.33\) m.
(e) M1 for substituting their optimal \(r\) into \(C(r)\), A1 for \(\$1540\) (accept 1542-1543).
(f) R1 for calculating \(C'(3) \approx 299\) (or stating it is greater than 0), A1 for concluding that cost would increase.

(a) At \(t=0\), \(D(t)\) is at its maximum of \(14.2\):
\(a(1) + d = 14.2 \Rightarrow a + d = 14.2\)
At its minimum, \(D(t)\) is \(6.8\):
\(-a + d = 6.8\)
Adding the equations:
\(2d = 21 \Rightarrow d = 10.5\)
Subtracting the equations:
\(2a = 7.4 \Rightarrow a = 3.7\) (as shown).

(b) The first low tide occurs when the cosine function first reaches its minimum value of \(-1\), which is when \(b t = \pi\).
Substitute \(t = 6.2\):
\(6.2 b = \pi \Rightarrow b = \frac{\pi}{6.2} \approx 0.5067 \approx 0.507\) (to 3 s.f.).

(c) At 09:00, \(t = 9\):
\(D(9) = 3.7 \cos(0.5067 \times 9) + 10.5 = 3.7 \cos(4.5603) + 10.5\)
Using a GDC in radian mode:
\(D(9) \approx 3.7 \times (-0.1534) + 10.5 \approx 9.93\) meters.

(d) We want to solve \(D(t) = 12.0\) for \(0 \le t \le 12\):
\(3.7 \cos(0.5067 t) + 10.5 = 12.0\)
\(\cos(0.5067 t) = \frac{1.5}{3.7} \approx 0.4054\)
First solution:
\(0.5067 t = \arccos(0.4054) \approx 1.1533\)
\(t_1 = \frac{1.1533}{0.5067} \approx 2.276\) hours (or 2 hours 17 minutes).
Second solution in the first cycle:
\(0.5067 t = 2\pi - 1.1533 \approx 5.1299\)
\(t_2 = \frac{5.1299}{0.5067} \approx 10.124\) hours (or 10 hours 7 minutes).

(e) The period of the tide is \(T = \frac{2\pi}{b} = \frac{2\pi}{\pi / 6.2} = 12.4\) hours.
Within the first cycle of \(12.4\) hours, the depth is \(\ge 12.0\) meters during two intervals:
1. From \(t = 0\) to \(t = 2.276\) (duration: \(2.276\) hours).
2. From \(t = 10.124\) to \(t = 12.4\) (duration: \(12.4 - 10.124 = 2.276\) hours).
Total safe time per full tidal cycle of 12.4 hours is \(2 \times 2.276 = 4.552\) hours.
Let's check the entire 24-hour period (\(0 \le t \le 24\)):
- Interval 1: \(0 \le t \le 2.276\) (duration \(2.276\) hours)
- Interval 2: \(10.124 \le t \le 14.676\) (centered around the second peak at \(t = 12.4\); duration \(4.552\) hours)
- Interval 3: From \(24 - 1.476 = 22.524 \le t \le 24\) (the third peak is at \(t = 24.8\), so the safe interval starts at \(24.8 - 2.276 = 22.524\) and is cut off at \(t = 24\); duration \(24 - 22.524 = 1.476\) hours).
Total safe time in the 24-hour day:
\(\text{Total} = 2.276 + 4.552 + 1.476 = 8.304 \approx 8.30\) hours.

(a) M1 for setting up \(a + d = 14.2\), M1 for setting up \(-a + d = 6.8\), A1 for showing step-by-step solution to arrive at \(a = 3.7, d = 10.5\).
(b) M1 for identifying that the minimum occurs when \(bt = \pi\), M1 for substituting \(t = 6.2\), A1 for \(b = 0.507\).
(c) M1 for substituting \(t = 9\) into their equation, A1 for \(9.93\).
(d) M1 for setting up \(D(t) = 12\), A1 for finding \(t_1 = 2.28\), M1 for symmetry / using \(2\pi - \theta\), A1 for \(t_2 = 10.1\).
(e) M1 for finding the second high tide at \(t=12.4\), M1 for identifying the safe intervals, M1 for setting up sum of lengths of the three intervals, A1 for \(8.30\) hours (accept \(8.3\)).

(a) Deposit = \(0.15 \times 350,000 = \$52,500\).
Loan amount = \(350,000 - 52,500 = \$297,500\).

(b) (i) Using financial GDC solver:
\(N = 25 \times 12 = 300\)
\(I\% = 4.8\)
\(PV = 297,500\)
\(FV = 0\)
\(P/Y = 12\)
\(C/Y = 12\)
Solving for PMT gives:
\(PMT \approx -1704.37\).
So, each monthly payment is \(\$1704.37\).

(ii) Total amount paid = \(1704.37 \times 300 = \$511,311.00\).

(iii) Total interest paid = \(511,311.00 - 297,500 = \$213,811.00\).

(c) (i) Remaining balance after 10 years (120 payments made, 180 remaining):
Using financial GDC solver with \(N = 180\):
\(I\% = 4.8\)
\(PMT = -1704.37\)
\(FV = 0\)
\(P/Y = 12\)
\(C/Y = 12\)
Solving for PV gives:
\(PV \approx 218,319.55\).
So the remaining balance is \(\$218,319.55\) (or \(\$218,000\) to 3 s.f.).

(ii) Under the early payment plan:
Total paid over the first 10 years = \(120 \times 1704.37 = \$204,524.40\).
Lump sum payment = \(\$218,319.55\).
Total paid under this scheme = \(204,524.40 + 218,319.55 = \$422,843.95\).
Total paid under the original 25-year plan = \(\$511,311.00\).
Interest saved = \(511,311.00 - 422,843.95 = \$88,467.05\) (or \(\$88,500\) to 3 s.f.).

(a) M1 for \(0.85 \times 350,000\) or equivalent subtraction, A1 for \(\$297,500\).
(b) (i) M1 for correct finance solver setup (N=300, I=4.8, PV=297500), A2 for \(\$1704.37\) (award A1 for \(1700\)).
(ii) M1 for their \(PMT \times 300\), A1 for \(\$511,311.00\).
(iii) M1 for subtraction of the principal loan, A1 for \(\$213,811.00\).
(c) (i) M1 for setting remaining payments to \(N=180\) (or \(N=120\) in alternative balance formulation), A1 for GDC solver inputs, A1 for \(\$218,319.55\) (accept \(\$218,320\) or \(\$218,000\)).
(ii) M1 for calculating total payments in first 10 years (\(\$204,524.40\)), M1 for adding the lump sum (\(\$422,843.95\)), M1 for finding the difference from \(\$511,311\), A1 for \(\$88,467.05\) (accept \(\$88,500\)).