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

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

(b) We require \(P(X < w) = 0.08\).
Using the inverse normal function on a graphic display calculator:
\(w \approx 133.138... \approx 133\) grams (to 3 significant figures).

(c) Let \(Y\) be the number of apples weighing more than 165 grams in a box of 20.
\(Y\) follows a binomial distribution, \(Y \sim B(20, 0.105649...)\).
We want to find \(P(Y = 3)\).
Using the binomial probability function on a graphic display calculator:
\(P(Y = 3) \approx 0.201\) (to 3 significant figures).

(a)
[M1] for setting up the normal distribution calculation (e.g., indicating correct mean and standard deviation).
[A1] for correct probability 0.106 (or 0.1056...).

(b)
[M1] for setting up the equation \(P(X < w) = 0.08\) or using inverse normal.
[A1] for \(w = 133\) grams (accept 133.1).

(c)
[M1] for recognizing the binomial distribution \(Y \sim B(20, p)\) where \(p\) is the answer from part (a).
[M1] for setting up the probability expression \(P(Y = 3)\) or \(\binom{20}{3} p^3 (1-p)^{17}\).
[A1] for correct probability 0.201 (accept answers in range 0.200 - 0.202 depending on rounding of \(p\)).

(a) Find the midpoint of \(AB\):
Midpoint \(= \left(\frac{2+8}{2}, \frac{8+10}{2}\right) = (5, 9)\).
Find the gradient of \(AB\):
\(m_{AB} = \frac{10-8}{8-2} = \frac{2}{6} = \frac{1}{3}\).
The gradient of the perpendicular bisector is the negative reciprocal:
\(m_{\perp} = -3\).
The equation is:
\(y - 9 = -3(x - 5) \implies y = -3x + 24\).

(b) To find the circumcenter, solve the simultaneous equations of the perpendicular bisectors:
\(y = -3x + 24\)
\(y = -0.5x + 9\)
Equating the two expressions for \(y\):
\(-3x + 24 = -0.5x + 9 \implies 15 = 2.5x \implies x = 6\).
Substitute \(x = 6\) back into either equation:
\(y = -3(6) + 24 = 6\).
Thus, the coordinates of the circumcenter are \((6, 6)\).

(c) The distance from the circumcenter \((6, 6)\) to clinic \(A(2, 8)\) is:
\(d = \sqrt{(6-2)^2 + (6-8)^2} = \sqrt{4^2 + (-2)^2} = \sqrt{16+4} = \sqrt{20} \approx 4.47\) km (to 3 significant figures).

(a)
[M1] for calculating the midpoint \((5, 9)\) or the gradient of \(AB\) \(\left(\frac{1}{3}\right)\).
[M1] for finding the perpendicular gradient \(-3\).
[A1] for the correct equation \(y = -3x + 24\).

(b)
[M1] for setting up simultaneous equations using the perpendicular bisectors of \(AB\) and \(BC\).
[A1] for correct coordinates \((6, 6)\).

(c)
[M1] for setting up the distance formula using \((6, 6)\) and \(A(2, 8)\) (or \(B\) or \(C\)).
[A1] for correct distance \(\sqrt{20}\) or \(4.47\) km (accept 4.472...).

(a) First, calculate the differences between the ranks, \(d = R_1 - R_2\), and their squares, \(d^2\):
- Painting A: \(d = 1 - 3 = -2 \implies d^2 = 4\)
- Painting B: \(d = 2 - 1 = 1 \implies d^2 = 1\)
- Painting C: \(d = 3 - 5 = -2 \implies d^2 = 4\)
- Painting D: \(d = 4 - 2 = 2 \implies d^2 = 4\)
- Painting E: \(d = 5 - 8 = -3 \implies d^2 = 9\)
- Painting F: \(d = 6 - 4 = 2 \implies d^2 = 4\)
- Painting G: \(d = 7 - 7 = 0 \implies d^2 = 0\)
- Painting H: \(d = 8 - 6 = 2 \implies d^2 = 4\)

Sum of \(d^2\): \(\sum d^2 = 4 + 1 + 4 + 4 + 9 + 4 + 0 + 4 = 30\).
Use the Spearman's rank correlation coefficient formula with \(n = 8\):
\(r_s = 1 - \frac{6 \sum d^2}{n(n^2 - 1)}\)
\(r_s = 1 - \frac{6(30)}{8(8^2 - 1)} = 1 - \frac{180}{8(63)} = 1 - \frac{180}{504} = 1 - 0.35714... = 0.642857... \approx 0.643\) (to 3 significant figures).

(b) The value of \(r_s \approx 0.643\) indicates a moderate to strong positive correlation between the critic's technique rankings and originality rankings. This means that paintings that are ranked higher in technique also tend to be ranked higher in originality.

(c) The calculated Spearman's rank correlation coefficient is \(r_s \approx 0.643\). Since \(0.643 > 0.619\) (the critical value), we reject the null hypothesis of no correlation. Therefore, there is a statistically significant positive correlation between the rankings at the 5% significance level.

(a)
[M1] for calculating the correct differences \(d\) or squared differences \(d^2\) (at least 4 correct values shown).
[M1] for evaluating the sum \(\sum d^2 = 30\) and substituting into the formula.
[A1] for \(r_s \approx 0.643\) (accept 0.6428... or \(\frac{9}{14}\)).

(b)
[R1] for identifying a "moderate to strong" (or "positive") correlation.
[R1] for interpreting the direction in context (e.g., "paintings with higher technique rank tend to have higher originality rank").

(c)
[R1] for comparing \(r_s\) to the critical value (e.g., \(0.643 > 0.619\)).
[A1] for concluding that the correlation is statistically significant.

(a) (i) The amplitude \(a\) is given by:
\(a = \frac{\text{Maximum} - \text{Minimum}}{2} = \frac{14.6 - 8.2}{2} = 3.2\).

(ii) The vertical shift \(d\) is given by:
\(d = \frac{\text{Maximum} + \text{Minimum}}{2} = \frac{14.6 + 8.2}{2} = 11.4\).

(iii) The period of the tide is twice the time from maximum to minimum depth. So Period \(= 2 \times 6 = 12\) hours.
Since \(\text{Period} = \frac{2\pi}{b}\), we have:
\(b = \frac{2\pi}{12} = \frac{\pi}{6}\).

(b) Substitute \(t = 4\) into the model:
\(H(4) = 3.2 \cos\left(\frac{\pi}{6} \times 4\right) + 11.4 = 3.2 \cos\left(\frac{2\pi}{3}\right) + 11.4 = 3.2(-0.5) + 11.4 = -1.6 + 11.4 = 9.8\) meters.

(c) We find when \(H(t) = 10\):
\(3.2 \cos\left(\frac{\pi}{6} t\right) + 11.4 = 10 \implies 3.2 \cos\left(\frac{\pi}{6} t\right) = -1.4\)
\(\cos\left(\frac{\pi}{6} t\right) = -0.4375\)
Using a GDC to find the inverse cosine in the interval \(0 \le t \le 6\):
\(\frac{\pi}{6} t \approx 2.02364\)
\(t \approx \frac{6}{\pi} \times 2.02364 \approx 3.8649... \approx 3.86\) hours.
Thus, the boat can no longer safely enter the harbor after 3.86 hours.

(a)
[A1] for \(a = 3.2\).
[A1] for \(d = 11.4\).
[M1] for recognizing that the period is 12 hours (or equating \(H(6) = 8.2\)).
[A1] for \(b = \frac{\pi}{6}\) (accept 0.524).

(b)
[A1] for \(9.8\) meters (or 9.80).

(c)
[M1] for setting up the inequality or equation \(H(t) = 10\).
[A1] for \(t \approx 3.86\) hours (accept 3.865).

(a) Let the height of the container be \(h\) meters.
The volume \(V\) of the container is given by:
\(V = \text{length} \times \text{width} \times \text{height} = (2x)(x)(h) = 2x^2 h\).
Since \(V = 36\):
\(2x^2 h = 36 \implies h = \frac{18}{x^2}\).

The container has an open top, so the surface area \(A\) consists of the base area and the area of the four side walls:
- Base Area \(= 2x \times x = 2x^2\)
- Side Walls Area \(= 2(x \times h) + 2(2x \times h) = 6xh\).

Substitute \(h = \frac{18}{x^2}\) into the surface area equation:
\(A = 2x^2 + 6x\left(\frac{18}{x^2}\right)\)
\(A = 2x^2 + \frac{108}{x}\) (as required).

(b) Differentiating \(A\) with respect to \(x\):
\(\frac{\text{d}A}{\text{d}x} = 4x - 108x^{-2} = 4x - \frac{108}{x^2}\).

(c) To find the minimum surface area, set \(\frac{\text{d}A}{\text{d}x} = 0\):
\(4x - \frac{108}{x^2} = 0\)
\(4x^3 = 108\)
\(x^3 = 27\)
\(x = 3\).
Therefore, the surface area is minimized when \(x = 3\) meters.

(a)
[M1] for expressing the height in terms of \(x\): \(h = \frac{18}{x^2}\) (or setting up volume equation \(2x^2h = 36\)).
[M1] for writing an expression for the surface area of an open box: \(A = 2x^2 + 2xh + 4xh\).
[A1] for substituting \(h\) and simplifying to show the given equation \(A = 2x^2 + \frac{108}{x}\).

(b)
[A1] for \(4x - \frac{108}{x^2}\).

(c)
[M1] for setting \(\frac{\text{d}A}{\text{d}x} = 0\).
[A1] for \(x = 3\).

(a) We use the compound interest formula:
\(FV = PV \times \left(1 + \frac{r}{100k}\right)^{kn}\)
where \(PV = 5000\), \(r = 4.5\), \(k = 12\) (compounded monthly), and \(n = 5\) years.

\(FV = 5000 \times \left(1 + \frac{4.5}{1200}\right)^{12 \times 5}\)
\(FV = 5000 \times (1.00375)^{60}\)
\(FV \approx 5000 \times 1.2517958... \approx 6258.979...\)

To the nearest cent, the amount in Leah's account is $6258.98.

(b) The monthly deposits in the piggy bank form an arithmetic series where:
- The first term, \(u_1 = 20\)
- The common difference, \(d = 5\)
- The number of terms, \(n = 60\)

Using the sum of an arithmetic series formula:
\(S_n = \frac{n}{2}[2u_1 + (n-1)d]\)
\(S_{60} = \frac{60}{2}[2(20) + (60-1)(5)]\)
\(S_{60} = 30[40 + 59 \times 5]\)
\(S_{60} = 30[40 + 295]\)
\(S_{60} = 30[335] = 10050\).

Leah will have saved $10,050 in her piggy bank after 5 years.

(a)
[M1] for substituting correct values into the compound interest formula (or setting up financial solver parameters on a GDC).
[A1] for showing correct substituted expression or intermediate calculation (e.g., 6258.98).
[A1] for the final correct answer to the nearest cent: $6258.98.

(b)
[M1] for identifying the series as arithmetic with \(u_1 = 20\), \(d = 5\), and \(n = 60\).
[M1] for substituting into the arithmetic sum formula.
[A1] for the correct final answer: $10050 (accept 10050.00).

(a) First, find the predicted electricity consumption for \(T = 25\,^{\circ}\text{C}\):
\(E_{\text{pred}} = 3.8(25) + 60.1 = 95 + 60.1 = 155.1\text{ kWh}\).

The residual is defined as:
\(\text{Residual} = \text{Observed} - \text{Predicted} = 150 - 155.1 = -5.1\text{ kWh}\).

(b) Substitute \(T = 18\) into the regression equation:
\(E = 3.8(18) + 60.1 = 68.4 + 60.1 = 128.5\text{ kWh}\).

(c) The estimation is reliable because \(18\,^{\circ}\text{C}\) lies within the range of the recorded temperatures (from \(15\,^{\circ}\text{C}\) to \(30\,^{\circ}\text{C}\)). This is an interpolation, which makes the estimation more trustworthy than extrapolation.

(d) The mean temperature will increase.
This is because the new temperature value of \(24.5\,^{\circ}\text{C}\) is greater than the current mean of \(22.4\,^{\circ}\text{C}\). Adding a data point above the current mean pulls the average value up.

(a)
[M1] for calculating the predicted value \(E_{\text{pred}} = 155.1\).
[A1] for correct residual of \(-5.1\) kWh.

(b)
[A1] for \(128.5\) kWh (accept 129 or 128.50).

(c)
[R1] for stating that the estimate is reliable.
[R1] for referring to the fact that 18 is within the data range (interpolation).

(d)
[A1] for stating "increase".
[R1] for the reasoning that \(24.5 > 22.4\) (the current mean).

(a) \(H_0\): Preferred subject and year group are independent (or "there is no association between preferred subject and year group").

(b) The expected value is calculated using:
\(E = \frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}}\)
For Year 10 and Science:
\(E = \frac{80 \times 80}{240} = \frac{6400}{240} = 26.6667... \approx 26.7\) (to 3 significant figures).

(c) \(\text{Degrees of Freedom} = (r - 1)(c - 1)\)
where \(r = 3\) and \(c = 3\).
\(\text{Degrees of Freedom} = (3 - 1)(3 - 1) = 2 \times 2 = 4\).

(d) Enter the observed contingency table into a matrix on the GDC and perform a \(\chi^2\) test of independence:
(i) \(\chi^2\text{ statistic} \approx 3.015... \approx 3.02\) (to 3 significant figures).
(ii) \(p\text{-value} \approx 0.555\) (to 3 significant figures).

(e) Since the \(p\)-value (0.555) is greater than the significance level (0.05), we fail to reject the null hypothesis \(H_0\). There is insufficient evidence to suggest that preferred subject and year group are not independent (we conclude that they are independent).

(a)
[A1] for stating independent (must specify preferred subject and year group in context).

(b)
[M1] for showing calculation \(\frac{80 \times 80}{240}\).
[A1] for showing correct intermediate value 26.6667... and concluding 26.7.

(c)
[A1] for 4.

(d)
[A1] for \(\chi^2 \approx 3.02\) (or 3.015).
[A1] for \(p \approx 0.555\).

(e)
[R1] for comparing \(p\)-value with 0.05 (or comparing \(\chi^2\) statistic 3.02 with critical value 9.488) and concluding to fail to reject \(H_0\) (or accept \(H_0\)).

(a) The volume of the box is given by \(V = x^2 h = 500\), which gives \(h = \frac{500}{x^2}\). The surface area of an open-topped box with a square base is \(A = x^2 + 4xh\). Substituting \(h\) into the area formula: \(A = x^2 + 4x\left(\frac{500}{x^2}\right) = x^2 + \frac{2000}{x}\).

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

(c) To find the minimum surface area, set the derivative to zero: \(2x - \frac{2000}{x^2} = 0 \implies 2x^3 = 2000 \implies x^3 = 1000 \implies x = 10\). Since \(\frac{\text{d}^2A}{\text{d}x^2} = 2 + \frac{4000}{x^3} > 0\) for \(x = 10\), this value minimizes the surface area.

(a) [3 marks]
M1 for expressing \(h\) in terms of \(x\): \(h = \frac{500}{x^2}\).
M1 for a correct expression for the surface area of an open box: \(A = x^2 + 4xh\).
A1 for substituting and obtaining the given expression: \(A = x^2 + \frac{2000}{x}\).

(b) [1 mark]
A1 for \(2x - \frac{2000}{x^2}\) (or equivalent).

(c) [2 marks]
M1 for setting their derivative to 0: \(2x - \frac{2000}{x^2} = 0\).
A1 for \(x = 10\).

(a) Let \(X\) be the mass of an apple, where \(X \sim N(150, \sigma^2)\). We are given \(P(X > 180) = 0.15\), which is equivalent to \(P(X \le 180) = 0.85\). Standardizing the variable gives \(P\left(Z \le \frac{180 - 150}{\sigma}\right) = 0.85\). Using a standard normal table or GDC, the \(z\)-score corresponding to a cumulative probability of \(0.85\) is \(z \approx 1.0364\). Therefore, \(\frac{30}{\sigma} = 1.0364 \implies \sigma = \frac{30}{1.0364} \approx 28.9\text{ g}\).

(b) We want to find \(P(130 < X < 160)\). Using \(\sigma \approx 28.945\) (or \(28.9\)) and the cumulative normal distribution function on a GDC with mean \(150\) and standard deviation \(28.945\): \(P(130 < X < 160) \approx 0.390\) (3 s.f.).

(a) [3 marks]
M1 for standardizing or writing a correct probability equation: \(P(X > 180) = 0.15\) or \(P(Z < z) = 0.85\).
A1 for \(z \approx 1.036\).
A1 for \(\sigma \approx 28.9\) (accept 28.9 to 29.0, or 28.945).

(b) [3 marks]
M1 for setting up the required probability expression \(P(130 < X < 160)\).
M1 for correct GDC input with mean 150 and their \(\sigma\).
A1 for \(0.390\) (accept 0.391 if using rounded \(\sigma = 28.9\)).

(a) The midpoint of \(AB\) is \(M_{AB} = \left(\frac{1+5}{2}, \frac{2+4}{2}\right) = (3, 3)\). The gradient of \(AB\) is \(m_{AB} = \frac{4-2}{5-1} = 0.5\). The gradient of the perpendicular bisector is \(m_{\perp} = -\frac{1}{0.5} = -2\). The equation of the perpendicular bisector is \(y - 3 = -2(x - 3) \implies y = -2x + 9\).

(b) The midpoint of \(BC\) is \(M_{BC} = \left(\frac{5+3}{2}, \frac{4+8}{2}\right) = (4, 6)\). The gradient of \(BC\) is \(m_{BC} = \frac{8-4}{3-5} = -2\). The gradient of the perpendicular bisector is \(m_{\perp} = -\frac{1}{-2} = 0.5\). The equation of the perpendicular bisector is \(y - 6 = 0.5(x - 4) \implies y = 0.5x + 4\).

(c) The equidistant point is the intersection of the two perpendicular bisectors: \(-2x + 9 = 0.5x + 4 \implies 2.5x = 5 \implies x = 2\). Substituting into either equation gives \(y = 5\). Thus, the coordinates are \((2, 5)\).

(a) [3 marks]
M1 for finding midpoint \((3, 3)\) and gradient \(0.5\) of \(AB\).
M1 for finding the negative reciprocal gradient \(-2\).
A1 for the correct equation: \(y = -2x + 9\).

(b) [2 marks]
M1 for finding midpoint \((4, 6)\) and gradient \(-2\) of \(BC\) to determine gradient of perpendicular bisector is \(0.5\).
A1 for the correct equation: \(y = 0.5x + 4\) (or equivalent).

(c) [2 marks]
M1 for attempting to solve the simultaneous equations.
A1 for the coordinates \((2, 5)\).

(a) Using the compound interest formula: \(FV = PV \left(1 + \frac{r}{k}\right)^{kt}\), where \(PV = 10\,000\), \(r = 0.045\), \(k = 12\), and \(t = 5\). \(FV = 10\,000 \left(1 + \frac{0.045}{12}\right)^{60} = 10\,000 (1.00375)^{60} \approx 12\,517.96\). Rounding to the nearest dollar, we get \(\$12\,518\).

(b) We want to find the number of years \(t\) such that \(FV \ge 20\,000\). Set up the equation: \(20\,000 = 10\,000 (1.00375)^{12t} \implies 2 = (1.00375)^{12t}\). Taking the natural logarithm of both sides: \(\ln(2) = 12t \ln(1.00375) \implies 12t = \frac{\ln(2)}{\ln(1.00375)} \approx 185.16\text{ months}\). Dividing by 12 gives \(t \approx 15.43\text{ years}\). Since we require the minimum number of complete years, we round up to the next integer, which is \(16\) years.

(a) [3 marks]
M1 for substituting correct values into the compound interest formula.
A1 for \(12\,517.96\).
A1 for rounding to the nearest dollar: \(\$12\,518\).

(b) [3 marks]
M1 for setting up the equation or inequality to double the value (e.g. \(20\,000 = 10\,000(1.00375)^{12t}\) or using TVM solver on GDC).
M1 for solving to find the number of months \(185.16\) or years \(15.43\).
A1 for \(16\) (must be a whole number of years).

(a) The initial state vector for a bike at B is \(\mathbf{s}_0 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}\). After 1 day, the state is \(\mathbf{s}_1 = \mathbf{T}\mathbf{s}_0 = \begin{pmatrix} 0.2 \\ 0.5 \\ 0.3 \end{pmatrix}\). After 2 days, the state is \(\mathbf{s}_2 = \mathbf{T}\mathbf{s}_1 = \begin{pmatrix} 0.6 & 0.2 & 0.1 \\ 0.3 & 0.5 & 0.4 \\ 0.1 & 0.3 & 0.5 \end{pmatrix} \begin{pmatrix} 0.2 \\ 0.5 \\ 0.3 \end{pmatrix} = \begin{pmatrix} 0.25 \\ 0.43 \\ 0.32 \end{pmatrix}\). The probability that the bike is at Station A after 2 days is the first element, which is 0.25. (b) The initial state vector is \(\mathbf{x}_0 = \begin{pmatrix} 150 \\ 100 \\ 50 \end{pmatrix}\). Using a GDC to compute \(\mathbf{x}_5 = \mathbf{T}^5 \mathbf{x}_0\), we get \(\mathbf{x}_5 \approx \begin{pmatrix} 86.286 \\ 123.510 \\ 90.204 \end{pmatrix}\). To the nearest whole number, the expected number of bikes is: Station A: 86, Station B: 124, Station C: 90. (c) Let the steady-state vector be \(\mathbf{x} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}\). We solve \(\mathbf{T}\mathbf{x} = \mathbf{x}\) and \(x + y + z = 1\). This yields the system: \(-0.4x + 0.2y + 0.1z = 0\), \(0.3x - 0.5y + 0.4z = 0\), and \(x + y + z = 1\). From the first equation, \(z = 4x - 2y\). Substituting into the second gives \(3x - 5y + 4(4x - 2y) = 0 \implies 19x - 13y = 0 \implies y = \frac{19}{13}x\). Thus, \(z = \frac{14}{13}x\). Using the sum equation: \(x + \frac{19}{13}x + \frac{14}{13}x = 1 \implies \frac{46}{13}x = 1 \implies x = \frac{13}{46} \approx 0.283\). Then \(y = \frac{19}{46} \approx 0.413\) and \(z = \frac{14}{46} \approx 0.304\). (d) In the long run, the proportion of bikes at Station C is \(\frac{14}{46}\). Let \(N\) be the total number of bikes. We require \(\frac{14}{46} N \ge 80 \implies N \ge \frac{80 \times 46}{14} \approx 262.86\). Thus, the minimum total number of bikes required is 263.

(a) M1 for attempting to find T^2 or multiplying T by the state vector [0; 1; 0]. A1 for finding the second-day state vector [0.25; 0.43; 0.32]. A1 for correct probability of 0.25. (b) M1 for writing down the matrix multiplication T^5 * [150; 100; 50]. A2 for finding the values before rounding [86.286, 123.510, 90.204] (award A1 if one error). A1 for correct rounding to 86, 124, 90. (c) M1 for setting up Tx = x. M1 for writing down at least two correct linear equations. M1 for using the constraint x + y + z = 1. M1 for attempting to solve the system. A1 for x = 13/46 (approx 0.283) and y = 19/46 (approx 0.413). A1 for z = 14/46 (approx 0.304). (d) M1 for setting up the inequality (14/46)*N >= 80. A1 for finding N >= 262.86. A1 for rounding up to the next integer to get 263.

(a) The bearing of C from A is \(045^\circ\) and B from A is \(105^\circ\), so \(\angle CAB = 105^\circ - 45^\circ = 60^\circ\). The bearing of A from B is \(105^\circ + 180^\circ = 285^\circ\). The bearing of C from B is \(315^\circ\). Thus, \(\angle ABC = 315^\circ - 285^\circ = 30^\circ\). (b) Since the sum of angles in \(\triangle ABC\) is \(180^\circ\), \(\angle ACB = 180^\circ - (60^\circ + 30^\circ) = 90^\circ\). In right-angled \(\triangle ABC\), \(AC = AB \cos(60^\circ) = 250 \times 0.5 = 125\text{ m}\). (c) In right-angled \(\triangle ACH\), \(\tan(12^\circ) = \frac{CH}{AC} \implies CH = 125 \tan(12^\circ) \approx 26.569 \approx 26.6\text{ m}\). (d) In right-angled \(\triangle BCH\), we first find \(BC = AB \sin(60^\circ) = 250 \times \frac{\sqrt{3}}{2} \approx 216.51\text{ m}\). Let \(\theta\) be the angle of elevation from B. \(\tan(\theta) = \frac{CH}{BC} = \frac{26.569}{216.51} \approx 0.1227 \implies \theta \approx 7.00^\circ\). (e) The shortest distance from C to the line segment AB is the perpendicular distance from C to AB. Let this distance be \(d\). In right-angled \(\triangle ACD\), \(d = AC \sin(60^\circ) = 125 \sin(60^\circ) \approx 108.25 \approx 108\text{ m}\).

(a) M1 for finding angle CAB = 105 - 45. A1 for showing CAB = 60. M1 for calculating the bearing of A from B as 285 (or using parallel lines). A1 for finding angle ABC = 30. (b) M1 for identifying that angle ACB = 90 (or using sine rule). M1 for writing down a valid trig ratio like AC = 250 * cos(60) or AC/sin(30) = 250/sin(90). A1 for AC = 125 m. (c) M1 for setting up tan(12) = CH/AC. A1 for CH = 125 * tan(12). A1 for CH = 26.6 m (or 26.57 m). (d) M1 for finding BC = 216.5 m (or 125*sqrt(3)). M1 for setting up tan(theta) = CH/BC. A1 for theta = 7.00 degrees (accept 7 degrees). (e) M1 for recognizing that the shortest distance is the perpendicular from C to AB. M1 for setting up d = 125 * sin(60). A1 for d = 108 m (accept 108.25 m).

(a) The value 5000 is the carrying capacity of the model. In context, it represents the maximum number of people in the town who can eventually hear the rumor. (b) At \(t = 0\), \(P(0) = 50\). Substituting these values: \(50 = \frac{5000}{1 + C e^0} \implies 1 + C = \frac{5000}{50} = 100 \implies C = 99\). (c) At \(t = 3\), \(P(3) = 414\). Substituting: \(414 = \frac{5000}{1 + 99 e^{-3k}} \implies 1 + 99 e^{-3k} = \frac{5000}{414} \approx 12.0773 \implies 99 e^{-3k} \approx 11.0773 \implies e^{-3k} \approx 0.111892 \implies -3k = \ln(0.111892) \approx -2.1902 \implies k \approx 0.730\) (to 3 s.f.). (d) The rate at which the rumor is spreading is given by the derivative \(P'(t)\). Using \(P(t) = \frac{5000}{1 + 99 e^{-0.730 t}}\) and calculating the derivative at \(t = 5\) using a GDC, we obtain \(P'(5) \approx 735.7 \approx 736\) people per day. (e) The rumor spreads fastest at the inflection point of the logistic curve, which occurs when the population is half of the carrying capacity: \(P(t) = 2500\). Setting up the equation: \(2500 = \frac{5000}{1 + 99 e^{-0.730 t}} \implies 1 + 99 e^{-0.730 t} = 2 \implies e^{-0.730 t} = \frac{1}{99} \implies t = \frac{\ln(99)}{0.730} \approx 6.29\) days. The number of people who know the rumor at this time is 2500.

(a) A1 for stating 5000 is the carrying capacity (or population limit). A1 for describing it as the maximum number of people who can hear the rumor. (b) M1 for substituting t = 0 and P = 50. M1 for solving the equation for C. A1 for C = 99. (c) M1 for substituting t = 3, P = 414, and C = 99 into the equation. M1 for rearranging to isolate e^(-3k). M1 for taking the natural logarithm of both sides. A1 for obtaining k = 0.730 (must show at least 4 s.f. like 0.7301 before rounding). (d) M1 for realizing that the rate is P'(t). M1 for using GDC to evaluate the derivative at t = 5. A1 for 736 people per day (accept 735.7). (e) M1 for stating that the maximum rate occurs when P = 2500 (half of carrying capacity). M1 for setting up the equation 2500 = 5000 / (1 + 99e^(-0.73t)). A1 for t = 6.29 days. A1 for P = 2500.

(a) The initial velocity is found by evaluating \(v(0)\): \(v(0) = 0^3 - 6(0)^2 + 9(0) + 2 = 2\text{ m s}^{-1}\). (b) Acceleration is the derivative of velocity: \(a(t) = v'(t) = 3t^2 - 12t + 9\). To find when acceleration is zero: \(3t^2 - 12t + 9 = 0 \implies 3(t^2 - 4t + 3) = 0 \implies 3(t-1)(t-3) = 0 \implies t = 1\text{ s or } t = 3\text{ s}\). (c) The local extrema occur where \(v'(t) = a(t) = 0\), which are at \(t = 1\) and \(t = 3\). Evaluating \(v(t)\) at these times: For \(t = 1\), \(v(1) = 1^3 - 6(1)^2 + 9(1) + 2 = 6\text{ m s}^{-1}\), giving local maximum point \((1, 6)\). For \(t = 3\), \(v(3) = 3^3 - 6(3)^2 + 9(3) + 2 = 2\text{ m s}^{-1}\), giving local minimum point \((3, 2)\). (d) Since the local minimum value of \(v(t)\) is positive (2) and the endpoints are positive, \(v(t) > 0\) for all \(t \ge 0\). The total distance is: \(\int_0^4 (t^3 - 6t^2 + 9t + 2) dt = \left[ \frac{t^4}{4} - 2t^3 + 4.5t^2 + 2t \right]_0^4 = (64 - 128 + 72 + 8) - 0 = 16\text{ m}\). (e) Displacement is the integral of velocity: \(s(t) = \int v(t) dt = 0.25t^4 - 2t^3 + 4.5t^2 + 2t + C\). Given \(s(0) = -5\), we find \(C = -5\). So, \(s(t) = 0.25t^4 - 2t^3 + 4.5t^2 + 2t - 5\). At \(t = 5\): \(s(5) = 0.25(625) - 2(125) + 4.5(25) + 2(5) - 5 = 156.25 - 250 + 112.5 + 10 - 5 = 23.75\text{ m}\).

(a) M1 for attempting to substitute t = 0 into v(t). A1 for v(0) = 2. (b) M1 for differentiating v(t) to get a(t) = 3t^2 - 12t + 9. M1 for setting a(t) = 0. A1 for t = 1 and t = 3. (c) M1 for identifying t = 1 and t = 3 as the t-coordinates of the extrema. A1 for calculating v(1) = 6. A1 for calculating v(3) = 2. A1 for writing the correct coordinates (1, 6) and (3, 2) with correct max/min classification. (d) M1 for setting up the integral of v(t) from 0 to 4. M1 for correct integration to find the antiderivative [t^4 / 4 - 2t^3 + 4.5t^2 + 2t]. A1 for 16 m. (e) M1 for integrating v(t) to find s(t) with constant C. A1 for finding C = -5. A1 for s(t) = 0.25t^4 - 2t^3 + 4.5t^2 + 2t - 5. A1 for s(5) = 23.75 m.

(a) Let \(W \sim N(250, 5^2)\). We want \(P(W > 255)\). Standardizing: \(Z = \frac{255 - 250}{5} = 1\). Using a GDC, \(P(Z > 1) \approx 0.158655 \approx 0.159\). (b) Let \(X\) be the number of packets weighing more than 255 g in a box of 10. \(X \sim B(10, 0.158655)\). We want \(P(X \ge 2) = 1 - P(X \le 1)\). Using a GDC, \(P(X \le 1) \approx 0.5200\). Thus, \(P(X \ge 2) \approx 1 - 0.5200 = 0.480\). (c) We want to find \(w\) such that \(P(W < w) = 0.01\). Using the inverse normal function on a GDC with \(\mu = 250\) and \(\sigma = 5\), we find \(w \approx 238.37\text{ g}\). To the nearest gram, \(w = 238\text{ g}\). (d) Let \(\mu\) be the population mean weight of the packets. The hypotheses are: \(H_0: \mu = 250\) and \(H_1: \mu < 250\). For the sample of 12 packets, the mean is \(\bar{x} \approx 246.83\text{ g}\) and the sample standard deviation is \(s_{n-1} \approx 2.443\text{ g}\). Performing a one-sample t-test with \(df = 11\): The t-statistic is \(t = \frac{246.833 - 250}{2.443 / \sqrt{12}} \approx -4.49\). The one-tailed p-value is \(p \approx 0.000451\) (or \(4.51 \times 10^{-4}\)). (e) Since the p-value \((0.000451) < 0.05\) (the significance level), we reject the null hypothesis \(H_0\). There is strong evidence to conclude that the mean weight of the coffee packets has decreased (it is significantly less than 250 g).

(a) M1 for setting up standard normal calculation or normalcdf(255, infinity, 250, 5). A1 for 0.159 (or 0.1587). (b) M1 for identifying the binomial distribution X ~ B(10, 0.1587). M1 for attempting to find 1 - P(X <= 1) using GDC. A1 for 0.480. (c) M1 for setting up the equation P(W < w) = 0.01. M1 for using inverse normal. A1 for 238 g (must be to the nearest gram). (d) A1 for correct null and alternative hypotheses (H0: mu = 250, H1: mu < 250). M1 for calculating the sample mean 246.83 and sample standard deviation 2.44. M1 for setting up or running the t-test on GDC. A1 for p-value = 0.000451 (accept any value rounding to 0.00045). (e) R1 for comparing the p-value with the significance level (0.00045 < 0.05). A1 for deciding to reject H0. R1 for interpreting the result in context (concluding that there is evidence that the mean weight of packets has decreased). A1 for a coherent concluding statement.