Skill Align ACT BSSS Mathematical Methods T Booklet 2 Practice Assessment Pack 0 - 2026 Edition
Original Skill Align Mathematical Methods T practice-assessment Pack 0, Booklet 2; not an official ACT BSSS examination and not endorsed by ACT BSSS or the ACT Government.
- Paper
- Booklet 2 Showcase
- Reading
- 10 minutes
- Writing
- 65 minutes
- Assessment
- 50 marks
Suggested Skill Align Booklet 2 conditions: 10 minutes reading time and 65 minutes working time. An approved graphics calculator or CAS may be used. Offline stored notes and programs and locally supplied reference material may be used only when authorised by the administering college. Internet access and external communication are prohibited. ACT BSSS assessment is school based. The administering college may vary timing, technology, stored-note, program and reference-material conditions. No public PDF download is supplied with Pack 0.
Booklet 2
Booklet 2 assesses technology-supported modelling, calculus, probability and inference, with all calculator output justified mathematically. Answer all questions. Show complete working, reasoning and interpretation. State exact values unless an approximation is requested, and include units and context where applicable. Suggested Skill Align Booklet 2 conditions: 10 minutes reading time and 65 minutes working time. An approved graphics calculator or CAS may be used. Offline stored notes and programs and locally supplied reference material may be used only when authorised by the administering college. Internet access and external communication are prohibited. ACT BSSS assessment is school based. The administering college may vary timing, technology, stored-note, program and reference-material conditions.
Question 6
10 marksQuestion 7
10 marksQuestion 8
10 marksQuestion 9
10 marksQuestion 10
10 marksWorked Solutions And Marking Guide
Question 6
(a) z=1.5.
(81-75) / 4=1.5.
(b) 0.0668.
Take the upper tail.
(c) 0.6826, so about 68.3 percent lie within four hours of the mean.
This interval is one standard deviation either side of the mean.
Detailed marking criteria
Part a (3 marks)
Award one mark for each numbered, task-specific statement in the indicative approach, to a maximum of 3. Award equivalent marks for independently valid alternative reasoning and results; do not require steps exclusive to another method. Do not award the final-result mark if the stated mathematical form is incomplete.
Mark-by-mark evidence:
- Uses z=(x-mu) / sigma with x=81,mu=75,sigma=4.
- Calculates (81-75) / 4.
- Obtains z=1.5.
Acceptable alternatives: For Q6(a), award up to 3 marks for an independently valid method that answers the stated task and reaches z=1.5. Map its reasoning and results to equivalent marks; do not require steps that belong exclusively to the indicative method. Retain any method explicitly required by the question.
Do not credit by itself: Using an invalid method or contradicting the task; in the indicative approach, check: Uses z=(x-mu) / sigma with x=81,mu=75,sigma=4. Giving the final result without establishing: Obtains z=1.5.
Part b (3 marks)
Award one mark for each numbered, task-specific statement in the indicative approach, to a maximum of 3. Award equivalent marks for independently valid alternative reasoning and results; do not require steps exclusive to another method. Apply consequential marking when a correct later method consistently uses an earlier incorrect value, unless the error materially simplifies the task. Do not award the final-result mark if the stated mathematical form is incomplete.
Mark-by-mark evidence:
- Recognises X>81 as the upper tail beyond z=1.5.
- Uses 1-P(Z<1.5).
- Obtains 1-0.9332=0.0668.
Acceptable alternatives: For Q6(b), award up to 3 marks for an independently valid method that answers the stated task and reaches 0.0668. Map its reasoning and results to equivalent marks; do not require steps that belong exclusively to the indicative method. Retain any method explicitly required by the question. Credit a standard-normal table or documented calculator distribution calculation with the correct event. Accept either correct standardisation or direct use of the stated mean and standard deviation.
Do not credit by itself: Using an invalid method or contradicting the task; in the indicative approach, check: Recognises X>81 as the upper tail beyond z=1.5. Giving the final result without establishing: Obtains 1-0.9332=0.0668.
Part c (4 marks)
Award one mark for each numbered, task-specific statement in the indicative approach, to a maximum of 4. Award equivalent marks for independently valid alternative reasoning and results; do not require steps exclusive to another method. Apply consequential marking when a correct later method consistently uses an earlier incorrect value, unless the error materially simplifies the task. For the final-result mark, include the requested units, context or justification.
Mark-by-mark evidence:
- Standardises 71 and 79 to z=-1 and z=1.
- Uses symmetry to calculate P(-1<Z<1)=2(0.8413)-1.
- Obtains 0.6826.
- Interprets this as about 68.3% of battery lives lying within four hours of the mean.
Acceptable alternatives: For Q6(c), award up to 4 marks for an independently valid method that answers the stated task and reaches 0.6826, so about 68.3 percent lie within four hours of the mean. Map its reasoning and results to equivalent marks; do not require steps that belong exclusively to the indicative method. Retain any method explicitly required by the question. Credit a standard-normal table or documented calculator distribution calculation with the correct event. Accept either correct standardisation or direct use of the stated mean and standard deviation.
Do not credit by itself: Using an invalid method or contradicting the task; in the indicative approach, check: Standardises 71 and 79 to z=-1 and z=1. Giving the final result without establishing: Interprets this as about 68.3% of battery lives lying within four hours of the mean.
Question 7
(a) 0.59.
118 / 200=0.59.
(b) Approximately 0.0348.
Use √(0.59(0.41) / 200).
(c) Approximately (0.522,0.658); the interval supports majority support if the sample is representative.
Use 0.59pm1.96(0.0348).
Detailed marking criteria
Part a (3 marks)
Award one mark for each numbered, task-specific statement in the indicative approach, to a maximum of 3. Award equivalent marks for independently valid alternative reasoning and results; do not require steps exclusive to another method. Do not award the final-result mark if the stated mathematical form is incomplete.
Mark-by-mark evidence:
- Identifies 118 successes from 200 observations.
- Calculates hat p=118 / 200.
- Obtains hat p=0.59.
Acceptable alternatives: For Q7(a), award up to 3 marks for an independently valid method that answers the stated task and reaches 0.59. Map its reasoning and results to equivalent marks; do not require steps that belong exclusively to the indicative method. Retain any method explicitly required by the question.
Do not credit by itself: Using an invalid method or contradicting the task; in the indicative approach, check: Identifies 118 successes from 200 observations. Giving the final result without establishing: Obtains hat p=0.59.
Part b (3 marks)
Award one mark for each numbered, task-specific statement in the indicative approach, to a maximum of 3. Award equivalent marks for independently valid alternative reasoning and results; do not require steps exclusive to another method. Apply consequential marking when a correct later method consistently uses an earlier incorrect value, unless the error materially simplifies the task. For the final-result mark, carry unrounded values until the final stated accuracy.
Mark-by-mark evidence:
- Uses √(hat p(1-hat p) / n).
- Substitutes √(0.59(0.41) / 200).
- Obtains an estimated standard error of approximately 0.0348.
Acceptable alternatives: For Q7(b), award up to 3 marks for an independently valid method that answers the stated task and reaches Approximately 0.0348. Map its reasoning and results to equivalent marks; do not require steps that belong exclusively to the indicative method. Retain any method explicitly required by the question. Credit an equivalent standard-error calculation only when the sample proportion, complement and sample size are substituted correctly.
Do not credit by itself: Using an invalid method or contradicting the task; in the indicative approach, check: Uses √(hat p(1-hat p) / n). Giving the final result without establishing: Obtains an estimated standard error of approximately 0.0348.
Part c (4 marks)
Award one mark for each numbered, task-specific statement in the indicative approach, to a maximum of 4. Award equivalent marks for independently valid alternative reasoning and results; do not require steps exclusive to another method. Apply consequential marking when a correct later method consistently uses an earlier incorrect value, unless the error materially simplifies the task. For the final-result mark, carry unrounded values until the final stated accuracy.
Mark-by-mark evidence:
- Uses 0.59pm1.96(0.0348).
- Calculates the endpoints 0.522 and 0.658.
- States the approximate 95% interval (0.522,0.658).
- Concludes that the interval supports majority support only if the random sample is representative and observations are independent.
Acceptable alternatives: For Q7(c), award up to 4 marks for an independently valid method that answers the stated task and reaches Approximately (0.522,0.658); the interval supports majority support if the sample is representative. Map its reasoning and results to equivalent marks; do not require steps that belong exclusively to the indicative method. Retain any method explicitly required by the question. Credit an equivalent confidence-interval calculation only when the centre, margin, both endpoints and required interpretation are recorded.
Do not credit by itself: Using an invalid method or contradicting the task; in the indicative approach, check: Uses 0.59pm1.96(0.0348). Giving the final result without establishing: Concludes that the interval supports majority support only if the random sample is representative and observations are independent.
Question 8
(a) At t=4 minutes; 12 litres per minute.
Solve r'(t)=2-0.5t=0.
(b) 256 / 3 litres, about 85.3 litres.
Evaluate int_0⁸(8+2t-0.25t²),dt=[8t+t²-t³ / 12]_0⁸=256 / 3.
(c) 32 / 3 litres per minute, about 10.7.
Divide the accumulated volume by eight minutes.
Detailed marking criteria
Part a (3 marks)
Award one mark for each numbered, task-specific statement in the indicative approach, to a maximum of 3. Award equivalent marks for independently valid alternative reasoning and results; do not require steps exclusive to another method. Do not award the final-result mark if the stated mathematical form is incomplete.
Mark-by-mark evidence:
- Differentiates to obtain r'(t)=2-0.5t.
- Solves r'(t)=0 to obtain t=4 and verifies the maximum on 0leq tleq8.
- Calculates r(4)=12 L / min.
Acceptable alternatives: For Q8(a), award up to 3 marks for an independently valid method that answers the stated task and reaches At t=4 minutes; 12 litres per minute. Map its reasoning and results to equivalent marks; do not require steps that belong exclusively to the indicative method. Retain any method explicitly required by the question. Credit a derivative sign analysis or valid concavity argument only when it proves the greatest rate on the stated time interval.
Do not credit by itself: Using an invalid method or contradicting the task; in the indicative approach, check: Differentiates to obtain r'(t)=2-0.5t. Giving the final result without establishing: Calculates r(4)=12 L / min.
Part b (3 marks)
Award one mark for each numbered, task-specific statement in the indicative approach, to a maximum of 3. Award equivalent marks for independently valid alternative reasoning and results; do not require steps exclusive to another method. Apply consequential marking when a correct later method consistently uses an earlier incorrect value, unless the error materially simplifies the task. Do not award the final-result mark if the stated mathematical form is incomplete.
Mark-by-mark evidence:
- Writes int_0⁸(8+2t-0.25t²),dt.
- Uses the antiderivative 8t+t²-t³ / 12.
- Obtains 256 / 3 L, approximately 85.3 L.
Acceptable alternatives: For Q8(b), award up to 3 marks for an independently valid method that answers the stated task and reaches 256 / 3 litres, about 85.3 litres. Map its reasoning and results to equivalent marks; do not require steps that belong exclusively to the indicative method. Retain any method explicitly required by the question. Credit an algebraically equivalent bounded integral or antiderivative only when the integrand, bounds, sign and required units are preserved.
Do not credit by itself: Using an invalid method or contradicting the task; in the indicative approach, check: Writes int_0⁸(8+2t-0.25t²),dt. Giving the final result without establishing: Obtains 256 / 3 L, approximately 85.3 L.
Part c (4 marks)
Award one mark for each numbered, task-specific statement in the indicative approach, to a maximum of 4. Award equivalent marks for independently valid alternative reasoning and results; do not require steps exclusive to another method. Apply consequential marking when a correct later method consistently uses an earlier incorrect value, unless the error materially simplifies the task. For the final-result mark, include the requested units, context or justification.
Mark-by-mark evidence:
- Uses average rate =total volume / time.
- Calculates (256 / 3) / 8=32 / 3.
- States approximately 10.7 L / min.
- Interprets the value as the constant inflow rate that would add the same total volume over eight minutes.
Acceptable alternatives: For Q8(c), award up to 4 marks for an independently valid method that answers the stated task and reaches 32 / 3 litres per minute, about 10.7. Map its reasoning and results to equivalent marks; do not require steps that belong exclusively to the indicative method. Retain any method explicitly required by the question.
Do not credit by itself: Using an invalid method or contradicting the task; in the indicative approach, check: Uses average rate =total volume / time. Giving the final result without establishing: Interprets the value as the constant inflow rate that would add the same total volume over eight minutes.
Question 9
(a) A=x(60-2x)=60x-2x^2.
From 2x+y=60, substitute y=60-2x.
(b) x=15 metres and y=30 metres.
Set A'(x)=60-4x=0.
(c) 450 m²; A''(x)=-4<0.
Use the second derivative and substitute the dimensions.
Detailed marking criteria
Part a (3 marks)
Award one mark for each numbered, task-specific statement in the indicative approach, to a maximum of 3. Award equivalent marks for independently valid alternative reasoning and results; do not require steps exclusive to another method. Do not award the final-result mark if the stated mathematical form is incomplete.
Mark-by-mark evidence:
- Uses the fencing constraint 2x+y=60.
- Rearranges to y=60-2x.
- Substitutes into A=xy to obtain A(x)=60x-2x^2.
Acceptable alternatives: For Q9(a), award up to 3 marks for an independently valid method that answers the stated task and reaches A=x(60-2x)=60x-2x^2. Map its reasoning and results to equivalent marks; do not require steps that belong exclusively to the indicative method. Retain any method explicitly required by the question.
Do not credit by itself: Using an invalid method or contradicting the task; in the indicative approach, check: Uses the fencing constraint 2x+y=60. Giving the final result without establishing: Substitutes into A=xy to obtain A(x)=60x-2x^2.
Part b (3 marks)
Award one mark for each numbered, task-specific statement in the indicative approach, to a maximum of 3. Award equivalent marks for independently valid alternative reasoning and results; do not require steps exclusive to another method. Apply consequential marking when a correct later method consistently uses an earlier incorrect value, unless the error materially simplifies the task. Do not award the final-result mark if the stated mathematical form is incomplete.
Mark-by-mark evidence:
- Differentiates to obtain A'(x)=60-4x.
- Solves A'(x)=0 to obtain x=15 m.
- Uses y=60-2x to obtain y=30 m.
Acceptable alternatives: For Q9(b), award up to 3 marks for an independently valid method that answers the stated task and reaches x=15 metres and y=30 metres. Map its reasoning and results to equivalent marks; do not require steps that belong exclusively to the indicative method. Retain any method explicitly required by the question. Credit a vertex argument or derivative-based optimisation only when it respects the physical domain and establishes the maximum.
Do not credit by itself: Using an invalid method or contradicting the task; in the indicative approach, check: Differentiates to obtain A'(x)=60-4x. Giving the final result without establishing: Uses y=60-2x to obtain y=30 m.
Part c (4 marks)
Award one mark for each numbered, task-specific statement in the indicative approach, to a maximum of 4. Award equivalent marks for independently valid alternative reasoning and results; do not require steps exclusive to another method. Apply consequential marking when a correct later method consistently uses an earlier incorrect value, unless the error materially simplifies the task. For the final-result mark, include the requested units, context or justification.
Mark-by-mark evidence:
- Evaluates A(15)=450 m^2.
- Finds A''(x)=-4.
- Uses A''(15)<0 to establish a local maximum.
- Checks the physical interval 0leq xleq30 or its zero-area endpoints to establish the global maximum.
Acceptable alternatives: For Q9(c), award up to 4 marks for an independently valid method that answers the stated task and reaches 450 m²; A''(x)=-4<0. Map its reasoning and results to equivalent marks; do not require steps that belong exclusively to the indicative method. Retain any method explicitly required by the question. Credit a vertex argument or derivative-based optimisation only when it respects the physical domain and establishes the maximum.
Do not credit by itself: Using an invalid method or contradicting the task; in the indicative approach, check: Evaluates A(15)=450 m^2. Giving the final result without establishing: Checks the physical interval 0leq xleq30 or its zero-area endpoints to establish the global maximum.
Question 10
(a) f(x)=3x² / 8.
Differentiate the cumulative distribution function on its continuous interval.
(b) 13 / 32.
F(1.5)-F(0.5)=27 / 64-1 / 64=13 / 32.
(c) √3(4), approximately 1.59.
The median m satisfies F(m)=1 / 2, so m³ / 8=1 / 2 and m³=4.
Detailed marking criteria
Part a (3 marks)
Award one mark for each numbered, task-specific statement in the indicative approach, to a maximum of 3. Award equivalent marks for independently valid alternative reasoning and results; do not require steps exclusive to another method. Do not award the final-result mark if the stated mathematical form is incomplete.
Mark-by-mark evidence:
- Uses f(x)=F'(x) on 0<x<2.
- Differentiates x³ / 8.
- Obtains f(x)=3x² / 8 on the support.
Acceptable alternatives: For Q10(a), award up to 3 marks for an independently valid method that answers the stated task and reaches f(x)=3x² / 8. Map its reasoning and results to equivalent marks; do not require steps that belong exclusively to the indicative method. Retain any method explicitly required by the question. Credit exact integration, a valid CDF calculation, or symmetry only when that method applies to the stated support and event.
Do not credit by itself: Using an invalid method or contradicting the task; in the indicative approach, check: Uses f(x)=F'(x) on 0<x<2. Giving the final result without establishing: Obtains f(x)=3x² / 8 on the support.
Part b (3 marks)
Award one mark for each numbered, task-specific statement in the indicative approach, to a maximum of 3. Award equivalent marks for independently valid alternative reasoning and results; do not require steps exclusive to another method. Apply consequential marking when a correct later method consistently uses an earlier incorrect value, unless the error materially simplifies the task. Do not award the final-result mark if the stated mathematical form is incomplete.
Mark-by-mark evidence:
- Uses P(0.5<X<1.5)=F(1.5)-F(0.5).
- Calculates 27 / 64-1 / 64.
- Obtains 13 / 32.
Acceptable alternatives: For Q10(b), award up to 3 marks for an independently valid method that answers the stated task and reaches 13 / 32. Map its reasoning and results to equivalent marks; do not require steps that belong exclusively to the indicative method. Retain any method explicitly required by the question. Credit exact integration, a valid CDF calculation, or symmetry only when that method applies to the stated support and event.
Do not credit by itself: Using an invalid method or contradicting the task; in the indicative approach, check: Uses P(0.5<X<1.5)=F(1.5)-F(0.5). Giving the final result without establishing: Obtains 13 / 32.
Part c (4 marks)
Award one mark for each numbered, task-specific statement in the indicative approach, to a maximum of 4. Award equivalent marks for independently valid alternative reasoning and results; do not require steps exclusive to another method. Apply consequential marking when a correct later method consistently uses an earlier incorrect value, unless the error materially simplifies the task. For the final-result mark, retain the requested exact form rather than replacing it with a decimal.
Mark-by-mark evidence:
- Sets the median condition F(m)=1 / 2.
- Solves m³ / 8=1 / 2 to obtain m³=4.
- States the exact median m=√3(4).
- Gives the decimal approximation mapprox1.59 within the support.
Acceptable alternatives: For Q10(c), award up to 4 marks for an independently valid method that answers the stated task and reaches √3(4), approximately 1.59. Map its reasoning and results to equivalent marks; do not require steps that belong exclusively to the indicative method. Retain any method explicitly required by the question. Credit exact integration, a valid CDF calculation, or symmetry only when that method applies to the stated support and event.
Do not credit by itself: Using an invalid method or contradicting the task; in the indicative approach, check: Sets the median condition F(m)=1 / 2. Giving the final result without establishing: Gives the decimal approximation mapprox1.59 within the support.
Diagnostic Checklist
| Topic | Questions | Marks | Marks Lost | Action |
|---|---|---|---|---|
| Unit 4 — Continuous random variables and the normal distribution — Standardise 81 hours | Q6(a) | 3 | ___ | Q6(a): practise this worked approach, or an independently valid equivalent method: Uses z=(x-mu) / sigma with x=81,mu=75,sigma=4; then Calculates (81-75) / 4; then Obtains z=1.5. |
| Unit 4 — Continuous random variables and the normal distribution — Using P(Z<1.5)=0.9332, find P(X>81) | Q6(b) | 3 | ___ | Q6(b): practise this worked approach, or an independently valid equivalent method: Recognises X>81 as the upper tail beyond z=1.5; then Uses 1-P(Z<1.5); then Obtains 1-0.9332=0.0668. |
| Unit 4 — Continuous random variables and the normal distribution — Using P(Z<1)=0.8413, find P(71<X<79) and interpret | Q6(c) | 4 | ___ | Q6(c): practise this worked approach, or an independently valid equivalent method: Standardises 71 and 79 to z=-1 and z=1; then Uses symmetry to calculate P(-1<Z<1)=2(0.8413)-1; then Obtains 0.6826; then Interprets this as about 68.3% of battery lives lying within four hours of the mean. |
| Unit 4 — Interval estimates for proportions — Find p | Q7(a) | 3 | ___ | Q7(a): practise this worked approach, or an independently valid equivalent method: Identifies 118 successes from 200 observations; then Calculates hat p=118 / 200; then Obtains hat p=0.59. |
| Unit 4 — Interval estimates for proportions — Find the estimated standard error | Q7(b) | 3 | ___ | Q7(b): practise this worked approach, or an independently valid equivalent method: Uses √(hat p(1-hat p) / n); then Substitutes √(0.59(0.41) / 200); then Obtains an estimated standard error of approximately 0.0348. |
| Unit 4 — Interval estimates for proportions — Construct an approximate 95 percent confidence interval and assess majority support | Q7(c) | 4 | ___ | Q7(c): practise this worked approach, or an independently valid equivalent method: Uses 0.59pm1.96(0.0348); then Calculates the endpoints 0.522 and 0.658; then States the approximate 95% interval (0.522,0.658); then Concludes that the interval supports majority support only if the random sample is representative and observations are independent. |
| Unit 3 — Further differentiation and applications — Find when the inflow rate is greatest and state that rate | Q8(a) | 3 | ___ | Q8(a): practise this worked approach, or an independently valid equivalent method: Differentiates to obtain r'(t)=2-0.5t; then Solves r'(t)=0 to obtain t=4 and verifies the maximum on 0leq tleq8; then Calculates r(4)=12 L / min. |
| Unit 3 — Integrals — Find the total volume added | Q8(b) | 3 | ___ | Q8(b): practise this worked approach, or an independently valid equivalent method: Writes int_0⁸(8+2t-0.25t²),dt; then Uses the antiderivative 8t+t²-t³ / 12; then Obtains 256 / 3 L, approximately 85.3 L. |
| Unit 3 — Integrals — Find and interpret the average inflow rate | Q8(c) | 4 | ___ | Q8(c): practise this worked approach, or an independently valid equivalent method: Uses average rate =total volume / time; then Calculates (256 / 3) / 8=32 / 3; then States approximately 10.7 L / min; then Interprets the value as the constant inflow rate that would add the same total volume over eight minutes. |
| Unit 3 — Further differentiation and applications — Show that the area is A(x)=60x-2x^2 | Q9(a) | 3 | ___ | Q9(a): practise this worked approach, or an independently valid equivalent method: Uses the fencing constraint 2x+y=60; then Rearranges to y=60-2x; then Substitutes into A=xy to obtain A(x)=60x-2x^2. |
| Unit 3 — Further differentiation and applications — Find the dimensions giving maximum area | Q9(b) | 3 | ___ | Q9(b): practise this worked approach, or an independently valid equivalent method: Differentiates to obtain A'(x)=60-4x; then Solves A'(x)=0 to obtain x=15 m; then Uses y=60-2x to obtain y=30 m. |
| Unit 3 — Further differentiation and applications — Find the maximum area and justify it is a maximum | Q9(c) | 4 | ___ | Q9(c): practise this worked approach, or an independently valid equivalent method: Evaluates A(15)=450 m²; then Finds A''(x)=-4; then Uses A''(15)<0 to establish a local maximum; then Checks the physical interval 0leq xleq30 or its zero-area endpoints to establish the global maximum. |
| Unit 4 — Continuous random variables and the normal distribution — Derive the probability density function for 0<x<2 | Q10(a) | 3 | ___ | Q10(a): practise this worked approach, or an independently valid equivalent method: Uses f(x)=F'(x) on 0<x<2; then Differentiates x³ / 8; then Obtains f(x)=3x² / 8 on the support. |
| Unit 4 — Continuous random variables and the normal distribution — Find P(0.5<X<1.5) | Q10(b) | 3 | ___ | Q10(b): practise this worked approach, or an independently valid equivalent method: Uses P(0.5<X<1.5)=F(1.5)-F(0.5); then Calculates 27 / 64-1 / 64; then Obtains 13 / 32. |
| Unit 4 — Continuous random variables and the normal distribution — Find the median of X, giving an exact value and a decimal approximation | Q10(c) | 4 | ___ | Q10(c): practise this worked approach, or an independently valid equivalent method: Sets the median condition F(m)=1 / 2; then Solves m³ / 8=1 / 2 to obtain m³=4; then States the exact median m=√3(4); then Gives the decimal approximation mapprox1.59 within the support. |