Part a (1 mark)

Identify first term 5000 and ratio 0.92.

Acceptable alternatives: Accept an equivalent mathematically valid method that demonstrates the same assessable decision.

Part a (1 mark)

Obtains the correct final result V_n=5000(0.92)^(n-1), and V_6=3295.41.

Part b (1 mark)

Apply ten years of declining-balance depreciation: 5000(0.92)¹⁰approx2171.94.

Acceptable alternatives: Accept an equivalent mathematically valid method that demonstrates the same assessable decision.

Part b (1 mark)

Obtains the correct final result 2171.94.

Part a (1 mark)

Represents the quantities in part a with a valid successive percentage change relationship.

Acceptable alternatives: Accept an equivalent mathematically valid method that demonstrates the same assessable decision.

Part a (1 mark)

Substitutes the values stated in part a consistently into that relationship.

Part a (1 mark)

Obtains the correct final result 17.5%.

Part a (1 mark)

From 13:47 to 15:47 is 2 hours.

Acceptable alternatives: Accept an equivalent mathematically valid method that demonstrates the same assessable decision.

Part a (1 mark)

Obtains the correct final result 2 hours 45 minutes.

Part a (1 mark)

Standardise 78 to obtain z=1.

Acceptable alternatives: Accept an equivalent mathematically valid method that demonstrates the same assessable decision.

Part a (1 mark)

Obtains the correct final result 0.1587.

Part b (1 mark)

Use x=72+1.282(6).

Acceptable alternatives: Accept an equivalent mathematically valid method that demonstrates the same assessable decision.

Part b (1 mark)

Obtains the correct final result 79.692, approximately 79.7.

Part a (1 mark)

Gives the correct result 16 minutes.

Acceptable alternatives: Accept an equivalent mathematically valid method that demonstrates the same assessable decision.

Part b (1 mark)

The total is 110 and 110 / 5 = 22.

Acceptable alternatives: Accept an equivalent mathematically valid method that demonstrates the same assessable decision.

Part b (1 mark)

Obtains the correct final result 22 minutes.

Part a (1 mark)

The route totals 6 + 4 + 5 = 15 minutes.

Acceptable alternatives: Accept an equivalent mathematically valid method that demonstrates the same assessable decision.

Part a (1 mark)

Represents the quantities in part a with a valid minimum route relationship.

Part a (1 mark)

Obtains the correct final result A-C-D-E, 15 minutes.

Part a (1 mark)

Interpret 2.4 using kilometres and minutes.

Acceptable alternatives: Accept an equivalent mathematically valid method that demonstrates the same assessable decision.

Part a (1 mark)

States a conclusion supported by the preceding work: Each additional kilometre is associated with 2.4 additional predicted minutes; the prediction is 39 minutes.

Part a (1 mark)

Use 240 / N = 45 / 180.

Acceptable alternatives: Accept an equivalent mathematically valid method that demonstrates the same assessable decision.

Part a (1 mark)

Obtains the correct final result 960 turtles.

Part a (1 mark)

Represents the quantities in part a with a valid without replacement relationship.

Acceptable alternatives: Accept an equivalent mathematically valid method that demonstrates the same assessable decision.

Part a (1 mark)

Obtains the correct final result 6 / 11.

Part a (1 mark)

Gives the correct result Each additional millimetre of rainfall is associated with 8.5 kL less predicted water use.

Acceptable alternatives: Accept an equivalent mathematically valid method that demonstrates the same assessable decision.

Part b (1 mark)

Gives the correct result 208 kL.

Acceptable alternatives: Accept an equivalent mathematically valid method that demonstrates the same assessable decision.

Part c (1 mark)

Residual = observed - predicted = 219 - 208.

Acceptable alternatives: Accept an equivalent mathematically valid method that demonstrates the same assessable decision.

Part c (1 mark)

The magnitude is close to 1 and the sign is negative.

Part c (1 mark)

The linear relationship may not continue beyond the data.

Part c (1 mark)

States a conclusion supported by the preceding work: 11 kL. There is a strong negative linear association. It is extrapolation well outside the observed range.

Part a (1 mark)

(6 / 2)[3.2 + 2(4.8+5.6+4.4) + 2.8] = 106.8.

Acceptable alternatives: Accept an equivalent mathematically valid method that demonstrates the same assessable decision.

Part a (1 mark)

Represents the quantities in part a with a valid trapezoidal rule relationship.

Part a (1 mark)

Obtains the correct final result 106.8 square metres.

Part b (1 mark)

106.8(25) = 2670.

Acceptable alternatives: Accept an equivalent mathematically valid method that demonstrates the same assessable decision.

Part b (1 mark)

Obtains the correct final result 2670 cubic metres.

Part c (1 mark)

Gives the correct result 2670 kL.

Acceptable alternatives: Accept an equivalent mathematically valid method that demonstrates the same assessable decision.

Part a (1 mark)

The orange probability is 3 / 15.

Acceptable alternatives: Accept an equivalent mathematically valid method that demonstrates the same assessable decision.

Part a (1 mark)

The expected number is 80(3 / 15)=16.

Part a (1 mark)

Obtains the correct final result Expected orange results 16; conditional probability frac7(12).

Part a (1 mark)

Gives the correct result 30 L / min.

Acceptable alternatives: Accept an equivalent mathematically valid method that demonstrates the same assessable decision.

Part b (1 mark)

1.65 kL = 1650 L, and 1650 / 30 = 55.

Acceptable alternatives: Accept an equivalent mathematically valid method that demonstrates the same assessable decision.

Part b (1 mark)

Obtains the correct final result 55 minutes.

Part a (1 mark)

P(A or B) = 0.48 + 0.37 - 0.19 = 0.66.

Acceptable alternatives: Accept an equivalent mathematically valid method that demonstrates the same assessable decision.

Part a (1 mark)

Obtains the correct final result 0.34.

Part a (1 mark)

The midpoint of the roots is (-5 + 3) / 2 = -1.

Acceptable alternatives: Accept an equivalent mathematically valid method that demonstrates the same assessable decision.

Part a (1 mark)

Obtains the correct final result x = -1.

Part a (1 mark)

Use 11500(1 + 0.039 / 12)⁴⁸, approximately 13438.10.

Acceptable alternatives: Accept an equivalent mathematically valid method that demonstrates the same assessable decision.

Part a (1 mark)

Represents the quantities in part a with a valid monthly compounding relationship.

Part a (1 mark)

Obtains the correct final result AUD 13,438.10.

Part a (1 mark)

Substitute B_0=20000.

Acceptable alternatives: Accept an equivalent mathematically valid method that demonstrates the same assessable decision.

Part a (1 mark)

States a conclusion supported by the preceding work: B_1=19,500; the 600 repayment exceeds the first month's 100 interest.

Part b (1 mark)

Iterate the recurrence to identify the accumulated repayment series.

Acceptable alternatives: Accept an equivalent mathematically valid method that demonstrates the same assessable decision.

Part b (1 mark)

Sum the finite geometric series.

Part b (1 mark)

States a conclusion supported by the preceding work: B_n=120000-100000(1.005)^n.

Part a (1 mark)

Quarterly rate is 0.045 / 4 and there are 12 quarters.

Acceptable alternatives: Accept an equivalent mathematically valid method that demonstrates the same assessable decision.

Part a (1 mark)

Obtains the correct final result 25000(1.01125)^12.

Part b (1 mark)

Monthly rate is 0.044 / 12 and there are 36 months.

Acceptable alternatives: Accept an equivalent mathematically valid method that demonstrates the same assessable decision.

Part b (1 mark)

Obtains the correct final result 25000(1.0036667)³⁶, approximately.

Part c (1 mark)

Gives the correct result Evaluate both expressions and choose the larger final value.

Acceptable alternatives: Accept an equivalent mathematically valid method that demonstrates the same assessable decision.

Part a (1 mark)

Gives the correct result 15%.

Acceptable alternatives: Accept an equivalent mathematically valid method that demonstrates the same assessable decision.

Part b (1 mark)

64000(0.85)³ = 39304.

Acceptable alternatives: Accept an equivalent mathematically valid method that demonstrates the same assessable decision.

Part b (1 mark)

Obtains the correct final result AUD 39,304.

Part a (1 mark)

Use BC² = 52² + 68² - 2(52)(68)cos(104 degrees).

Acceptable alternatives: Accept an equivalent mathematically valid method that demonstrates the same assessable decision.

Part a (1 mark)

Represents the quantities in part a with a valid cosine rule relationship.

Part a (1 mark)

Obtains the correct final result Approximately 95.1 m.

Part a (1 mark)

Use 58000(0.84)⁵ for the five-year value.

Acceptable alternatives: Accept an equivalent mathematically valid method that demonstrates the same assessable decision.

Part a (1 mark)

Calculate 16% of that value for the sixth-year loss.

Part a (1 mark)

Represents the quantities in part a with a valid declining-balance depreciation relationship.

Part a (1 mark)

States a conclusion supported by the preceding work: Value after 5 years approximately AUD 24,256.29; sixth-year depreciation approximately AUD 3,881.01.

Part a (1 mark)

Two times [√3 / 4](10²) = 50sqrt(3).

Acceptable alternatives: Accept an equivalent mathematically valid method that demonstrates the same assessable decision.

Part a (1 mark)

Obtains the correct final result 50sqrt(3) square metres, approximately 86.6.

Part b (1 mark)

86.6 / 20 x 7 is approximately 30.3 minutes.

Acceptable alternatives: Accept an equivalent mathematically valid method that demonstrates the same assessable decision.

Part b (1 mark)

Obtains the correct final result 30 minutes.

Part a (1 mark)

Gives the correct result 3.6 mm.

Acceptable alternatives: Accept an equivalent mathematically valid method that demonstrates the same assessable decision.

Part b (1 mark)

Subtract and add 3.6 mm to 240 mm.

Acceptable alternatives: Accept an equivalent mathematically valid method that demonstrates the same assessable decision.

Part b (1 mark)

Obtains the correct final result 236.4 mm to 243.6 mm inclusive.

Part a (1 mark)

Gives the correct result 160.

Acceptable alternatives: Accept an equivalent mathematically valid method that demonstrates the same assessable decision.

Part b (1 mark)

92 = 160a^5.

Acceptable alternatives: Accept an equivalent mathematically valid method that demonstrates the same assessable decision.

Part b (1 mark)

Obtains the correct final result Approximately 0.895.

Part a (1 mark)

Use the mean plus or minus one standard deviation for the interval.

Acceptable alternatives: Accept an equivalent mathematically valid method that demonstrates the same assessable decision.

Part a (1 mark)

Calculate 12.4 + 1.282(1.8) = 14.7076 for the cutoff.

Part a (1 mark)

State that the z-score calculations rely on a normal model.

Part a (1 mark)

Represents the quantities in part a with a valid normal distribution relationship.

Part a (1 mark)

Obtains the correct final result Interval 10.6 kg to 14.2 kg; upper-ten-percent cutoff approximately 14.7 kg; parcel masses are approximately normally distributed.