QCAA General Maths EA study materials
QCAA General Maths EA study resources

QCAA General Maths EA: Unit 3: Bivariate Data, Sequences and Earth Geometry

QCAA General Maths EA · Work through your exam, one topic at a time.

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Unit 3: Bivariate Data, Sequences and Earth Geometry

Core concepts

Unit 3 links statistics, recursion and measurement. For bivariate data, you must be able to construct a scatterplot, describe association (direction, form, strength) and interpret the correlation coefficient r (between −1 and 1; sign gives direction, magnitude gives strength). Know that correlation measures linear association only, and that a strong r does not prove causation. Be able to find the least-squares regression line y = a + bx, use it for prediction, and distinguish interpolation from extrapolation. Understand residuals (actual minus predicted) and how an outlier or influential point distorts both r and the line.

For sequences, distinguish arithmetic (common difference d, linear growth) from geometric (common ratio r, exponential growth). Work fluently with the recursive forms tₙ₊₁ = tₙ + d and tₙ₊₁ = r·tₙ, and the closed forms tₙ = t₁ + (n−1)d and tₙ = t₁rⁿ⁻¹. Recognise simple first-order linear recurrence relations tₙ₊₁ = a·tₙ + b, whose long-term behaviour depends on |a|.

For earth geometry, use the great-circle distance formula d = 111.2 × angular distance (degrees) between two points on the same meridian or parallel, and understand time zones via longitude (15° per hour).

Traps

  • Confusing correlation with causation, or claiming r = 0.9 means 90% of variation is explained (that is r²).
  • Using a regression line far outside the data range (extrapolation) and treating the answer as reliable.
  • Mixing up n and n−1 in sequence formulas, or starting a geometric series at the wrong term.
  • Forgetting that points on different parallels are not separated by a simple 111.2 km per degree along the parallel unless you account for the parallel's radius.

Worked rule of thumb

A geometric sequence has t₁ = 5, r = 1.2. Find t₆. Using tₙ = t₁rⁿ⁻¹: t₆ = 5(1.2)⁵ ≈ 5 × 2.4883 ≈ 12.44. Always check whether the question asks for a term (tₙ) or a sum.