IB Math AA SL SL — All Flashcards

a × 10ᵏ, with 1 ≤ a < 10 and k a whole number. Example: 4.53 × 10⁴.

1 ≤ a < 10 (at least 1, less than 10). So 7 × 10³ ✓ but 12 × 10³ ✗.

Positive. Example: 52 000 = 5.2 × 10⁴.

Negative. Example: 0.0007 = 7 × 10⁻⁴.

Count how many places the point moves to leave one non-zero digit in front. Left → positive, right → negative.

7.3 × 10⁴.

6.1 × 10⁻⁵ — the ᴇ symbol means × 10.

The coefficient 45.3 is not between 1 and 10. Correct: 4.53 × 10⁷.

Add them. Example: 10³ × 10⁴ = 10⁷.

Subtract them. Example: 10⁸ ÷ 10³ = 10⁵.

Multiply the exponents. Example: (10⁴)² = 10⁸.

Multiply the coefficients, add the powers of ten, then re-normalise. Example: (3×10⁴)(2×10³) = 6×10⁷.

4 × 10⁶ — square the coefficient (2² = 4) and double the exponent.

(3×10²)³ = 27×10⁶ = 2.7 × 10⁷ cm³.

5 × 10⁻⁴ — 0.5 = 5 × 10⁻¹, so subtract 1 from the exponent.

2.7 × 10⁷ — 27 = 2.7 × 10¹, so add 1 to the exponent.

A sequence where each term differs from the previous one by a constant amount, the common difference d. Example: 4, 7, 10, 13 has d = 3.

The constant gap between consecutive terms: d = uₙ − uₙ₋₁. Subtract any term from the next. Example: in 9, 5, 1 the difference is d = −4.

Both use uₙ = u₁ + (n − 1)d. Want a VALUE (u₂₀)? Put n = 20 and compute. Want a POSITION ('which term = 100')? Set the formula equal to that value and solve for n. Spot value-vs-position first.

You start at u₁ and add d only on each step after the first, so reaching the nth term takes (n − 1) steps. Example: u₅ = u₁ + 4d.

Divide the difference by the number of steps between them: (41 − 17) ÷ (7 − 3) = 24 ÷ 4 = 6.

Set uₙ = u₁ + (n − 1)d equal to the value and solve for n. Example: 4 + (n−1)5 = 99 ⇒ n = 20.

Substitute n = 1 for the first term (u₁ = 16); the coefficient of n is the common difference (d = −4).

When the differences are equal: u₂ − u₁ = u₃ − u₂. The middle term is the average of its neighbours.

Set u₂ − u₁ = u₃ − u₂ and solve: (k + 1) = (3k − 5) ⇒ k = 3.

A sequence is the list of terms (3, 7, 11, …); a series is their sum (3 + 7 + 11 + …).

Spot u₁ first. If you want a TERM ('the 6th row'), it's n − 1 jumps (row 1 = 0 jumps). If u₁ is a starting amount and you want the value 'after n years/steps', it's n jumps. Always ask: how many times do I add d to get there?

Go straight to Sₙ = (n/2)(2u₁ + (n − 1)d): S₂₀ = 10(14 + 19×4) = 900. Trap: do not waste time finding u₂₀ first — the u₁-and-d form needs only what you are given.

It is a TOTAL, so use the sum: set Sₙ > 500 and solve for n, then round UP to the next whole number (on Paper 2, scan the GDC table of Sₙ). Spot 'total/altogether' ⇒ Sₙ, not uₙ.

Know u₁ and d → use (n/2)(2u₁ + (n − 1)d). Know u₁ and the last term → use (n/2)(u₁ + uₙ).

u₁ = S₁ — substitute n = 1 into the sum. Example: Sₙ = 2n² + 3n ⇒ u₁ = 5.

uₙ = Sₙ − Sₙ₋₁ — the running total up to n minus the running total up to n − 1.

Its sum is a quadratic in n with no constant term (Sₙ = an² + bn). The common difference is 2a.

Use S₅ = (5/2)(u₁ + u₅): 70 = (5/2)(u₁ + 20) ⇒ u₁ + 20 = 28 ⇒ u₁ = 8.

Pairing the first and last terms gives a constant total u₁ + uₙ, and there are n/2 such pairs, so Sₙ = (n/2)(u₁ + uₙ).

u₆ = S₆ − S₅ gives a term; combined with the sum formula you can solve for u₁ and d.

S₈ = (8/2)(10 + 45) = 4 × 55 = 220.

uₙ is a single term (the nth one); Sₙ is the total of the first n terms: Sₙ = u₁ + u₂ + … + uₙ.

Add up. Substitute the index from the lower limit to the upper limit into the expression and sum the results. Example: Σ r=1→4 of (2r+1) = 3+5+7+9 = 24.

The lower limit (below Σ) is where the index starts; the upper limit (above Σ) is where it stops. Both endpoints are included.

Upper limit − lower limit + 1. Example: Σ r=2→9 has 9 − 2 + 1 = 8 terms.

When the summand is linear in the index (like 3r + 2). The common difference equals the coefficient of the index.

First term 5, last term 32, n = 10; then S = (10/2)(5 + 32) = 185.

u₁ = the summand at the lower limit; d = the coefficient of the index. Then use the arithmetic sum formula.

Counting terms as (upper − lower) and forgetting the +1, or assuming the index starts at 1.

On Paper 2 use sum(seq(expression, index, lower, upper)). On Paper 1 you must use the arithmetic sum formula by hand.

Look for a quantity that changes by the same amount each step (a fixed raise, a fixed number per row). Then u₁ = the start and d = the constant change.

u₁ = 20 and d = 4. The nth value is uₙ = 20 + (n − 1)4.

At the last term that is still positive or zero — find where uₙ = 0. Adding later negative terms only shrinks the total.

Solve uₙ = 0 for n, then evaluate Sₙ at that position. Example: u₁ = 48, d = −3 ⇒ u₁₇ = 0 ⇒ S₁₇ = 408.

Graph Sₙ or scan a table of Sₙ and read off the largest value; the peak is at the term where uₙ = 0.

Set up an inequality with uₙ, solve for n, then round to the next whole number (n must be a positive integer).

90 − 7(n − 1) < 20 ⇒ n > 11 ⇒ n = 12; u₁₂ = 13.

A total or 'altogether' is a sum, so use Sₙ. A single 'nth' value is a term uₙ.

n counts terms (rows, years, balls), which only come in whole numbers; round a decimal n to the appropriate integer and check.

A sequence where each term is the previous one multiplied by a constant, the common ratio r. Example: 2, 6, 18, 54 has r = 3.

The constant multiplier: r = uₙ ÷ uₙ₋₁. Divide any term by the one before. Example: 12 ÷ 4 = 3.

'by 8%' multiplies ⇒ geometric, r = 1.08, uₙ = u₁rⁿ⁻¹. 'by 8' adds ⇒ arithmetic, d = 8. Words like percent / ratio / times / doubles signal geometric; a fixed amount signals arithmetic.

You start at u₁ and multiply by r only on each step after the first — (n − 1) times. Example: u₅ = u₁ r⁴.

Divide the values, then take the (steps)-th root: 48 ÷ 6 = 8 over 3 steps, so r = ∛8 = 2.

When the number of steps between the two terms is even, r = ±(root of the value-ratio). An odd number of steps gives a unique r.

When the ratios are equal: u₂/u₁ = u₃/u₂, i.e. u₂² = u₁u₃ (middle squared = product of neighbours).

k² = 4 × 25 = 100, so k = 10.

Arithmetic ADDS the same d each step; geometric MULTIPLIES by the same r. 3, 6, 9 is arithmetic; 3, 6, 12 is geometric.

u₆ = 5 × 2⁵ = 160.

Because consecutive ratios are equal: u₂/u₁ = u₃/u₂. Cross-multiplying gives u₂² = u₁u₃.

Up to two — solve the quadratic and report both. Use any stated condition (e.g. all terms positive) to choose between them.

Count the ×r jumps from the start — that count is the power. The nth TERM is n − 1 jumps (term 1 = 0 jumps); 'after n bounces / years' is n jumps (the start is counted). E.g. dropped 6 m, after the 4th bounce = 6(½)⁴ = 0.375.

Look for a TOTAL — 'sum of', 'altogether', 'total saved' over terms that multiply by a constant ⇒ Sₙ = u₁(rⁿ − 1)/(r − 1). A single value ('the 8th term', 'value after 8 years') ⇒ uₙ = u₁rⁿ⁻¹.

Either works. Use (rⁿ − 1)/(r − 1) when r > 1 and (1 − rⁿ)/(1 − r) when 0 < r < 1 to keep the numbers positive.

u₁, r and n. If r isn't given, find it first from two terms.

u₁ = 3, r = 2: S₅ = 3(2⁵ − 1)/(2 − 1) = 3 × 31 = 93.

Set Sₙ > target and solve for n; on Paper 2 scan the GDC table of Sₙ and round up to the next whole number.

If r = 1 every term equals u₁, so the sum is just n × u₁ (and the formula would divide by zero).

S₄ = 6(1 − 0.5⁴)/(1 − 0.5) = 6(0.9375)/0.5 = 11.25.

Compare with u₁(rⁿ − 1)/(r − 1): r = 3 and u₁/(3 − 1) = 2 ⇒ u₁ = 4.

Use sum(seq(u₁ r^(x−1), x, 1, n)) or read a table of Sₙ. Round n up for 'smallest n' questions.

Substitute u₁ and r into Sₙ = u₁(rⁿ − 1)/(r − 1) and simplify until it matches. E.g. u₁ = 4, r = 3 → 4(3ⁿ − 1)/2 = 2(3ⁿ − 1).

Distance = the first drop + 2 × (sum of the rebound heights). The drop counts once; every rebound is travelled up and down. Use the finite geometric sum Sₙ for the rebound heights.

Each period the balance multiplies by r = 1 + (rate as a decimal). After n periods: balance = start × rⁿ.

Growth: r = 1 + x/100. Decay: r = 1 − x/100. E.g. 6% growth → 1.06; 15% decay → 0.85.

Solve rⁿ = 2 (logs) or use the GDC/TVM solver; round n up to the next whole period.

2000 × 1.06ⁿ > 4000 → 1.06ⁿ > 2 → n ≈ 11.9 → 12 years.

Enter I% = rate, PV = −start, PMT = 0, FV = target, P/Y = C/Y = periods per year, then solve for N. Money out is negative.

Depreciation is decay: r = 1 − rate (0 < r < 1), so the value shrinks by a fixed percentage each period.

Compound multiplies the growing balance by r each period (geometric); simple adds a fixed amount each period (arithmetic).

r = 0.85; 20 000 × 0.85⁴ ≈ $10 440.

Put them into FV = PV(1 + r/(100k))^(kn) and solve for n — take logs, or on Paper 2 scan the GDC table for when the balance first reaches FV. Spot which letter is the unknown before substituting.

The number of compounding periods per year: annual k = 1, half-yearly 2, quarterly 4, monthly 12.

Divide the annual rate by k and raise to (k × n) periods: FV = PV(1 + r/(100k))^(kn).

5000(1 + 0.04/4)^(4×3) = 5000(1.01)¹² ≈ $5634.13.

Compound multiplies the growing balance by (1 + rate) each period (geometric); simple adds a fixed amount each year (arithmetic).

Write the one-year amount as PV(1 + x)⁴ for quarterly (the power = the number of periods in the year; x = the per-period rate), expand with the binomial theorem, and substitute the small x.

Yes — for the same nominal rate, monthly beats quarterly beats annual, because interest compounds sooner.

Interest = FV − PV (the growth above the amount invested).

1 + r/(100k) — one plus the per-period rate as a decimal.

Compound decay — a value loses a fixed percentage each year, multiplying by r = 1 − rate (0 < r < 1).

Set PV × rⁿ < X with r = 1 − rate, then solve for n (logs or the GDC table) and round UP to the next whole year. It is 'first below', so you need the smallest whole n.

24 000 × 0.88⁵ ≈ $12 666.

1 − b as a percent. E.g. V = 5000(0.92)ᵗ loses 8% a year.

Growth multiplies by 1 + rate (> 1); depreciation multiplies by 1 − rate (< 1).

It keeps a fixed percentage each year, so it shrinks geometrically but never actually hits 0.

The yearly multiplier: 90% is kept, so 10% is lost each year.

2000 × 0.7² = 2000 × 0.49 = $980.

N (periods), I% (annual rate as a %), PV (present value), PMT (regular payment), FV (future value), P/Y and C/Y (periods per year).

Money you pay out (invest) is negative; money you receive is positive.

The compounding frequency: 1 annually, 2 half-yearly, 4 quarterly, 12 monthly. N = years × that frequency.

Enter N, PV (negative), PMT = 0, FV, P/Y = C/Y; leave I% blank and solve.

Leave N blank, fill I%, PV (negative), PMT = 0, FV, P/Y = C/Y; solve, then round N up (and ÷ frequency for years).

Months — divide by 12 to get years.

A part-period hasn't reached the target yet, so you need the next whole period.

On Paper 2 for any compound-interest problem — especially finding the rate or the time, which are awkward by hand.

The power you raise a to, to get b. a^x = b ⇔ x = log_a b. Example: log₂ 8 = 3 because 2³ = 8.

Keep the base: log_a b = x. The log equals the exponent.

a^x = b. The log value x is the exponent of the base a.

log₁₀ x — base 10.

log_e x — the natural logarithm, base e ≈ 2.718.

Taking log_a undoes raising a to a power: log_a(a^x) = x, and a^(log_a b) = b.

1, because e¹ = e.

log₅ 125 = 3.

Ask 'a to what power gives b?' — write b as a power of a; the exponent is the answer. E.g. log₃ 81 = 4 since 3⁴ = 81.

0 — because a⁰ = 1.

1 — because a¹ = a.

1/16 = 2⁻⁴, so log₂(1/16) = −4 (a reciprocal gives a negative power).

3 = 9^(1/2), so log₉ 3 = ½ (a root gives a fractional power).

−log_a b — a reciprocal flips the sign.

Type log or ln on the GDC; for other bases use the change-of-base rule (topic 1.7).

8^(1/3) = 2, so log₈ 2 = ⅓.

A chain of justified steps from what you know to the result. The marks are the reasons, not the final answer.

Start from the given (or one side) and work toward the target. Never start from the answer and work backwards.

= (equation) is true for particular values you solve for; ≡ (identity) is true for ALL values.

Even = 2k, odd = 2k + 1, where k is an integer.

n, n + 1, n + 2, … — start from n and add 1 each time.

(2a + 1) + (2b + 1) = 2a + 2b + 2 = 2(a + b + 1), which is even. Use different letters a, b.

Manipulate it until you can take out a factor of k: write it as k × (an integer).

It is a product of two consecutive integers, and one of any two consecutive integers is even.

Reusing one letter (e.g. 2k + 1 twice) forces the two numbers to be equal, which breaks a general proof.

Reach the target exactly, then conclude in words — "… = 2m, which is even, so …". That sentence is often the last mark.

3: n + (n + 1) + (n + 2) = 3(n + 1).

n, n + 1, n + 2 — they go up by 1. Use one starting letter.

n + (n+1) + (n+2) = 3n + 3 = 3(n + 1) = 3 × a whole number.

One of any two consecutive integers is even, and an even factor makes the product even.

Take out a factor of k: write it as k × (a whole number).

Show it always leaves the same remainder: k × (whole) + r with r ≠ 0.

No — the sum is 3(n² + 2n + 1) + 2, so it always leaves remainder 2.

3 × the middle integer (3n + 3 = 3(n + 1)).

No — examples never prove 'for all'. Use algebra (n, n + 1, …) and reason generally.

Name them with one letter (n, n + 1, n + 2), then add or multiply.

= (equation) is true for particular values you solve for; ≡ (identity) is true for ALL values, which you prove.

Transform ONE side (the busier one) step by step into the other. Never move terms across the ≡.

One value only checks that single case; an identity must hold for every x.

The busier side — the one with brackets/powers to expand or fractions to combine.

Expand all brackets, then collect like terms until it matches the target.

a² − 2ab + b² — don't forget the middle term −2ab.

Put the side over a common denominator, combine the numerators, then simplify/factor.

x(x + 1) — the product of the two distinct denominators.

Use the identity/result you just proved — don't re-derive it from scratch.

Expand: (x² + 6x + 9) − (x² − 6x + 9) = 12x. The x² and 9 cancel.

aᵐ × aⁿ = aᵐ⁺ⁿ (multiply→add); aᵐ ÷ aⁿ = aᵐ⁻ⁿ (divide→subtract); (aᵐ)ⁿ = aᵐⁿ (power of a power→multiply).

a⁰ = 1 for any a ≠ 0. Example: 7⁰ = 1.

A reciprocal: a⁻ⁿ = 1/aⁿ. Example: 2⁻³ = 1/8.

A root: a^(1/n) = ⁿ√a, and a^(m/n) = (ⁿ√a)ᵐ. Example: 8^(2/3) = (∛8)² = 4.

Cube root first (∛27 = 3), then square: 3² = 9.

√x = x^(1/2), and the reciprocal flips the sign: 1/√x = x^(−1/2).

Raise both sides to the reciprocal 3/2: a = 4^(3/2) = (√4)³ = 8.

No — the laws need the SAME base. 2³ × 3² = 8 × 9 = 72 must be done directly.

It is a quadratic in disguise: a²ˣ = (aˣ)², so substitute y = aˣ, solve the quadratic, then solve back for x.

Because y = aˣ and a power is always positive — only positive y can give a real x. Discard zero or negative roots.

Coefficients go UP as powers, + becomes ×, − becomes ÷: 2 log x + log y − log z = log(x²y/z). To go the other way (one log → several), read the laws right-to-left. Spot whether they want 'a single log' or 'expanded'.

No — that is the #1 trap. The laws only act on products, quotients and powers, never on a sum.

log_a 1 = 0 (since a⁰ = 1) and log_a a = 1 (since a¹ = a).

It brings an exponent down to the front as a coefficient: log_a(xᵐ) = m log_a x. Example: log 8 = log 2³ = 3 log 2.

2 ln 3 = ln 9; then ln 6 + ln 9 − ln 2 = ln(6 × 9 ÷ 2) = ln 27.

24 = 2³ × 3, so log 24 = 3 log 2 + log 3 = 3p + q.

Change of base: log₂ 50 = (log 50)/(log 2) ≈ 5.64 (any base b works: log_a x = (log_b x)/(log_b a)). Use it whenever the base is not 10 or e, or to combine logs of different bases.

Change to base 2: log₂32 ÷ log₂8 = 5 ÷ 3 = 5/3.

Change to base 10: log 8 ÷ log 3 = 3 log 2 ÷ log 3 = 3p/q.

Product then power law: log₂8 + log₂x³ = 3 + 3 log₂ x.

Write both sides as powers of the same base, then equate the exponents. E.g. 4ˣ = 8 → 2²ˣ = 2³ → 2x = 3 → x = 3/2.

Take logs of both sides; the power law brings x down: x log a = log b, so x = log b / log a.

The power law: log aˣ = x log a turns the unknown exponent into a coefficient you can divide out.

x log 5 = log 20, so x = log 20 / log 5 = log₅ 20 (≈ 1.86).

Convert to exponential form: expr = aᶜ, then solve. E.g. log₃(x − 1) = 2 → x − 1 = 9 → x = 10.

e⁰ = x² − 16, so x² = 17 and x = ±√17 (both keep the argument positive).

Combine them into a single log with the product or quotient law, then convert to exponential form and solve.

A logarithm needs a positive argument, so reject any solution that makes an argument ≤ 0.

k⁴ = 81, so k = ⁴√81 = 3 (take the positive root).

Enter each side as Y₁ and Y₂, graph, then use 2nd → TRACE → 5: intersect. Check for more than one crossing.

Only when |r| < 1 — the terms shrink toward 0. Then S∞ = u₁/(1 − r).

Write it as a GP: 0.47 + 0.0047 + … with u₁ = 0.47 and r = 0.01. Then S∞ = u₁/(1 − r) = 0.47/0.99 = 47/99. Each repeating block is the previous one × 0.01.

If |r| ≥ 1 the terms don't approach zero, so the running total grows without limit — there is no finite sum.

r = 8/12 = ⅔, so S∞ = 12/(1 − ⅔) = 12/(⅓) = 36.

25 = 10/(1 − r) → 1 − r = 0.4 → r = 0.6.

u₁ = S∞(1 − r) = 40 × 0.8 = 32.

Putting r in the denominator instead of (1 − r), or using S∞ when |r| ≥ 1.

State that |r| ≥ 1, so the terms do not approach zero and the total is unbounded.

The gap is S∞ − Sₙ = u₁rⁿ/(1 − r). Set it below the tolerance and solve (GDC table or logs), rounding n up.

Use the finite sum with n = 2m: S₂ₘ = u₁(r²ᵐ − 1)/(r − 1). The power law r²ᵐ = (r²)ᵐ usually simplifies it (e.g. 3²ᵐ = 9ᵐ).

Total = drop + 2 × S∞ of the rebound heights, where the rebounds are a GP with first term (drop × r). For r = ½ this is 3 × the drop.

Start and end every row with 1; each inside number is the sum of the two directly above it. Rows: 1 / 1 1 / 1 2 1 / 1 3 3 1.

The coefficients of (a + b)ⁿ. E.g. row 3 (1, 3, 3, 1) → (a + b)³ = a³ + 3a²b + 3ab² + b³.

Take r = 3 factors counting down from 8 on top, r! on the bottom: ⁸C₃ = (8×7×6)/(3×2×1) = 56. The big factorials cancel — never expand them in full. (ⁿCᵣ = n!/(r!(n − r)!).)

5!/(2! 3!) = (5 × 4)/(2 × 1) = 10.

Type n, then MATH → ▶ (PRB) → 3: nCr, then r, then ENTER. E.g. 10 nCr 4 = 210.

Small n (about ≤ 6) and you want the WHOLE expansion → Pascal's triangle is fastest. Large n, or you only need ONE term/coefficient → use ⁿCᵣ (the general term ⁿCᵣaⁿ⁻ʳbʳ) and skip the rest.

n + 1 terms. E.g. (x + 2)⁹ has 10 terms.

The power of a falls from n to 0; the power of b rises from 0 to n; in every term the two powers sum to n.

Both equal 1 (the first and last coefficient of every row). The row is symmetric: ⁿCᵣ = ⁿCₙ₋ᵣ.

ⁿCᵣ is the entry in row n, position r (counting from 0). Row 5 = ⁵C₀, ⁵C₁, …, ⁵C₅ = 1, 5, 10, 10, 5, 1.

Multiply each coefficient ⁿCᵣ by a falling power of a and a rising power of b: aⁿ + ⁿC₁aⁿ⁻¹b + … + bⁿ.

Coeffs 1, 4, 6, 4, 1 with rising powers of 3: x⁴ + 12x³ + 54x² + 108x + 81.

Raise the WHOLE term: (3x)² = 9x², not 3x². Always bracket the term before squaring/cubing.

Use −b as the second term; even powers come out +, odd powers −. So the signs alternate: + − + − …

1 + 4(−2x) + 6(−2x)² + 4(−2x)³ + (−2x)⁴ = 1 − 8x + 24x² − 32x³ + 16x⁴.

Take r = 0, 1, 2, 3 in turn (the lowest powers of x) and stop — no need for the whole expansion.

1 + ¹⁰C₁x + ¹⁰C₂x² = 1 + 10x + 45x².

(1 + rate)ⁿ is the compound-interest factor; expanding it gives the value (or a quick approximation) by hand.

The two powers in every term sum to n, and there are n + 1 terms in total.

Two errors — the sign and the 3. Here a = 3x (not x) and b = −2 (not 2). Correct: ⁵C₃ (3x)²(−2)³ = 10 × 9 × (−8) x² = −720x². Always raise the WHOLE bracket term — number, variable and sign — to its power. (General term: ⁿCᵣaⁿ⁻ʳbʳ.)

Use the general term ⁿCᵣ aⁿ⁻ʳ bʳ; set the exponent of x equal to the power you want, solve for r, then compute that one term.

Write the general term, find the r that gives that power of x, and compute the coefficient (raising the whole coefficient/sign to the power).

The power of x is 0. Set the exponent of x to 0, solve for r, then compute that term.

Write that coefficient via the general term, set it equal to the given value, and solve. E.g. (x+k)⁷ coeff x⁵ = 63 → 21k² = 63 → k = ±√3.

An even power of the unknown (e.g. k²) gives two values. Check for a restriction like 'k > 0' before keeping both.

Use the simplest coefficient: ⁿC₂ = n(n − 1)/2 gives a quadratic in n; solve for the positive integer.

Use ⁿC₁k = (x coefficient) and ⁿC₂k² = (x² coefficient); eliminate k and solve for n, then k.

r = 2: ⁶C₂(2x)⁴(−3)² = 15 × 16 × 9 = 2160.

Form both equations and divide one by the other to eliminate a variable.

m = (y₂ − y₁)/(x₂ − x₁) = rise ÷ run. Subtract the coordinates in the same order top and bottom.

m > 0 uphill, m < 0 downhill, m = 0 horizontal (y = c), vertical lines (x = a) have no gradient.

Gradient–intercept y = mx + c; point–gradient y − y₁ = m(x − x₁); general ax + by + d = 0.

m is the gradient; c is the y-intercept (where the line crosses the y-axis).

Rearrange to y = mx + c — the gradient is m = −a/b.

Use point–gradient form y − y₁ = m(x − x₁), then expand and tidy.

Find the gradient m = (y₂ − y₁)/(x₂ − x₁) first, then use point–gradient form with either point.

Set x = 0 (or, in y = mx + c, read off c).

Set y = 0 and solve for x.

Vertical: x = a (gradient undefined). Horizontal: y = b (gradient 0).

When they have the same gradient: m₁ = m₂ (with different y-intercepts).

When their gradients multiply to −1: m₁m₂ = −1, i.e. m₂ = −1/m₁.

Take the negative reciprocal — flip the fraction and change the sign. E.g. ⅔ → −3/2.

Write 5 as 5/1; the perpendicular gradient is −1/5.

Use the SAME gradient, then point–gradient form y − y₁ = m(x − x₁).

Use the negative-reciprocal gradient, then point–gradient form.

The line through the midpoint of a segment, perpendicular to it (negative-reciprocal gradient).

((x₁ + x₂)/2, (y₁ + y₂)/2) — average each coordinate.

The line perpendicular to the tangent at a point; its gradient is −1/(tangent gradient). Used in calculus.

They are perpendicular, but a vertical line (x = a) has no gradient, so the product rule can't be applied — state it separately.

Find the value(s) of x that make both sides equal.

Rearrange so one side is 0, then find the roots.

You can lose the solution x = 0; factor instead.

The x-intercepts (zeros) of y = f(x).

The x-coordinates of the intersection points of the two graphs.

Graph both sides and use the GDC intersect tool (no neat algebra).

Paper 1: analytic (algebra). Paper 2: graphical / GDC.

Substitute it back — both sides should be equal.

The 'zero' tool (or read the x-intercepts).

Translates the graph up by k (down if k < 0) — an outside change, as expected.

Translates the graph RIGHT by a — inside changes move the opposite way.

To get the same output, x must be 3 bigger, so the graph sits 3 to the right.

Left 2 (inside +2 is the opposite of its sign).

Top a (right), bottom b (up).

(5, 6) — right 2, up 1.

Yes — every feature slides by the same vector.

Inside f acts on x (left/right, opposite sign); outside f acts on y (up/down, as written).

Stretches the graph vertically by factor a (every y ×a).

Stretches the graph horizontally by factor 1/b (the reciprocal).

Squash horizontally by factor 1/2 (the graph narrows).

Reflects the graph in the x-axis (y-coordinates flip sign).

Reflects the graph in the y-axis (x-coordinates flip sign).

(2, 15) — multiply y by 3.

(−2, 5) — negate x.

The x-intercepts (their y is 0, so 0 × a = 0).

Outside the function affects y; inside affects x (reciprocal for stretch, opposite for shift/flip).

A vertical stretch by a, then a translation k up — both on the y-values.

Multiply by 2 first (stretch), then subtract 1 (translate).

Stretch before translate: ×2 then −1, not −1 then ×2.

(1, 12) — multiply y by 3, x unchanged.

(3, 8) — right 1, up 5.

A translation 2 right and 5 up (vector (2, 5)).

A reflection in the x-axis, then a translation 4 up.

The inside ones: stretch by 1/b, then translate right h.

Translation / stretch (scale factor) / reflection (in which axis), with direction and amount.

The output of function f for input x. f(3) means substitute x = 3 into the rule.

No — it's 'f of x', the function applied to x. The brackets hold the input.

Each input gives exactly ONE output. (Different inputs may share an output.)

Replace every x with a (in brackets for negatives/expressions), then simplify.

(−3)² − 4(−3) = 9 + 12 = 21.

Set the rule equal to k and solve for x (output → input).

Yes — e.g. f(x) = x² gives f(2) = f(−2) = 4. So f(x) = k may have several solutions.

Go up from x = a to the curve, then across to the y-axis.

Read across from y = k to the curve, then down to the x-axis (there may be several x).

Substitute the whole expression: 3(2a) − 5 = 6a − 5.

The set of all allowed inputs (x-values).

The set of all possible outputs (y-values).

How far the graph extends left ↔ right (the x-extent).

How far the graph extends down ↕ up (the y-extent).

No dividing by zero (denominator ≠ 0) and no even root of a negative (under √ ≥ 0). Also a log argument must be > 0.

x ≠ 3 — the denominator can't be zero.

x ≥ 2 — what's under the root must be ≥ 0 (0 is allowed).

y ≥ k — the vertex (h, k) is the minimum.

y > 0 — always positive, approaching but never reaching 0.

All real numbers, x ∈ ℝ.

It undoes f: if f(a) = b then f⁻¹(b) = a. Inputs and outputs swap.

No — f⁻¹ is the inverse function (reverses f), not the reciprocal.

y = x. Each point (a, b) on f becomes (b, a) on f⁻¹.

Write y = f(x), swap x and y, then solve for y — that's f⁻¹(x).

Swap: x = 2y + 3 ⇒ y = (x − 3)/2, so f⁻¹(x) = (x − 3)/2.

They swap: domain of f⁻¹ = range of f; range of f⁻¹ = domain of f.

On the line y = x — solve f(x) = x to find where.

Pick a point: f(a) = b should give f⁻¹(b) = a. Or check f(f⁻¹(x)) = x.

Its domain is f's range, which can be limited (e.g. √x has range y ≥ 0, so its inverse x² is restricted to x ≥ 0).

It maps to itself — which is why f and f⁻¹ meet on y = x.

A sketch shows the correct SHAPE plus the KEY FEATURES labelled — not every point plotted exactly.

Axis intercepts, turning points (max/min), asymptotes, and correct end-behaviour — each labelled.

Set x = 0.

Set y = 0 and solve (factor, formula, or GDC).

From the leading coefficient: a > 0 opens up (minimum), a < 0 opens down (maximum).

A line the curve approaches but doesn't reach — drawn as a dashed guide line.

Where the denominator equals zero.

Graph it on the GDC, then transfer the shape to paper with intercepts, turning points and asymptotes labelled (values read off the GDC).

Not exact, but key points must be in the right relative positions and labelled with their values.

Vertical asymptote x = 2, horizontal asymptote y = 1, two branches approaching them.

Intercepts, maximum/minimum points, asymptotes, increasing/decreasing intervals, symmetry, and behaviour as x → ±∞.

An x-intercept — a value of x where f(x) = 0 (also called a root).

y-intercept: set x = 0. x-intercept (zero/root): set y = 0.

Point = coordinates (a, b); value = the y-coordinate b; 'where' = the x-coordinate a. Read the question.

Local = highest in its neighbourhood; global = highest over the whole graph.

A line x = a the curve shoots toward (±∞) — where a denominator is 0.

The value y approaches as x → ±∞ (the curve levels off).

As x increases, y increases — the graph goes up from left to right.

Decreasing for x < 0, increasing for x > 0 — it turns at the vertex (x = 0).

Graph it on the GDC and use the maximum/minimum tool to read the coordinates.

It lies on both curves, so f(x) = g(x) there; the shared value is the y-coordinate.

Set f(x) = g(x), bring everything to one side, solve for x, then substitute back for y.

Graph both functions on the GDC and use the intersect tool — once per crossing.

The horizontal line y = k.

The x-intercepts (zeros) — where the graph meets the x-axis.

Usually not — substitute each x back into a function to get the y-coordinate of the point.

Yes — e.g. a line can cut a parabola twice; find every crossing.

x² + 1 = 2x + 1 ⇒ x² − 2x = 0 ⇒ x = 0 or 2 ⇒ (0, 1) and (2, 5).

x³ − 2x = 1 (i.e. x³ − 2x − 1 = 0) — the intersection IS the solution.

f(g(x)) — apply the inner function g first, then f. Read right-to-left.

Not in general — order matters; the inner function changes the result.

Compute g(a) first, then put that value into f. Work inside-out.

Replace every x in f with the whole expression g(x) (in brackets), then simplify.

f(x²) = 2x² + 1.

g(2x + 1) = (2x + 1)² = 4x² + 4x + 1.

Form the composite expression, set it equal to k, and solve for x.

Compose with the unknown f, then match coefficients to the given result.

Doing the functions in the wrong order, or forgetting brackets when substituting (e.g. (2x+1)²).

Write y = f(x), swap x and y, solve for y. (Geometrically, reflect in y = x.)

Swap x and y, multiply up to clear the fraction, gather the y-terms, factor out y, then divide.

x = 5y − 2 ⇒ y = (x + 2)/5, so f⁻¹(x) = (x + 2)/5.

f(f⁻¹(x)) = x (and f⁻¹(f(x)) = x) — they undo each other.

They swap: domain of f⁻¹ = range of f; range of f⁻¹ = domain of f.

x² isn't one-to-one over all x; restricting to x ≥ 0 makes it invertible, giving f⁻¹(x) = √x.

Swap, multiply up, gather y: f⁻¹(x) = (3x + 1)/(x − 2).

No — f⁻¹ is the inverse function; 1/f is the reciprocal.

Standard ax²+bx+c, factored a(x−p)(x−q), vertex a(x−h)²+k.

a > 0 opens up (minimum); a < 0 opens down (maximum).

It's c — the constant term (set x = 0).

Set each bracket to zero: a(x − p)(x − q) gives x = p and x = q.

The turning point (h, k) and the max/min value k.

x = 4 and x = −1 (watch the sign on (x + 1)).

Exactly midway between the two x-intercepts.

Direction (sign of a), x-intercepts, y-intercept, and the vertex.

Factored form, a(x − p)(x − q).

x = −b/(2a) for y = ax² + bx + c — also midway between the x-intercepts.

Vertex form a(x − h)² + k, which shows the vertex (h, k) directly.

Halve b, square it for the bracket (x + b/2)², then adjust the constant to keep it equal.

It's k, reached at x = h: minimum if a > 0, maximum if a < 0.

(x − 3)² + 2, so the vertex is (3, 2).

(h, k) — note (x − 3)² means h = +3.

Write y = a(x − h)² + k, substitute the point to find a.

Value = k (a number); point = (h, k) (coordinates).

x = −(−6)/(2·1) = 3.

Set it to = 0, write as two brackets, then set each bracket to zero.

x = (−b ± √(b² − 4ac)) / (2a), for ax² + bx + c = 0.

Rearrange the equation so one side is 0.

Square-root both sides with ±: x − h = ±√n, then solve.

A square has two roots; dropping ± loses a solution.

The quadratic formula.

On Paper 2 — use the equation solver or read the x-intercepts.

(x − 2)(x − 3) = 0 ⇒ x = 2 or 3.

From ax² + bx + c = 0 (set to zero first); keep their signs.

Δ = b² − 4ac — the expression under the root in the quadratic formula.

Two distinct real roots; the graph cuts the x-axis twice.

One repeated real root; the graph touches the x-axis (tangent).

No real roots; the graph misses the x-axis.

Set line = curve, form a quadratic = 0, then set Δ = 0 (one intersection).

Δ = 0.

Δ < 0.

Δ = k² − 36 = 0 ⇒ k = 6.

No — just compute Δ and read its sign.

Rearrange to one side, then find the roots (solve = 0).

Between the roots.

Outside the roots: x < p or x > q.

Between the roots (the reverse of an upward one).

Two inequalities joined by 'or': x < p or x > q.

A single chain: p ≤ x ≤ q (or p < x < q).

Use ≤/≥ (closed) when equality is included; </> (open) when not.

Reverse the inequality sign.

Roots −2, 3; upward → outside: x < −2 or x > 3.

A hyperbola with two branches (top-right and bottom-left), hugging both axes.

Domain x ≠ 0, range y ≠ 0.

x = 0 (vertical) and y = 0 (horizontal).

None — it never reaches either axis.

Vertical x = h, horizontal y = k.

Right by 3, so the vertical asymptote is x = 3.

Raises the horizontal asymptote to y = k.

Draw the asymptotes, find any intercepts, then draw the two branches hugging the asymptotes.

Division by zero is undefined — that's the vertical asymptote.

Where the denominator is zero: cx + d = 0.

y = a/c — the ratio of the leading coefficients.

Where the numerator = 0 (a fraction is zero only when its top is zero).

At x = 0: y = b/d.

x = 4 (denominator zero).

y = 2 (leading coefficients 2/1).

Dividing top and bottom by x, the b and d terms vanish, leaving a/c.

One — the linear denominator has a single zero.

Draw the asymptotes, plot the x- and y-intercepts, then draw the two branches.

(0, 1), because a⁰ = 1.

Horizontal asymptote y = 0; range y > 0 (always positive).

a > 1 grows; 0 < a < 1 decays. Both pass through (0, 1).

It lifts to y = c (the curve levels off at c, not 0).

At x = 0: k·1 + c = k + c.

The initial value (at t = 0).

The per-period factor: b > 1 growth, b < 1 decay.

Solve A₀·bᵗ = target with logs, or graph and use intersect on the GDC.

No — it's always positive, never reaching the x-axis.

(1, 0), because logₐ1 = 0.

Vertical asymptote x = 0; domain x > 0.

y = aˣ — they're reflections in y = x.

All real numbers (y ∈ ℝ).

x = 0 isn't in the domain (log 0 is undefined).

No — it keeps increasing (slowly) forever.

x = h (the inside must be positive: x > h).

x > 2.

They swap (inverses): (0,1)↔(1,0), asymptote y=0↔x=0, domains/ranges swap.

d = √((x₂−x₁)² + (y₂−y₁)² + (z₂−z₁)²) — Pythagoras with a z-term.

((x₁+x₂)/2, (y₁+y₂)/2, (z₁+z₂)/2) — average each coordinate.

Distance squares the gaps and roots; midpoint averages the coordinates.

No — each gap is squared, so the sign disappears.

B = 2M − A (each coordinate).

√(4 + 9 + 36) = √49 = 7.

√(l² + w² + h²).

Average the two endpoints — you should recover the midpoint.

V = ⁴⁄₃πr³.

A = 4πr².

V = ⅓πr²h — one third of the cylinder.

πrl, where l is the slant height = √(r² + h²).

V = πr²h; A (closed) = 2πr² + 2πrh.

⅓ × base area × perpendicular height.

Add the volumes of the parts (subtract for a hole).

Don't count the join between two pieces; only exposed faces.

⁴⁄₃πr³ = 36π ⇒ r³ = 27 ⇒ r = 3.

Spot a right-angled triangle inside the solid and use SOH-CAH-TOA.

The angle between the line and its projection (shadow) on the plane.

A face or base diagonal — found with Pythagoras.

a√2 (= √(a² + a²)).

a√3 (= √(a² + a² + a²)).

A diameter subtends a right angle (90°) at any point on the circle.

It's much easier to apply trig to a flat 2D triangle than to the 3D picture.

Two steps: a length by Pythagoras, then an angle by trig.

sin θ = opp/hyp, cos θ = adj/hyp, tan θ = opp/adj.

The longest side, opposite the right angle.

Pick the ratio linking the angle, the wanted side and a known side; rearrange for the unknown.

Form the ratio, then take the inverse (sin⁻¹, cos⁻¹, tan⁻¹).

When you have two sides and need the third with no angle involved.

10 sin 30° = 5.

tan⁻¹(3/4) ≈ 36.9°.

Calculator in the wrong mode (degrees vs radians), or mislabelling opp/adj.

√(25 + 144) = 13.

a/sinA = b/sinB = c/sinC (side over the sine of its opposite angle).

a² = b² + c² − 2bc·cosA, with A opposite a.

cos A = (b² + c² − a²)/(2bc).

When you have a side with its opposite angle, plus one more side or angle.

For SAS (two sides + included angle → third side) or SSS (three sides → an angle).

Flip it: sinA/a = sinB/b, so the unknown sine is on top.

When A = 90°, cosA = 0 and a² = b² + c².

The cosine rule — it usually gives you a pair to then use the sine rule.

a² = 49 + 81 − 2·7·9·½ = 67 ⇒ a ≈ 8.19.

½ab·sinC, where C is the angle between sides a and b.

The included angle — the one between the two sides you use.

Set ½ab·sinC = Area, solve for sin C, then take sin⁻¹ (watch for the obtuse solution).

sin C = sin(180° − C), so an acute and an obtuse angle can give the same area.

½(6)(8)sin30° = 12.

Find it first (cosine rule from SSS, or sine rule), then use ½ab·sinC.

Yes, when C = 90° (sin 90° = 1) — it reduces to ½ × base × height.

Using a non-included angle, or forgetting the factor of ½.

The angle measured upward from the horizontal to a point above you.

The angle measured downward from the horizontal to a point below you.

The horizontal — never the vertical.

The depression angle down to an object equals the elevation angle from the object back up (alternate angles).

tan θ = height/horizontal distance (height opposite, distance adjacent).

h = 50 tan 30° ≈ 28.9 m.

Add the eye height to the triangle's height for the true total.

The sine or cosine rule.

Clockwise from North, written with three digits (e.g. 045°, 250°).

East 090°, South 180°, West 270° (North is 000°/360°).

With three digits: 007°.

The reverse direction — add 180° (if under 180°) or subtract 180° (if 180° or more).

070° + 180° = 250°.

135° (halfway between S 180° and E 090°, clockwise from N).

Find the interior angle at the turn, then use the cosine rule (two legs + included angle).

You must use the North lines at the turning point — the interior angle usually involves a 180° relationship.

The angle at the centre of a circle whose arc length equals the radius.

2π (and π in a half circle, 180°).

Multiply by π/180.

Multiply by 180/π.

π/6, π/4, π/3, π/2.

60 × π/180 = π/3.

3π/4 × 180/π = 135°.

sin/cos of a radian angle need radian mode; a mode mismatch gives wrong values.

Radians.

s = rθ, with θ in radians.

A = ½r²θ, with θ in radians.

Radians — convert from degrees first if needed.

θ = s/r.

Arc + two radii = rθ + 2r.

Sector area minus triangle area: ½r²θ − ½r²sinθ.

½r²sinθ (two sides r, included angle θ).

½(36)(1.5) = 27.

θ = 15/5 = 3 radians.

(cos θ, sin θ): cos is x, sin is y.

sin 30° = ½, cos 30° = √3/2.

sin 45° = cos 45° = √2/2 (= 1/√2).

sin 60° = √3/2, cos 60° = ½.

1/√3, 1, √3.

Positive ratios by quadrant: All (Q1), Sin (Q2), Tan (Q3), Cos (Q4).

sin θ — supplementary angles share the same sine.

−cos θ.

sin²θ = 1 − 4/9 = 5/9 ⇒ sin θ = √5/3.

Using the sine rule to find an ANGLE (two sides + a non-included angle, SSA).

Because sin θ = sin(180° − θ) — an acute and an obtuse angle share the same sine.

Subtract the acute sin⁻¹ value from 180°.

No — only finding an angle with the sine rule can give two answers.

No — cos⁻¹ gives a single angle in 0°–180°.

Add it to the known angle; keep it only if the total is under 180°.

B ≈ 36.9° or 180° − 36.9° = 143.1°.

Only 50°, since 70° + 130° = 200° > 180°.

sin²θ + cos²θ = 1, for every angle θ.

sin²θ = 1 − cos²θ.

cos²θ = 1 − sin²θ.

sin θ = ±√(1 − cos²θ); pick the sign from the quadrant.

sin²θ = 1 − 4/9 = 5/9 ⇒ sin θ = √5/3.

cos²θ.

sin²θ.

√ gives only the magnitude; the quadrant decides + or −.

Replace 1 − sin²θ or 1 − cos²θ with the other square, then cancel.

sin 2θ = 2 sin θ cos θ (not 2 sin θ!).

cos²θ − sin²θ = 1 − 2sin²θ = 2cos²θ − 1.

1 − 2sin²θ.

2cos²θ − 1.

2(3/5)(4/5) = 24/25.

2(16/25) − 1 = 7/25.

(cos²−sin²)(cos²+sin²) = cos 2θ.

Writing sin 2θ = 2 sin θ (dropping cos θ).

Find sin θ and cos θ first (often via the Pythagorean identity), then substitute.

−1 ≤ y ≤ 1.

360° (2π radians).

180° (π radians).

tan = sin/cos, so it blows up where cos x = 0 (90°, 270°, …).

No — it's unbounded (range all reals), so amplitude doesn't apply.

Amplitude = (max − min)/2.

cos x = sin(x + 90°) — same wave shifted left 90°.

At x = 0° and x = 360° (cos starts at its max).

The amplitude (vertical stretch).

360°/b (or 2π/b) — b divides the period.

Shifts the wave vertically; the midline is y = d.

The horizontal (phase) shift — right by c.

a = (max − min)/2.

d = (max + min)/2.

b = 2π/period (or 360°/period).

360°/2 = 180° (or 2π/2 = π).

max = d + a, min = d − a.

sin, cos and tan are periodic, so they hit the same value repeatedly.

Use x and 180° − x (then add periods if needed).

x and 360° − x.

x and x + 180°.

Solve for 2x over the doubled interval, find all solutions, then divide each by 2.

Up to four (the doubled interval 0°–720° gives twice as many).

Let s = sin x, factor (2s+1)(s−1)=0, then solve sin x = each value.

Use cos²x = 1 − sin²x (or vice versa) to get one ratio, then it's a quadratic.

Graph each side and use intersect (or graph the difference and find zeros) over the interval.

Every individual or item you want to know about — the whole group the study is about.

The part of the population you actually collect data from.

Data collected from the whole population (everyone).

It is cheaper, faster, or the test is destructive (so a census is impossible).

When it represents the population — chosen fairly and large enough.

One that over- or under-represents part of the population, so its results don't generalise.

A larger sample helps only if it is chosen fairly; a huge but unfair sample is still biased.

Destructive testing — e.g. measuring how long bulbs last, which destroys each bulb tested.

A parameter describes the population; a statistic is calculated from a sample and estimates the parameter.

Simple random, systematic, stratified, quota, convenience.

Every member of the population has an equal chance of being chosen (e.g. drawing lots or random numbers).

Order the population and take every k-th member after a random start.

k = population size ÷ sample size.

Split the population into groups (strata) and sample each in proportion to its size.

(group size ÷ population) × sample size.

Fill fixed numbers from each group, but choose the members non-randomly.

Choose whoever is easiest or first available.

Quota and convenience — the members are not chosen randomly, so the sample is often unrepresentative.

Substitute the known value into the line (y = ax + b for y from x).

The regression line of y on x.

Predicting a value inside the range of the original data — generally reliable.

Predicting a value outside the range of the data — generally unreliable.

The relationship may not continue outside the data range.

When it is interpolation AND the correlation is strong (|r| close to 1).

Yes — if the correlation is weak, even interpolation is unreliable.

Whether it's interpolation/extrapolation, and the strength of r.

No — predicting outside the data is unreliable regardless of r.

P(A | B) = P(A ∩ B) ÷ P(B).

The probability of A given that B has already occurred.

The given event — the one after the '|'.

Find P(A ∩ B) from the addition rule first.

The total of the given group B (e.g. all students), not the whole sample.

A conditional probability (given the first-stage outcome).

Use P(cause ∩ effect) ÷ P(effect), with P(effect) the total over all paths.

If P(A | B) = P(A), then A and B are independent.

P(A ∩ B) = P(A | B) × P(B).

z = (x − μ)/σ.

How many standard deviations x is above (z > 0) or below (z < 0) the mean.

The value equals the mean.

The value is below the mean.

They put values from different normal distributions on a common (standardised) scale.

The one with the larger z-value (further above its own mean).

Z ~ N(0, 1) — mean 0 and standard deviation 1.

No — it is a count of standard deviations, so it is unitless.

z = 2.

Given a left-tail probability P(X < x), it returns the value x.

invNorm(area, μ, σ), where area is the left-tail probability.

The left (lower) tail — the area to the left of x.

Use the left area 1 − p: x = invNorm(1 − p, μ, σ).

Each tail is (1 − c)/2; use those areas in invNorm.

Find z = invNorm(p, 0, 1), then σ = (x − μ)/z.

Find z = invNorm(p, 0, 1), then μ = x − zσ.

invNorm needs σ to return x directly; with σ unknown you must work through the standardised z.

Negative — a left-tail probability below 0.5 gives a negative z.

Each data value (or class) together with its frequency — how many times it occurs.

The data value that occurs most often (the highest frequency).

Add up all the frequencies.

A display of grouped continuous data using touching bars.

The frequency of that class.

The data is continuous, so the classes are adjacent intervals with no gaps.

The class (interval) with the greatest frequency — the tallest bar.

Histograms show continuous data (bars touch); bar charts show categories (bars have gaps).

The mode is the data value itself, not the frequency written beside it.

A running total of the frequencies up to the top of each class.

At the upper boundary of its class.

A smooth increasing S-shaped curve (an ogive).

Read across from a cumulative frequency of n/2, down to the data axis.

Read across from n/4 (Q1) and 3n/4 (Q3).

IQR = Q3 − Q1.

Subtract the cumulative frequency at a from the cumulative frequency at b.

The value below which 90% of the data lie — read across from 0.9n.

Read across from (100 − X)% of n, since the curve counts values below a level.

Minimum, lower quartile Q1, median, upper quartile Q3, maximum.

From Q1 to Q3 (the middle 50% of the data), with the median marked inside.

Maximum − minimum.

Q3 − Q1 — the spread of the middle 50%.

About 25% (a quarter) in each of the four sections.

A value is an outlier if it is below Q1 − 1.5·IQR or above Q3 + 1.5·IQR.

Find IQR, then the fences Q1 − 1.5·IQR and Q3 + 1.5·IQR; compare the value with them.

Compare the medians (centre) and the IQRs or ranges (spread).

Range uses the extremes (max − min); IQR uses the quartiles (Q3 − Q1), so it ignores outliers.

Add all the values and divide by how many there are (Σx ÷ n).

The middle value when the data is put in order.

The value that occurs most often.

Σfx ÷ Σf — multiply each value by its frequency, add, then divide by the total frequency.

The total frequency Σf, not the number of different values.

Total = mean × n.

The median.

The mean.

For categorical (non-numeric) data, where you can't add or order values.

Use Σfx ÷ Σf with x = the class midpoints.

(lower boundary + upper boundary) ÷ 2.

The exact values within each class are unknown, so midpoints are used to represent them.

The class with the greatest frequency.

The class containing the (n/2)-th value, found from the running (cumulative) total.

No — they are often different classes; find each separately.

Set Σfx ÷ Σf = the given mean (x = midpoints) and solve for the unknown frequency.

Σf, the total frequency.

Enter the midpoints as the data list and the frequencies as the frequency list, then run 1-Var Stats.

It is the middle value (for odd n) or the mean of the two middle values (for even n).

The median of the lower half of the ordered data.

The median of the upper half of the ordered data.

No — leave the median out of both halves before finding the quartiles.

Maximum − minimum.

IQR = Q3 − Q1, the spread of the middle 50%.

The IQR uses only the quartiles, so extreme values barely change it.

The mean of the two middle values.

Enter the data in a list and run 1-Var Stats — it lists Q1, Med and Q3.

How far values typically lie from the mean — the spread of the data.

The data is clustered close to the mean.

The standard deviation squared (σ²).

Enter the data in a list and run 1-Var Stats; read σx.

σx — the population standard deviation used in the IB syllabus.

Values in one list, frequencies in another, then run 1-Var Stats with both lists.

The mean increases by c; the standard deviation is unchanged.

Both are multiplied by |k|.

Take the square root: σ = √variance.

Paired (x, y) data plotted as points, revealing any relationship between the variables.

By its direction (positive/negative/none) and its strength (strong/weak).

As one variable increases, the other tends to increase too.

From −1 to 1 inclusive.

The direction of the correlation (positive or negative).

The strength — near 1 is strong, near 0 is weak.

The points lie exactly on a straight line (perfect linear correlation).

No — correlation does not imply causation; a third factor may explain it.

Linear only — a strong curved pattern can still give a small r.

The best-fit line y = ax + b used to predict y from x.

Enter the two lists and run linear regression (LinReg) — it gives a and b.

The change in y for each 1-unit increase in x.

The predicted value of y when x = 0.

The mean point (x̄, ȳ).

Substitute the known mean into y = ax + b (the mean point is on the line).

Solve the line of y on x and the line of x on y simultaneously — they meet at (x̄, ȳ).

The regression line of y on x.

The regression line of x on y.

Favourable outcomes ÷ total outcomes: P(A) = n(A)/n(U).

Between 0 and 1 inclusive.

P(A′) = 1 − P(A).

Use the complement: 1 − P(none).

36 (6 × 6 ordered outcomes).

Yes — they are different ordered outcomes.

Multiply the probabilities along the chain.

After each draw the totals reduce — one fewer item and one fewer of the drawn type.

With replacement the probabilities stay the same each draw; without, they change.

n × P, where P is the probability of the event each trial.

Yes — it is a long-run average, not a single count.

Find P first (from a sample space, proportion or table), then multiply by n.

The average number of occurrences you'd expect over many repeats.

Multiply the average per trial (expected value) by the number of trials n.

60 × 1/6 = 10.

40 × 0.25 = 10.

No — it is a long-run average, so a single run may differ.

100 × 1/2 = 50.

The union — elements in A or B (or both).

The intersection — elements in both A and B.

The complement — elements not in A.

The intersection (the 'both' region), then work outward.

Only A = n(A) − n(A ∩ B).

Region count ÷ total in the universal set.

P(A ∪ B) = P(A) + P(B) − P(A ∩ B).

So the overlap (in both A and B) isn't counted twice.

P(A) + P(B), because P(A ∩ B) = 0.

The probability of each outcome at that stage.

Multiply the probabilities along the branches of that path.

1.

Find each path (multiply along it) and add the matching paths.

The second-stage probabilities use reduced totals (one fewer item, one fewer of that type).

With replacement they repeat each stage; without, they change.

1 − P(none).

(3/5)(3/5) = 9/25.

(3/5)(2/4) = 3/10.

One event happening doesn't change the probability of the other.

P(A ∩ B) = P(A) × P(B).

Check whether P(A ∩ B) equals P(A) × P(B).

The events cannot both happen, so P(A ∩ B) = 0.

P(A) + P(B).

No — if they can't co-occur, knowing one occurred changes the other's probability, so they are dependent.

P(A) + P(B) − P(A)·P(B).

Substitute P(A ∩ B) = P(A)·P(B) into the addition rule and solve.

0.6 × 0.5 = 0.3.

A variable that takes separate values, each with a probability.

1.

Set the sum of all probabilities equal to 1 and solve.

E(X) = Σ x·P(X = x).

The expected (mean) value — the long-run average of X.

Yes — it's an average, so it can be a non-outcome value like 2.7.

When the expected net gain E(X) = 0.

Let X be the net gain, set E(X) = 0, and solve for the prize.

The net gain — include any cost or loss.

Fixed number n of independent trials, two outcomes each, with constant success probability p.

P(X = k) — the probability of exactly k successes.

P(X ≤ k) — the probability of at most k successes.

P(X = k) = ⁿCₖ pᵏ (1−p)ⁿ⁻ᵏ.

1 − P(X ≤ k − 1) = 1 − binomcdf(n, p, k − 1).

binomcdf(n, p, b) − binomcdf(n, p, a − 1).

1 − P(X = 0).

The probability of success changes between trials, so p is not constant.

Round up, since you need to reach the target probability with a whole number of trials.

E(X) = np.

Var(X) = np(1 − p).

√(np(1 − p)).

The expected number of successes in n trials.

Using np or np² instead of np(1 − p) — you must multiply by both p and (1 − p).

variance ÷ mean = 1 − p, so p = 1 − variance/mean.

n = mean ÷ p.