Cuba 12th Grade Math Final Exam common mistakes
Cuba 12th Grade Math Final Exam · Understand the traps and how to avoid them.
When asked for the relative position of two lines or a line and a plane in space, the candidate computes the determinant of the three direction/connecting vectors and stops, declaring 'they intersect' or 'they are skew' from that single number without checking the parallel/contained cases separately.
The candidate collapses the classification into one test, forgetting that the determinant being zero covers three different situations (parallel distinct, coincident, coplanar-intersecting), and that the determinant being nonzero only rules out coplanarity.
What to do instead
Use a decision tree and write it out each time: compare direction vectors first (proportional or not). If proportional, test whether a point of one satisfies the other (coincident vs parallel distinct). If not proportional, compute the mixed product; zero means they intersect in a point (find it), nonzero means they are skew. For line–plane: check n·v; if zero, test whether the line's point lies in the plane (contained vs parallel). State the case name explicitly in the answer.
When computing a distance from a point to a plane or between skew lines, the candidate substitutes into the formula without normalising, using n·PQ instead of |n·PQ|/|n|, or forgetting the absolute value, and reports a negative distance or one scaled by the norm of the normal vector.
The formula is recalled as 'dot product over something' and the denominator (the modulus of the normal, or the cross product for skew lines) is dropped because it feels like a formality; the absolute value is dropped because the arithmetic 'came out positive anyway' in practice exercises.
What to do instead
Memorise the distance formula as a fraction with a visible denominator: d = |Ax₀+By₀+Cz₀+D| / √(A²+B²+C²), and for skew lines d = |(P₂−P₁)·(v₁×v₂)| / |v₁×v₂|. After computing, sanity-check the magnitude against the coordinates (a distance cannot exceed the distance between the two named points). Write the absolute value bars in the very first line of working, before substituting numbers, so they cannot be omitted later.
In a probability problem, the candidate uses the binomial formula for a situation where draws are without replacement from a small population, or treats events as independent when the question describes sampling without replacement, and multiplies probabilities that should be conditional.
The binomial is the most recent and most drilled model, so any 'number of successes in n trials' phrasing triggers it, and the distinction between with and without replacement is buried in a single word of the statement that the candidate reads past.
What to do instead
Before choosing a model, answer two questions in writing: are the trials independent (with replacement, or a large population), and is the number of trials fixed? If sampling is without replacement from a small population, use the hypergeometric/combinatorial count, not the binomial. When events are not independent, write the conditional probability P(B|A) explicitly and use the multiplication rule P(A∩B)=P(A)P(B|A). Circle the words 'with/without replacement' and 'independent' in the statement before starting.
When working with the normal distribution, the candidate uses the raw value x in the standard normal table instead of standardising with z=(x−μ)/σ, or misuses the symmetry/complement to obtain a probability greater than one or a negative probability, and in inference problems confuses the standard deviation of the sample with the standard error σ/√n.
The standardisation step is treated as an optional preliminary, and the table is consulted from memory of 'look up the number'; the sample-versus-population distinction is blurred because both are called 'standard deviation' in the statement.
What to do instead
Write the standardisation as an explicit line every time: P(X≤a)=P(Z≤(a−μ)/σ), and only then consult the table. Draw the bell curve and shade the required region before reading the table, so the complement or symmetry is decided visually. In inference, write the standard error as σ/√n in the first line and label it 'standard error', keeping σ (population) and s (sample) visually distinct. Check that every probability lies in [0,1].
The candidate interprets a p-value as the probability that the null hypothesis is true, or concludes that a non-significant result proves the null hypothesis, and describes a confidence interval as having a 95% chance of containing the true parameter for this particular sample.
The conditional direction of the p-value is reversed in ordinary speech ('the probability the result is due to chance'), and textbook phrasing is compressed. The correct statement is conditional on the null hypothesis and concerns data, not hypotheses, which is the opposite of what intuition supplies.
What to do instead
Write out the definition in full sentences with the conditional made explicit: the p-value is the probability of data at least as extreme as observed, assuming the null model. Practise rewriting wrong interpretations into correct ones, and phrase confidence intervals as a property of the procedure over repeated sampling, not of the single computed interval.
The candidate computes P(A|B) as P(A)P(B), or uses P(A|B) = P(A∩B)/P(B) but substitutes P(A) for P(A∩B) when events are not independent, and applies the multiplication rule for independent events to dependent sampling without replacement.
The formula P(A∩B) = P(A)P(B) is over-learned from independence examples and gets applied as a default. Condition notation is also read backwards: candidates compute P(B|A) when P(A|B) is asked, especially in Bayes-type word problems where the given and target events are described in prose.
What to do instead
Before computing, write the two events in words and label which is given ('given B') and which is the target. Use the general rule P(A∩B) = P(A|B)P(B) unless independence is explicitly justified, and for without-replacement problems draw the tree with changing denominators. For Bayes problems, write the partition and total probability denominator explicitly before dividing.
The candidate carries out a hypothesis test and states the conclusion as 'the null hypothesis is true' or 'proven', or interprets a p-value as the probability that H₀ is true, rather than reporting a decision about evidence at the stated significance level.
Everyday language treats 'not rejected' as 'confirmed', and the p-value's definition (probability of data at least as extreme, given H₀) is easily inverted in memory to 'probability H₀ is true'. The candidate also forgets to state the test statistic, degrees of freedom and the decision rule in context.
What to do instead
Use a fixed conclusion sentence: 'At the 5% significance level there is (insufficient/sufficient) evidence to reject H₀; we cannot conclude H₀ is true, only that it is not contradicted.' Write the p-value definition once in the margin as a reminder. Always report the test statistic, its distribution and degrees of freedom, and phrase the conclusion in terms of the original variables.
The candidate computes the number of arrangements of a multiset or a committee with repeated roles by treating identical items as distinguishable (using n! instead of n!/(n₁!n₂!…)), or adds counts where the cases are not mutually exclusive, producing double counting.
The multiplication principle is learned first and applied by default, and the division by factorials for identical objects feels like an arbitrary patch. Overlap between cases is hard to see without a systematic case structure, and small examples are rarely enumerated by hand to check.
What to do instead
Before computing, label each object as distinct or identical and decide whether order matters and whether repetition is allowed, writing the four-way classification explicitly. For case-based counting, define cases by a single discriminating property so they partition the set, then verify with a small analogue (three or four items) enumerated by hand or by listing.
The candidate solves a trigonometric equation such as sin(2θ) = 1/2 by writing only θ = 15° or a single principal value, ignoring the second solution in the period, the general solution with the period added, and any restriction on the interval given in the problem.
Inverse-trig buttons on a calculator return exactly one value, so the candidate equates 'the answer' with 'the calculator output'. Reference-angle symmetry and periodicity require deliberate reasoning that the calculator conceals, and the doubled angle in sin(2θ) makes the period of the solution set half of what intuition suggests.
What to do instead
Solve in two stages: first find all solutions for the inner argument over a full period using reference angles and quadrant analysis, then divide the entire solution set (including the period term) by the coefficient of θ and filter to the required interval. Always sketch the unit circle or one period of the graph to count how many solutions the interval should contain before listing them.
The candidate applies the Law of Sines to a triangle given two sides and an included angle (SAS) or three sides (SSS), or uses the Law of Cosines when a simple right-triangle ratio applies, and in SSA cases reports only the acute solution and misses the obtuse one.
Both laws look interchangeable in formula sheets, and the decision rule (which information is given) is not tied to a concrete procedure. The ambiguous SSA case is under-practised because most textbook problems are constructed to have a single clean answer.
What to do instead
Use a decision checklist keyed to the given data: ASA/AAS or a known side–opposite-angle pair means Law of Sines; SAS or SSS means Law of Cosines. For SSA, always compute the candidate angle and then its supplement, test both against the angle sum, and state how many triangles exist. Check the final triangle by verifying the angle sum and re-deriving one side with the other law.