| Translate every observation into a model equation Write equations that can be used to find the constants. | Because [first input-output observation] and [second observation], the equations are [full model with first pair substituted] and [full model with second pair substituted]. | The setup shows which observations constrain the parameters. A time shift changes the exponent or logarithm argument even when it does not change elapsed time between observations. |
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| Solve, store, and check the fitted constants Find the constants as decimal approximations. | [Eliminate/subtract/divide] to obtain [one-parameter equation]. Thus [constants], to the requested precision. Substitution reproduces [given observations]. | Elimination uses the model structure instead of guessing. Stored precision prevents an early approximation from changing a later root or threshold. |
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| Compute a secant slope and interpret its units Average rate of change, followed by a computation or contextual interpretation. | The average rate is [shown difference quotient] [output units per input unit]. From [start] to [end], [quantity] [increased/decreased] by an average of [magnitude] [output units] per [input unit]. | The quotient is the slope of the line through the two endpoint observations. It compresses the net change over the interval into one constant rate. |
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| Use the secant estimate the question requests Use the average rate from the previous part to estimate. | Using the secant through [endpoint pair], the estimate is [anchor output] + [rate] times ([target input] - [anchor input]) = [estimate], in [units]. | The average rate fixes the slope, and an endpoint fixes the line. A nonlinear curve can share those endpoints and still differ everywhere between them. |
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| Explain secant comparisons and changing average rates A legacy B(iii) asks whether estimates are above/below the model, or whether later average rates are greater/less. | The estimates lie on the secant through [endpoints]. On [relevant span], [model] is [concave up/down], so [curve-versus-line or ordered-slope relationship]. Therefore [requested comparison]. | Concavity describes how slopes change. A chord averages those slopes between its endpoints, which also controls its position relative to the curve. |
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| Turn a stopping rule into a threshold and domain A model is usable until a boundary, or current official Q2 C(i) and C(ii). | Solve [model equals boundary] to obtain [relevant T]. The model applies for [interval], because [start] and [why predictions after T violate the contextual rule]. Include or exclude equality according to the condition. | The equation finds where the rule changes status; the inequality and context determine which side and whether equality belongs in the domain. |
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| Find the range on the contextual interval Possible output values during opening hours or a stated validity period. | On the contextual domain [input interval], [endpoint/vertex/branch checks] give [output range]. These are continuous model predictions in [units]; actual counts are whole numbers where appropriate. | An unrestricted range can include values at times that never occur in the scenario. Restricting the domain changes which extrema and outputs matter. |
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| Explain confidence using the observed interval A question compares the reliability of predictions at two different times. | There is more confidence at [input] because it lies within the observed interval [bounds]. The other prediction extrapolates beyond the data and assumes [contextual pattern] continues. | A model is calibrated using observations in a particular region. Predictions outside it rely on additional untested persistence assumptions. |
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| Define subtraction order before interpreting error Residuals, model-versus-estimate differences, or a maximum permitted prediction error. | Using [stated convention], the difference is [first value] - [second value] = [result]. Its sign means [which value is greater]; its magnitude is [amount] [units]. | Sign has meaning only after the subtraction order is fixed. Magnitude gives the size of the discrepancy; sign tells its direction. |
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| Explain why an exponential becomes a semi-log line A legacy subpart asks about an exponential model on a semi-log graph. | Because the logarithm of the output equals [linear expression in time], log-output is linear in time. The semi-log graph is a line with [signed slope], consistent with [base condition]. | Logarithms convert multiplication into addition, turning a constant multiplicative change into a constant additive change in log-output. |
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