Worked solutions and error analysis

Work. The product −3(x − 2) is −3x + 6, not −3x − 6. The corrected equation is 10 − 3x = 19, so −3x = 9 and x = −3. The problem is the sign in distribution, before the final division.

Answer: x = −3

Check. 4 − 3(−3 − 2) = 4 + 15 = 19.

Key takeaway. Replacement habit: write the two products (−3)x and (−3)(−2) before combining constants.

Work. The discount was 20% of the original price p, so the paid fraction is 80%: 0.80p = 72. Thus p = 90. Adding 20% of the reduced price uses the wrong base.

Answer: $90

Check. 20% of 90 is 18, and 90 − 18 = 72.

Key takeaway. Replacement habit: label 100% before converting a percent statement into an equation.

Work. The group means cannot receive equal weight because the group sizes differ. The combined total is 10(60) + 30(80) = 3000, across 40 students. The mean is 3000/40 = 75.

Answer: 75

Check. Three-quarters of the students belong to the mean-80 group, so the combined mean should be closer to 80.

Key takeaway. Replacement habit: reconstruct each group total before dividing by the combined count.

Work. The original denominator excludes x = 3. Simplification is valid only for x ≠ 3. The simplified equation yields the excluded input, so the original equation has no solution.

Answer: No solution.

Check. The original expression at 3 is 0/0, which is undefined, not 6.

Key takeaway. Replacement habit: write exclusions above the first algebraic transformation and audit every candidate against them.

Work. Division by x assumes x ≠ 0 and may discard a solution. Move all terms to one side: x² − 5x = x(x − 5) = 0. The zero-product property gives x = 0 or 5.

Answer: x = 0 or x = 5

Check. Both values satisfy x² = 5x.

Key takeaway. Replacement habit: factor before dividing by an expression that might equal zero.

Work. The squared expression becomes zero at x = −2, not at 2. In a(x − h)² + k, the inner expression is x − h; here h = −2 and k = −5.

Answer: (−2, −5)

Check. Substitution gives y(−2) = −5, the minimum. At x = 2, the output is 43.

Key takeaway. Replacement habit: find the input that makes the squared term zero instead of reading its sign mechanically.

Work. First test k = 1: the equation becomes 2x + 1 = 0, with one solution. For k ≠ 1, it is quadratic and has one real root when 4 − 4(k − 1) = 0, which gives k = 2. Both cases count.

Answer: k = 1 or k = 2

Check. At k = 1, x = −1/2; at k = 2, (x + 1)² = 0.

Key takeaway. Replacement habit: classify the equation degree before applying a degree-specific rule.

Work. The left side is nonnegative, so x ≥ 0. The squared equation x + 6 = x² has roots −2 and 3, but −2 violates the sign condition.

Answer: x = 3

Check. $\sqrt{\text{9}}$ = 3, whereas at −2 the original sides are 2 and −2.

Key takeaway. Replacement habit: capture sign restrictions before squaring, then verify the surviving candidate.

Work. The random selection is from the 18 seniors, not all 40 members. Six favorable seniors out of 18 possible seniors gives 6/18 = 1/3.

Answer: $\frac{\text{1}}{\text{3}}$

Check. 6/40 would describe selecting a senior officer from all club members.

Key takeaway. Replacement habit: state the selection group in words before writing a fraction.

Work. Both each observation and the mean increase by 100. Their differences remain x_i − $\overline{x}$. Since standard deviation depends on these differences, it does not change.

Answer: The standard deviation is unchanged.

Check. Data {1, 3} and {101, 103} have identical distances from their respective means.

Key takeaway. Replacement habit: reason about deviations from the new mean, not the sizes of the new observations.

Work. A volume has three length dimensions. Each is scaled by 5/2, so the volume factor is (5/2)³ = 125/8. The factor 5/2 applies only to corresponding lengths.

Answer: $\frac{\text{125}}{\text{8}}$

Check. Cubes with sides 2 and 5 have volumes 8 and 125.

Key takeaway. Replacement habit: write the units of the requested quantity and match their exponent to the scale factor.

Work. The formula s = rθ requires radians. Convert 60° = π/3 radians, then s = 12(π/3) = 4π. Or take 60/360 of circumference 24π.

Answer: 4π

Check. The arc is one-sixth of the circumference, so it cannot exceed the full circumference.

Key takeaway. Replacement habit: label angle units next to the variable in the formula.

Work. An ordered pair is written (x, y), so the requested input is 4. The calculation may be perfectly correct; the error occurs when transferring the result to the requested quantity.

Answer: 4

Check. Label the coordinates explicitly as x = 4, y = 9 before entering the response.

Key takeaway. Replacement habit: rewrite the target above the final answer line; tool output is not automatically the answer.

Work. The number of boxes b must be an integer with 24b ≥ 250. Thus b ≥ 250/24, and the smallest admissible integer is 11, not the nearest integer.

Answer: 11 boxes

Check. Ten boxes hold 240 chairs; eleven hold 264.

Key takeaway. Replacement habit: translate the operational requirement before deciding whether to round up or down.

Work. The study did not randomly assign app use, so confounding could explain the association. Volunteers also need not represent all students. The report overstates both causation and generalizability.

Answer: Neither the causal claim nor the schoolwide generalization is justified by this design alone.

Check. Motivation might influence both app use and grades; willingness to volunteer can also be related to academic behavior.

Key takeaway. Replacement habit: audit sampling and assignment separately before approving a statistical claim.