7th Grade Mathematics Semester 2 Final Comprehensive Review Study Guide

Linear and Proportional Relationships

• Alida’s Babysitting Earnings (Problem 1):     * Data Table:         * Hours: $1, 2, 3, 4$         * Earnings in Dollars ($y$): $5, 10, 15, 20$     * Proportional Relationship: The relationship between hours worked and earnings is proportional.     * Explanation: The graph forms a straight line that passes directly through the origin $(0,0)$.

• Constant of Proportionality (Slope) (Problem 2):     * Definition: The constant of proportionality is synonymous with the slope of the line.     * Calculation: Using the formula $\text{Slope} = \frac{\text{rise}}{\text{run}}$.     * Execution: Based on the graph points (such as $(0,0)$ and $(2,1)$), the slope is calculated as $\frac{1}{2}$ or $0.5$.

• Interpreting Points in a Relationship (Problem 3):     * Context: The graph displays "Meat Consumption" with pounds of meat consumed on the $y$-axis and number of people on the $x$-axis.     * Point $(0,0)$ meaning: This point represents zero people eating $0$ pounds of meat.     * Point $(5,10)$ meaning: This point represents that $5$ people eat $10$ pounds of meat.

• Proportional Tables (Problem 4):     * Criteria: A table shows a proportional relationship if the ratio $\frac{y}{x}$ is constant for all pairs.     * Example (Table B): Pairs $(2, 8)$, $(3, 12)$, $(4, 16)$, and $(5, 20)$. Ratios: $\frac{8}{2} = 4, \frac{12}{3} = 4, \frac{16}{4} = 4, \frac{20}{5} = 4$. This is proportional.

• Proportional Prediction with Ingredients (Problem 14):     * Data: $1.5$ cups of flour makes $6$ pies; $1.75$ cups of flour makes $7$ pies.     * Unit Rate: $\frac{6}{1.5} = 4$ pies per cup of flour.     * Application: To find the number of pies from $8$ cups of flour: $8 \times 4 = 32$ pies.

Percentages, Interest, and Discounts

• Modeling Proportions for Word Problems (Problem 5):     * Scenario: In a survey, $40\%$ of motorcycle owners have more than one vehicle. There were $540$ people surveyed.     * Setup: To find the number of owners ($x$), the valid proportion is $\frac{x}{540} = \frac{40}{100}$.

• Calculating Gratuity (Problem 6):     * Parameters: Dinner bill is $\$45.00$ with a tip rate of $12\%$.     * Calculation: $45 \times 0.12 = 5.4$.     * Result: A tip of $\$5.40$.

• Percent Change (Problem 7):     * Formula: $\text{Percent Change} = \frac{\text{Change}}{\text{Original Amount}} \times 100$     * Application: Entries increased from $25$ to $45$.     * Steps: $\text{Change} = 45 - 25 = 20$. $\text{Ratio} = \frac{20}{25} = 0.8$.     * Result: $80\%$ increase.

• Percentage of a Number (Problem 8):     * Context: Soccer camp attendance was $80$ last year. This year it is $145\%$ of last year.     * Calculation: $80 \times 1.45 = 116$.     * Result: $116$ players are at camp.

• Percent Discount (Problem 9):     * Context: A hoodie cost $\$60$ originally and was sold for $\$45$.     * Calculation: Total discount is $\$15$. Percent discount = $\frac{15}{60} = \frac{1}{4} = 25\%$.     * Result: $25\%$ off.

• Simple Interest (Problem 10):     * Formula: $I = prt$     * Variables: Principal ($p$) = $\$1,345$; Time ($t$) = $4$ years; Rate ($r$) = $3.5\%$ ($0.035$).     * Equation: $1345 \times 4 \times 0.035 = 188.3$.     * Result: $\$188.30$ in simple interest earned.

• Evaluating Percentage Calculation Methods (Problem 11):     * Valid methods for a $30\%$ discount:         * Multiply the original price by $0.30$         * Multiply the original price by $30\%$         * Move the decimal point one place to the left ($10\%$) then triple it (3 \times 10\% = 30\%\).     * Invalid Method: Setting up a proportion of the original price to the discount equal to $30$ over $100$ (it should be discount to original price).

Statistics and Data Analysis

• Biased vs. Unbiased Sampling (Problem 12):     * To determine the most popular club at school, surveying students attending an anime club meeting results in a biased sample because the population is limited to a convenience group already interested in a specific club.

• Box and Whisker Plots (Problem 13):     * Example: Blueberry Pies Sold per Day.     * Data Points: Plot shows Minimum = $1$, Q1 = $2$, Median = $4$, Q3 = $5$, Maximum = $10$.     * Key Fact: Each section of a box plot represents exactly $25\%$ of the data.     * Interpretation: The median number of pies sold is $4$.

• Inference from Comparative Box Plots (Problem 15):     * Comparison: Math Test 1 vs. Test 2.     * Observations: The medians for both tests are identical (both at $12$). The range for Test 1 was greater than Test 2.

Probability

• Likelihood of Events (Problem 16):     * Event $m$ is marked at $\frac{1}{8}$ on a probability scale from $0$ (Impossible) to $1$ (Certain).     * Category: Unlikely.

• Experimental vs. Expected Growth (Problem 17):     * Experiment: Rolling a $6$-sided die $60$ times.     * Expected Value: Every number should ideally appear $10$ times ($\frac{60}{6} = 10$).     * Observations: Roll results: 1:$10$, 2:$8$, 3:$10$, 4:$4$, 5:$14$, 6:$14$.     * Finding: We would have expected more $5s$ and $6s$ based on experimental data compared to theoretical probability.

• Theoretical Probability Predictions (Problem 18):     * Event: Rolling a multiple of two ($2, 4, 6$) on a $6$-sided cube.     * Probability: $P(\text{Multiple of } 2) = \frac{3}{6} = \frac{1}{2}$.     * Prediction for 200 Trials: $200 \times \frac{1}{2} = 100$ times.

• Spinner Probability (Problem 19):     * Results: Green: $23$, Red: $14$, Blue: $13$ (after $50$ spins).     * Interpretation: Green occurs approximately twice as often as Red or Blue. Spinner 1 (which shows half green and quarters for red and blue) is the logical source.

Geometry: Formulas and Applications

• Scale Drawings (Problem 20):     * Scale: $1\,\text{inch} = 20\,\text{feet}$.     * Measured height of model: $10.5\,\text{inches}$.     * Actual calculation: $10.5 \times 20 = 210\,\text{feet}$.

• Triangle Conditions (Problem 21):     * Interior angles of any triangle must sum to exactly $180^{\circ}$.     * Evaluating Set (B): $50^{\circ} + 60^{\circ} + 70^{\circ} = 180^{\circ}$ (This is a valid triangle).

• Cross Sections (Problem 22):     * A vertical cross section of a rectangular prism results in a rectangle or a square.

• Circumference of a Circle (Problem 23):     * Formula: $C = 2\pi r$     * Values: $r = 60\,\text{m}, \pi = 3.14$.     * Calculation: $2 \times 3.14 \times 60 = 376.8\,\text{m}$.

• Area of a Circle (Problem 24):     * Formula: $A = \pi r^2$     * Values: $d = 18\,\text{in} \rightarrow r = 9\,\text{in}$.     * Calculation: $3.14 \times (9^2) = 3.14 \times 81 = 254.34$.     * Rounded: $254.3\,\text{in}^2$.

• Area of Composite Figures (Problem 25):     * Composition: A rectangle ($20 \times 8$) and a semi-circle ($r = 4$).     * Rectangle Area: $20 \times 8 = 160$.     * Semi-circle Area: $\frac{1}{2} \pi r^2 = 0.5 \times 3.14 \times 4^2 = 0.5 \times 3.14 \times 16 = 25.12$.     * Total: $160 + 25.12 = 185.12$.

• Volume of Solids (Problem 28 and 29):     * Rectangular Tank height (Problem 28): $V = lwh$. $23400 = 36 \times 26 \times h$. $23400 = 936h$. $h = \frac{23400}{936} = 25\,\text{in}$.     * Triangular Prism Volume (Problem 29): $V = (\text{area of base}) \times \text{length} = (\frac{1}{2} \times 24 \times 7) \times 22 = 84 \times 22 = 1848\,\text{m}^3$.

• Surface Area (Problem 30):     * For a rectangular prism with dimensions $10\,\text{cm}, 5\,\text{cm}, \text{and } 2\,\text{cm}$:     * Front/Back: $10 \times 2 \times 2 = 40\,\text{cm}^2$.     * Top/Bottom: $10 \times 5 \times 2 = 100\,\text{cm}^2$.     * Note sum: $140\,\text{cm}^2$ (excluding side faces).

Solving for Unknown Variables in Geometry

• Supplementary Angles (Problem 26):     * Angles on a straight line sum to $180^{\circ}$.     * Equation: $7x + 145 = 180$.     * Solution: $7x = 35 \rightarrow x = 5$.

• Complementary Angles (Problem 27):     * Angles forming a right angle sum to $90^{\circ}$.     * Equation: $48 + x = 90$.     * Solution: $x = 42$.