Relative Frequencies

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Relative Frequencies · Grade 11 · Worksheet 3

1. A sound engineer is analyzing the decibel level of a concert. The sound intensity I in watts per square meter is given by the formula L = 10 log(I/I₀), where L is the sound level in decibels and I₀ = 10⁻¹² W/m² is the reference intensity. If the sound level at the concert is 115 dB, what is the sound intensity I in watts per square meter? Round your answer to 3 decimal places. Answer: 0.316 Solution: Write the given formula: L = 10 log(I/I₀) Substitute the known values: 115 = 10 log(I/10⁻¹²) Divide both sides by 10: 11.5 = log(I/10⁻¹²) Convert from logarithmic to exponential form: I/10⁻¹² = 10¹¹·⁵ Multiply both sides by 10⁻¹²: I = 10¹¹·⁵ × 10⁻¹² Simplify using exponent rules: I = 10¹¹·⁵⁻¹² =…
   Full step-by-step solution

   Step 1: Write the given formula: L = 10 log(I/I₀) Step 2: Substitute the known values: 115 = 10 log(I/10⁻¹²) Step 3: Divide both sides by 10: 11.5 = log(I/10⁻¹²) Step 4: Convert from logarithmic to exponential form: I/10⁻¹² = 10¹¹·⁵ Step 5: Multiply both sides by 10⁻¹²: I = 10¹¹·⁵ × 10⁻¹² Step 6: Simplify using exponent rules: I = 10¹¹·⁵⁻¹² = 10⁻⁰·⁵ Step 7: Calculate 10⁻⁰·⁵ = 1/10⁰·⁵ = 1/√10 ≈ 1/3.1623 ≈ 0.316 Step 8: Round to 3 decimal places: 0.316 The answer is 0.316.

2. Emma is analyzing a two-way table that shows the relationship between participation in an advanced mathematics program and performance on a standardized test. The table includes 200 students. She calculates that the relative frequency of students who both participated in the program and scored in the top quartile is 0.35. If the school district has 1,500 students in Grade 11, what is the best estimate for the number of students who would both participate in the advanced program and score in the top quartile? Answer: 525 Solution: Identify the given relative frequency from the sample: 0.35. This relative frequency means that 35% of the sample students both participated in the program and scored in the top quartile.
   Full step-by-step solution

   Step 1: Identify the given relative frequency from the sample: 0.35. Step 2: This relative frequency means that 35% of the sample students both participated in the program and scored in the top quartile. Step 3: To estimate the number in the larger population of 1500 students, multiply the total population by the relative frequency: 1500 * 0.35. Step 4: Perform the multiplication: 1500 * 0.35 = 525. Step 5: The best estimate for the number of students who would both participate and score in the top quartile is 525.

3. Liam is analyzing data from a local library to understand reading habits. He finds that out of 450 library members surveyed, 180 members borrowed at least one non-fiction book last month. Based on this relative frequency, if the library has 5,000 members total, how many members would be expected to borrow at least one non-fiction book in a given month? Answer: 2000 Solution: Calculate the relative frequency of members who borrowed non-fiction books from the sample. Relative frequency = Number of members who borrowed non-fiction / Total sample size = 180 / 450.
   Full step-by-step solution

   Step 1: Calculate the relative frequency of members who borrowed non-fiction books from the sample. Relative frequency = Number of members who borrowed non-fiction / Total sample size = 180 / 450. Step 2: Simplify the fraction: 180/450 = 18/45 = 2/5 = 0.4. Step 3: Apply this relative frequency to the total library membership. Expected number = Relative frequency × Total members = 0.4 × 5000. Step 4: Perform the multiplication: 0.4 × 5000 = 2000. The answer is 2000.

4. Kaia surveyed 220 Grade 11 students about their preferred method of note-taking. 132 students preferred digital note-taking. What is the relative frequency of students who prefer digital note-taking, and what does this value mean in context? Answer: 0.6 Solution: Identify the number of students who prefer digital note-taking: 132 Identify the total number of students surveyed: 220 Calculate the relative frequency: 132 ÷ 220 Simplify the fraction: 132/220 = 3/5 = 0.6 Interpretation: The relative frequency of 0.6 means that 60% of the Grade 11 students…
   Full step-by-step solution

   Step 1: Identify the number of students who prefer digital note-taking: 132 Step 2: Identify the total number of students surveyed: 220 Step 3: Calculate the relative frequency: 132 ÷ 220 Step 4: Simplify the fraction: 132/220 = 3/5 = 0.6 Step 5: Interpretation: The relative frequency of 0.6 means that 60% of the Grade 11 students surveyed prefer digital note-taking. This represents the proportion of the total group that has this preference.

5. Emma is a botanist studying the germination rates of two varieties of native seeds, Kowhai and Manuka, under controlled conditions. She planted 75 Kowhai seeds and 45 Manuka seeds. After 21 days, 63 of the Kowhai seeds and 27 of the Manuka seeds had germinated. What is the relative frequency of germination for the Manuka seeds? Interpret this value in the context of the problem. Answer: 0.6 Solution: Identify the number of Manuka seeds that germinated: 27. Identify the total number of Manuka seeds planted: 45. Calculate the relative frequency: 27 / 45.
   Full step-by-step solution

   Step 1: Identify the number of Manuka seeds that germinated: 27. Step 2: Identify the total number of Manuka seeds planted: 45. Step 3: Calculate the relative frequency: 27 / 45. Step 4: Simplify the fraction: 27/45 = 3/5. Step 5: Convert to decimal: 3/5 = 0.6. Interpretation: The relative frequency of 0.6 means that 60% of the Manuka seeds germinated under the given conditions. This indicates the probability of a randomly chosen Manuka seed germinating is 0.6, based on this sample.

6. A research team is studying the decay of a radioactive isotope used in medical imaging. The initial mass of the sample is 120 grams, and its half-life is 8 days. The mass remaining after t days is modeled by the function M(t) = 120 * (1/2)^(t/8). Determine the time when only 15 grams of the isotope will remain. Answer: 24 days Solution: M(t) = 120 * (1/2)^(t/8) where M(t) is the mass in grams after t days. We want to find t when M(t) = 15 grams. Set up the equation.
   Full step-by-step solution

   We are given the decay model: M(t) = 120 * (1/2)^(t/8) where M(t) is the mass in grams after t days. We want to find t when M(t) = 15 grams. Step 1: Set up the equation. 15 = 120 * (1/2)^(t/8) Step 2: Divide both sides by 120 to isolate the exponential term. 15 / 120 = (1/2)^(t/8) Simplify 15/120: both divisible by 15 → 1/8. So: 1/8 = (1/2)^(t/8) Step 3: Express 1/8 as a power of 1/2. 1/8 = 1/(2^3) = (1/2)^3 So: (1/2)^3 = (1/2)^(t/8) Step 4: Since the bases are the same (and between 0 and 1), we can equate the exponents. 3 = t/8 Step 5: Solve for t. t = 3 * 8 t = 24 Step 6: Interpret the result. It will take 24 days for the 120-gram sample to decay to 15 grams. Final answer: 24 days