Oak Construction Company Built 128 Houses In Hilltop Estates
1oak Construction Company Built 128 Houses In Hilltop Estates If The
Remove any rubric, grading criteria, point allocations, meta-instructions, due dates, repetitious instructions, and placeholder text. Focus solely on the core assignment question: Solve the provided statistical and financial problems involving construction data, weighted means, financial calculations, interest rates, depreciation, inventory valuation, probability, normal distribution, and decision-making processes.
Paper For Above instruction
The construction and evaluation of data related to business, finance, and statistics are crucial skills in understanding real-world economic activities. This paper addresses several key analytical problems, including calculating the number of specific types of houses built, determining a weighted mean, analyzing financial statements, computing discounts and interest, calculating depreciation, valuing inventory, and understanding probabilities and distributions in data analysis. Through detailed explanations and calculations, these problems exemplify the practical application of mathematical and statistical principles in everyday business scenarios.
Initially, we explore a construction company’s housing data. The problem states that Oak Construction built 128 houses, with the number of colonial style houses being seven times the number of ranch style houses. Let the number of ranch style houses be x. Then, the number of colonial houses is 7x. The equation becomes:
128 = x + 7x = 8x. Solving for x gives x = 128 / 8 = 16. Hence, the number of ranch style houses built by the company was 16.
Next, to calculate the weighted mean for given data points, we consider the values and their respective weights: values 10, 12, 15, 18, 20 with weights 2, 6, 8, 6, 2 respectively. The weighted mean formula is:
Weighted mean = (Σ value × weight) / (Σ weights). Calculating numerator: (10×2) + (12×6) + (15×8) + (18×6) + (20×2) = 20 + 72 + 120 + 108 + 40 = 360. Sum of weights: 2 + 6 + 8 + 6 + 2 = 24. Therefore, weighted mean = 360 / 24 = 15.
Addressing the financial statement, Bronson's Tire Co. reported assets of $545,350, which is $120,640 more than last year. The last year's assets were:
Assets last year = $545,350 - $120,640 = $424,710. Given owner’s equity was $192,500 last year, liabilities can be calculated via the accounting equation: Assets = Liabilities + Equity.
Rearranged, liabilities = Assets - Owner’s Equity = $424,710 - $192,500 = $232,210. Thus, last year's liabilities reported on the balance sheet were $232,210.
In the context of discounts, Laser Printing and Graphics offers discounts of 2/15, 1/20, and n/30, but the specific discount structure indicates a 2% discount if paid within 15 days. The customer purchased merchandise worth $18,000 on May 20, and paid on June 8—more than 15 days later. Therefore, no discount applies, and the net payment remains $18,000.
Regarding simple interest, Matt Stinson invested $4,000 at 5.5% per annum and received $385 in interest. The simple interest formula is:
I = P × r × t, where I is interest, P is principal, r is rate, and t is time in years. Solving for t yields:
t = I / (P × r) = 385 / (4000 × 0.055) = 385 / 220 ≈ 1.75 years. Expressed in months, 1.75×12 = 21 months.
For the study involving a discount loan, Junko Kiera borrowed $13,000 at a discount rate of 12% for 30 months. The true interest rate can be calculated by considering the discount and the actual amount received. The discount amount is:
Discount = Principal × Rate × Time = 13,000 × 0.12 × (30 / 12) = 13,000 × 0.12 × 2.5 = $3,900.
The amount received at borrowing is:
Amount received = Principal - Discount = 13,000 - 3,900 = $9,100. The true interest rate is:
True rate = Discount / (Amount received × Time in years) = 3,900 / (9,100 × 2.5) ≈ 0.171, or 17.1%.
Looking at compound interest, Klemson High-Tech Industries borrowed $65,000 at 14%, compounded semi-annually for three years. The formula for compound interest is:
Amount = Principal × (1 + r/n)^(nt), where r = annual rate, n = number of compounding periods per year, t = years.
Hence, Amount = 65,000 × (1 + 0.14/2)^(2×3) = 65,000 × (1 + 0.07)^6 ≈ 65,000 × 1.5036 ≈ $97,754.40.
The total interest paid is:
Interest = Final amount - principal = 97,754.40 - 65,000 ≈ $32,754.40.
The future value of an ordinary annuity paying 10% quarterly for nine years with quarterly deposits of $450 uses the future value of an annuity formula:
FV = P × [(1 + i)^n - 1]/i, where i = interest rate per period, n = total number of periods.
Interest rate per quarter: 10% / 4 = 2.5% = 0.025. Total periods: 9 years × 4 = 36. quarters.
FV = 450 × [(1 + 0.025)^36 - 1] / 0.025 ≈ 450 × (2.454 - 1) / 0.025 ≈ 450 × 1.454 / 0.025 ≈ 450 × 58.16 ≈ $26,172.
Regarding depreciation, an asset costing $75,000 with salvage value $2,800 over five years has annual depreciation expense under the straight-line method as:
Depreciation = (Cost - Salvage value) / Useful life = ($75,000 - $2,800) / 5 ≈ $72,200 / 5 = $14,440 annually.
The inventory valuation via the average cost method involves calculating the total cost of all inventory purchases and dividing by total units available. The purchases are:
- 30 jetskis at $4,000 each
- 100 jetskis at $3,000 each
- 20 jetskis at $3,500 each
- 50 jetskis at $2,500 each
Correlation between total units purchased and their costs yields a total inventory value of $230,625 for 75 jetskis remaining, based on the average cost method calculations, matching the preferred options.
Analyzing statistical statements, the true statement is: "The stem in a stem-and-leaf display is the leading digit," which accurately describes the structure of stem-and-leaf plots.
The weighted mean wages for the three groups earning $8, $9, and $12 per hour with respective counts can be calculated as:
Weighted mean = (8×3 + 9×6 + 12×1) / (3 + 6 + 1) = (24 + 54 + 12) / 10 = 90 / 10 = $9.
In probability classification, events are equally likely under the classical probability model, which assumes each outcome has an equal chance.
The probability of exactly 6 out of 8 voters intending to vote for the incumbent, given a 60% likelihood per voter, is computed using the binomial probability formula:
P = C(8,6) × 0.6^6 × 0.4^2 ≈ 28 × 0.0467 × 0.16 ≈ 0.21.
The listing of all possible outcomes and their probabilities is called a probability distribution.
Both binomial and Poisson distributions are classified as discrete distributions, applicable to scenarios with countable outcomes.
Regarding the normal distribution, it is not true that the points of the curve meet the X-axis at z = –3 and z = 3; instead, the curve approaches the axis asymptotically and does not touch it, but in standard normal distribution, about 99.7% of data lie within z-scores of -3 and 3.
The z-score for an income of $1,100, given a mean of $1,000 and standard deviation of $100, is:
Z = (X - μ) / σ = (1100 - 1000) / 100 = 1.00.
In decision-making processes, "seasonal indexes" are not considered components; components include alternatives, payoffs, and states of nature.
Applying probabilities to a payoff table results in the expected monetary value, a key decision-making criterion in probabilistic frameworks.
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