Aa criterion : if two corresponding angles of two triangles are congruent then the two triangles are similar
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Since we are given the margin of error and it's equal to ±0.1 feet, and we know the surveyed distance, we can calculate the maximum and minimum distance. We must remember that margin of errors usually involves and maximum and minimum margin of a measure, and it means that the real measure will not be greater or less than the values located at the margins.
We know that the surveyed distance is 1200 feet with a margin of error of ±0.1 feet, so, we can calculate the maximum and minimum distances that the reader could assume in the following way:
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Given:
μ = 500 days, the population mean
σ = 60 days, the population standard deviation
Therefore
μ + σ = 560
μ - σ = 440
μ + 2σ = 620
μ - 2σ = 380
μ + 3σ = 680
μ - 3σ = 320
The figure shown below illustrates the normal distribution
About 68% of the total area lies in x = (μ-σ, μ+σ)
About 95% of the total area lies in x = (μ-2σ, μ+2σ)
About 99.7% of the total area lies in x = (μ-3σ, μ+3σ).
μ = 500 days, the population mean
σ = 60 days, the population standard deviation
Therefore
μ + σ = 560
μ - σ = 440
μ + 2σ = 620
μ - 2σ = 380
μ + 3σ = 680
μ - 3σ = 320
The figure shown below illustrates the normal distribution
About 68% of the total area lies in x = (μ-σ, μ+σ)
About 95% of the total area lies in x = (μ-2σ, μ+2σ)
About 99.7% of the total area lies in x = (μ-3σ, μ+3σ).
We have been given that on the day of his 18th birthday Harry decided to start saving money regularly . Starting on that day, he could save 30.00 on the same date every month. We are asked to find the amount saved by the day before Harry's 60th birthday.
First of all, we will find years from 18 years to 60 years.
We know that 1 year equals 12 months.
To find total amount saved, we will multiply 504 months by amount saved per month.
Therefore, Harry would have saved by the day before his 60th birthday.