Key Concepts:
Significant Figures:
Significant figures include all nonzero digits, any zeros between nonzero digits, and trailing zeros in a decimal number.Which zeros are significant?
Zeros between nonzero digits
(Captive Zeros) are significant.Leading zeros
(before the first nonzero digit) are not significant.Trailing zeros in a decimal number
are significant.Trailing zeros in a whole number without a decimal
are not significant.Decimal Places:
The number of digits after the decimal point.Rounding:
Adjusting a number to a given decimal place or significant figure.Absolute Error:
The absolute error is the difference between the measured (or estimated) value and the true value. It represents the size of the error without considering its direction (positive or negative). \[ \text{Absolute Error} = |\text{Measured Value} - \text{True Value}| \]Percentage Error:
The relative difference between an estimated value and the actual value. \[ \text{Percentage Error} = \left( \frac{| \text{Approximate Value} - \text{Exact Value} |}{\text{Exact Value}} \right) \times 100 \]Estimation Techniques:
Approximating calculations to simplify operations, such as using rounding to estimate sums and products.Solving Equations Correct to n Decimal Places:
Finding numerical solutions to equations with a specified accuracy.Examples:
Example 1: Rounding a Number to Significant Figures
Round 0.00723456 to 3 significant figures.Solution:
The first three significant figures are 7, 2, and 3. Therefore, rounding to 3 significant figures: \[ 0.00723456 \approx 0.00723. \]Example 2: Calculating Percentage Error
An estimated value for a quantity is 48.3, while the actual value is 50. Find the percentage error.Solution:
Step 1: Compute the absolute error. \[ | 48.3 - 50 | = 1.7. \] Step 2: Compute the percentage error. \[ \text{Percentage Error} = \left( \frac{1.7}{50} \right) \times 100 = 3.4\%. \] Thus, the percentage error is **3.4\%**.Example 3: Solving an Equation Correct to 2 Decimal Places
Solve \(x^2 - 2x - 3 = 0\) correct to 2 decimal places.Solution:
Using the quadratic formula: \[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-3)}}{2(1)} \] \[ x = \frac{2 \pm \sqrt{4 + 12}}{2} = \frac{2 \pm \sqrt{16}}{2} \] \[ x = \frac{2 \pm 4}{2}. \] Thus, the two roots are: \[ x = \frac{2 + 4}{2} = 3.00, \quad x = \frac{2 - 4}{2} = -1.00. \] The solutions correct to **2 decimal places** are **\(x = 3.00\) and \(x = -1.00\)**.