To calculate the variance of the numbers 6, 7, 8, 9, we use the formula for variance: \[ \sigma^2 = \frac{1}{n} \sum_{i=1}^n (x_i - \mu)^2, \] where \( \mu \) is the mean of the numbers and \( n \) is the number of observations.
Step 1: Calculate the mean \( \mu \) The numbers are 6, 7, 8, and 9. First, find the mean: \[ \mu = \frac{6 + 7 + 8 + 9}{4} = \frac{30}{4} = 7.5. \] Step 2: Calculate the squared differences from the mean Now, we calculate the squared differences from the mean for each number:
Step 3: Find the variance The variance is the average of the squared differences: \[ \sigma^2 = \frac{2.25 + 0.25 + 0.25 + 2.25}{4} = \frac{5}{4} = 1.25. \]
Thus, the variance is \( \frac{5}{4} \). Therefore, the correct answer is option (E).
\[ f(x) = \begin{cases} x\left( \frac{\pi}{2} + x \right), & \text{if } x \geq 0 \\ x\left( \frac{\pi}{2} - x \right), & \text{if } x < 0 \end{cases} \]
Then \( f'(-4) \) is equal to:If \( f'(x) = 4x\cos^2(x) \sin\left(\frac{x}{4}\right) \), then \( \lim_{x \to 0} \frac{f(\pi + x) - f(\pi)}{x} \) is equal to:
Let \( f(x) = \frac{x^2 + 40}{7x} \), \( x \neq 0 \), \( x \in [4,5] \). The value of \( c \) in \( [4,5] \) at which \( f'(c) = -\frac{1}{7} \) is equal to:
The general solution of the differential equation \( \frac{dy}{dx} = xy - 2x - 2y + 4 \) is:
\[ \int \frac{4x \cos \left( \sqrt{4x^2 + 7} \right)}{\sqrt{4x^2 + 7}} \, dx \]