What is the domain of each of the following functions in interval notation? Match the functions in the column on the left with ONE interval from the column on the right. There will be leftover answer choices. h(x)= – 3x + 1 A. v g(x) =/3x– 1 6x v (x) = Зх-1 K(x) =/-3x- 1 E. (-00,00) F.
Continuous Probability Distributions
Probability distributions are of two types, which are continuous probability distributions and discrete probability distributions. A continuous probability distribution contains an infinite number of values. For example, if time is infinite: you could count from 0 to a trillion seconds, billion seconds, so on indefinitely. A discrete probability distribution consists of only a countable set of possible values.
Normal Distribution
Suppose we had to design a bathroom weighing scale, how would we decide what should be the range of the weighing machine? Would we take the highest recorded human weight in history and use that as the upper limit for our weighing scale? This may not be a great idea as the sensitivity of the scale would get reduced if the range is too large. At the same time, if we keep the upper limit too low, it may not be usable for a large percentage of the population!
![**Domain of Functions in Interval Notation**
**Objective:** Determine the domain for each function and match it with the appropriate interval on the right. Note that there are extra choices.
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**Functions:**
1. \( h(x) = -3x + 1 \)
2. \( g(x) = \sqrt{3x - 1} \)
3. \( j(x) = \frac{6x}{3x - 1} \)
4. \( k(x) = \sqrt{-3x - 1} \)
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**Interval Options:**
- A. \(\left[ -\frac{1}{3}, \frac{1}{3} \right]\)
- B. \((-\infty, -\frac{1}{3}]\)
- C. \((-\infty, \frac{1}{3}]\)
- D. \([\frac{1}{3}, \infty)\)
- E. \((-\infty, \infty)\)
- F. \((-\infty, \frac{1}{3}) \cup (\frac{1}{3}, \infty)\)
- G. \([-\frac{1}{3}, \infty)\)
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**Explanation:**
- **\( h(x) = -3x + 1 \):** A linear function with no restrictions, so the domain is all real numbers: \( E. \ (-\infty, \infty) \).
- **\( g(x) = \sqrt{3x - 1} \):** The expression inside the square root must be non-negative: \(3x - 1 \geq 0\), so \(x \geq \frac{1}{3}\). Therefore, the domain is: \( D. \ [\frac{1}{3}, \infty) \).
- **\( j(x) = \frac{6x}{3x - 1} \):** Denominator cannot be zero, so \(3x - 1 \neq 0\). Solving gives \(x \neq \frac{1}{3}\), so the domain is all real numbers except \(x = \frac{1}{3}\): \( F. \ (-\infty, \frac{1}{3}) \cup (\frac{1}{3}, \infty) \).
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