The image shows three multiple-choice math problems. I will address each one by analyzing and solving them individually:

Problem 19:

Statement: "Nine points are spaced at intervals of one unit around a 3x3 square, as shown. Two of the 9 points are chosen at random. What is the probability that the points are one unit apart?"

Solution:

1. There are a total of 9 points.
2. The total number of ways to select two points is $\binom{9}{2} = \frac{9 \times 8}{2} = 36$.
3. Now, we need to find the number of ways to choose two points that are one unit apart. These will be points that are directly adjacent (horizontally or vertically).
4. In a 3x3 grid, there are:
   • 6 horizontal pairs of adjacent points (3 rows × 2 pairs per row),
   • 6 vertical pairs of adjacent points (3 columns × 2 pairs per column).
5. Therefore, there are a total of $6 + 6 = 12$ pairs that are one unit apart.
6. The probability is the ratio of favorable outcomes to the total outcomes: $\text{Probability} = \frac{12}{36} = \frac{1}{3}$ Answer: (B) $\frac{1}{3}$

Problem 20:

Statement: "The student in Ms. Chen’s math class received a positive integer score. There are eight scores that are each one of the eight consecutive integers, and no two students received the same score. What is the maximum possible number of students in the class?"

Solution:

1. Let the scores be represented as $n, n+1, n+2, \dots, n+7$, where $n$ is the smallest score.

2. The sum of these eight consecutive integers is $8n + (0+1+2+...+7) = 8n + 28$.

3. Since the sum represents the total number of students (a positive integer), the minimum possible value for $n$ is 1.

4. To maximize the number of students, $n = 1$, yielding scores of $1, 2, 3, \dots, 8$.

5. Therefore, the maximum possible number of students in the class is 8.

   Answer: (C) 8

Problem 21:

Statement: "A long, thin wire is bent in the shape of a semicircle of radius 8 as shown by the dashed curve. A 60° central angle is formed at the bottom of the wire by two straight segments. What is the length of the straight segments of the wire?"

Solution:

1. The wire forms a semicircle with a radius of 8 units. The central angle at the bottom of the semicircle is 60°.

2. The straight segments form an isosceles triangle where the two legs are the radii of the circle (8 units each).

3. We can calculate the base of the isosceles triangle formed by these two segments using trigonometry:

   • The angle between the radii is 60°.
   • The base length $b$ can be calculated using the formula for a chord: $b = 2 \times r \times \sin\left(\frac{\theta}{2}\right)$ where $r = 8$ and $\theta = 60^\circ$. So, $b = 2 \times 8 \times \sin(30^\circ) = 16 \times 0.5 = 8 \text{ units}$

4. Thus, the length of the straight segment is 8 units.

   Answer: (C) 8

Do you need more details or have any questions?

Related Questions:

1. What is the formula to calculate combinations, and how is it applied in probability problems?
2. How do you find the sum of an arithmetic series?
3. What is the relationship between the radius of a circle and the chord formed by a central angle?
4. How can you calculate the number of adjacent pairs in different geometric shapes?
5. What is the significance of consecutive integers in statistical problems?

Tip:

For geometry problems involving circles, always consider using trigonometric relationships to simplify complex chord or angle calculations.