Here are the answers to the questions presented:
Question 8
- This question does not provide sufficient context or an equation to solve for \( x \). Therefore, please provide more context or the specific equation to solve.
Question 9 To find \( x \), set the equation based on the lengths of the segments provided: \[ SR + ST = QR + TU \ 12 + 6 = 9 + (8x - 12) \ 18 = 9 + 8x - 12 \ 18 = -3 + 8x \ 18 + 3 = 8x \ 21 = 8x \ x = \frac{21}{8} = 2.625 \text{ (not a provided answer)} \] It seems the values provided do not solve this.
Question 10 A) If \( HI \cong IJ \) and \( m\angle KIJ = 35^\circ \), then:
- Since \( HI \) and \( IJ \) are congruent, \( m\angle KGH \) should also be \( 35^\circ \) based on previous angle properties unless more information or relationships are given.
- It seems more info is needed; otherwise, each angle may not have been given properly.
B) Without more information about angle relationships or the arrangement, the measure of \( HI \) cannot be determined.
Question 12 Given that the distance halfway around Mercury (circumference) = 4,761 mi., to find the diameter we can use the formula: \[ \text{Diameter} = \frac{\text{Circumference}}{\pi} \ \text{Diameter} = \frac{4761 \text{ mi}}{\pi} \approx 1515 \text{ miles} \ \text{(using } \pi \approx 3.14\text{)} \]
Answer: 1515 miles.
Question 13 The length of the arc \( L \) can be calculated using the formula: \[ L = r\theta \ \text{Diameter} = 94 \text{ km} \implies r = \frac{94}{2} = 47 \text{ km} \ \text{Central angle } = \frac{3 \pi}{2} \text{ radians} \] Then, \[ L = 47 \cdot \frac{3 \pi}{2} = \frac{141 \pi}{2} \text{ km} \]
Answer: 141π/2.
Question 14 The area of the entire pie: \[ A = \pi r^2 = \pi \left(\frac{10}{2}\right)^2 = 25\pi \text{ in}^2 \ \text{The angle of the eaten part } = 225^\circ \text{ out of } 360^\circ \text{ gives the area.} \] Eaten area = \(\frac{225}{360} \cdot 25\pi \approx 15.625\pi \text{ in}^2 \)
Answer: 15.625π in².
Question 18 A) The complement of the event is pulling out numbers other than 4, 9, or 3 from tiles numbered 1 to 14. B) The events are dependent because the tiles were not returned to the bag after each draw.
Question 19 Each number has half odd and half even. Therefore, the probability of getting odd once is \( \frac{4}{8} = \frac{1}{2}. \) For three trials: \( P = \left(\frac{1}{2}\right)^3 = \frac{1}{8}. \)
Answer: \(\frac{1}{8}\).
Question 21 Using the principle of inclusion-exclusion: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \ = \frac{98}{200} + \frac{62}{200} - \frac{40}{200} = \frac{120}{200} = \frac{3}{5}. \]
Answer: \(\frac{3}{5}\).
Question 22 A) This is a combination situation as the order of players does not matter. B) The number of ways to choose 3 players from 17: \[ C(17, 3) = \frac{17 \times 16 \times 15}{3 \times 2 \times 1} = 680. \]
Answers: Combination; 680 possibilities.
Feel free to ask for clarifications or further assistance on any of the questions!