answer:- First, recognize that the triangle is isosceles with sides 10, 10, and 12, where 12 is the base. - Draw the altitude from the vertex opposite the base, splitting the triangle into two congruent right triangles. The altitude splits the base into two segments of length 6 each. - Use the Pythagorean theorem to find the length of the altitude ( h ): [ h^2 + 6^2 = 10^2 implies h^2 = 100 - 36 = 64 implies h = 8. ] - Calculate the area of the triangle: [ text{Area} = frac{1}{2} times 12 times 8 = 48. ] Therefore, the area of the triangle is (boxed{48}).

answer:The children in the Euler family have ages 6, 6, 6, 6, 8, 8, 16. To find the mean age, we calculate: 1. Sum up the ages: 6 + 6 + 6 + 6 + 8 + 8 + 16 = 56 2. Count the children: There are 7 children. 3. Divide the total age by the number of children: frac{56}{7} = 8. Thus, the mean age of the children in the Euler family is boxed{8}.

answer:First, we need to calculate the total weight of the beef James bought. Since each pack is 4 pounds and he bought 5 packs, we multiply 4 pounds by 5: 4 pounds/pack * 5 packs = 20 pounds Next, we need to calculate the total cost of the beef. The price of beef is 5.50 per pound, so we multiply the total weight of the beef (20 pounds) by the price per pound (5.50): 20 pounds * 5.50/pound = 110 Therefore, James paid boxed{110} for the beef.

answer:1. **Initial Setup**: Celia, Alice, and Betsy each hold two cards from the set {9, 10, J, Q, K, A} of Spades. Celia's goal is to collect all six cards by asking Alice or Betsy for cards she does not have. The game ends when one team has all six cards. 2. **Case Analysis**: We start by assuming Celia has the 9 and 10. Celia will ask Alice for the Jack (J). 3. **Case 1: Alice does not have the Jack (J)**: - Probability: ( frac{1}{2} ) - If Alice does not have the J, Betsy must have it. It is now Alice's turn. - Alice must guess Celia's cards correctly to prevent Celia from winning. Alice has a ( frac{1}{3} ) chance of guessing correctly. - Therefore, the probability that Celia wins in this case is: [ frac{1}{2} times left(1 - frac{1}{3}right) = frac{1}{2} times frac{2}{3} = frac{1}{3} ] 4. **Case 2: Alice has the Jack (J)**: - Probability: ( frac{1}{2} ) - Celia now has the J and must decide whether to ask Alice or Betsy next. - If Celia asks Betsy and fails, Alice's remaining card will be known, and Betsy can use this information to win. If Celia succeeds, she must guess the remaining cards correctly. 5. **Subcase 2a: Celia guesses Alice's remaining card correctly**: - Probability: ( frac{1}{3} ) - Celia wins immediately. - Contribution to Celia's winning probability: [ frac{1}{2} times frac{1}{3} = frac{1}{6} ] 6. **Subcase 2b: Celia guesses Alice's remaining card incorrectly**: - Probability: ( frac{2}{3} ) - Betsy now has the Queen (Q), and Alice must guess Betsy's remaining card correctly. - Alice has a ( frac{1}{3} ) chance of guessing correctly. - Contribution to Celia's winning probability: [ frac{1}{2} times frac{2}{3} times left(1 - frac{1}{3}right) = frac{1}{2} times frac{2}{3} times frac{2}{3} = frac{2}{9} ] 7. **Combining Probabilities**: - Total probability of Celia winning: [ frac{1}{3} + frac{1}{6} + frac{2}{9} = frac{6}{18} + frac{3}{18} + frac{4}{18} = frac{13}{18} ] The final answer is ( boxed{ 100p + q = 100 times 13 + 18 = 1318 } )