answer:To calculate the probability of drawing a spade or a king from a well-shuffled pack of 52 cards, we need to consider the number of spades and the number of kings in the deck, while also accounting for the overlap (the King of Spades, which is both a spade and a king). There are 13 spades in a deck of cards (one for each rank: Ace through King). There are also 4 kings in the deck (one for each suit: spades, hearts, diamonds, and clubs). However, since the King of Spades is counted in both groups, we need to subtract it from the total to avoid double-counting. So the total number of favorable outcomes is the number of spades (13) plus the number of kings (4) minus the overlap (1 King of Spades): Total favorable outcomes = 13 spades + 4 kings - 1 King of Spades Total favorable outcomes = 13 + 4 - 1 Total favorable outcomes = 16 The total number of possible outcomes is the total number of cards in the deck, which is 52. Now, the probability of drawing a spade or a king is the number of favorable outcomes divided by the total number of possible outcomes: Probability = Total favorable outcomes / Total number of possible outcomes Probability = 16 / 52 To simplify the fraction, we can divide both the numerator and the denominator by 4: Probability = 4 / 13 So, the probability of drawing a spade or a king from a well-shuffled pack of 52 cards is boxed{4/13} .

answer:1. Given an equilateral triangle triangle ABC with side length 1, point P is such that PA perp to the plane ABC and PA = frac{sqrt{6}}{4}. 2. Let point A' be the point symmetric to A with respect to the plane PBC. Consequently, both AA' will be perpendicular to PBC and A'A = 2 times PA = 2 times frac{sqrt{6}}{4} = frac{sqrt{6}}{2}. 3. Let O be the intersection of AA' with the plane PBC. Since O lies on line PD where D is the midpoint of BC. As BC is a side in the equilateral triangle, D = left( frac{1}{2} right). 4. In the right triangle PAD, we have (AD = frac{BC sqrt{3}}{2} = frac{sqrt{3}}{2} ). To find PD, we use the Pythagorean theorem: [ PD = sqrt{PA^2 + AD^2} = sqrt{left(frac{sqrt{6}}{4}right)^2 + left(frac{sqrt{3}}{2}right)^2} = sqrt{frac{6}{16} + frac{3}{4}} = sqrt{frac{6}{16} + frac{12}{16}} = sqrt{frac{18}{16}} = sqrt{frac{9}{8}} = frac{3 sqrt{2}}{4} ] 5. Using projections, the length of AO can be computed using the formula for the perpendicular from a point to a line: [ AO = frac{PA cdot AD}{PD} = frac{frac{sqrt{6}}{4} cdot frac{sqrt{3}}{2}}{frac{3 sqrt{2}}{4}} = frac{sqrt{18}}{4cdot3} = frac{3}{6} = frac{1}{2} ] Since (A^prime) is symmetric to (A), (A^prime A = 2A O = 1). 6. Since A and A^prime are reflections with respect to the plane PBC, [ A^prime B = AB = 1, quad text{and} quad A^prime C = AC = 1 ] Thus making A^prime ABC a regular tetrahedron where all edges are equal. 7. Conclusively, the dihedral angle or the angle between line A^prime C and AB is equivalent to the angle between the edges of a regular tetrahedron opposite its shared face, which is 90°. [ boxed{90^circ} ]

answer:1. Define the friends' individual catches as a, b, c, and d. When pairs of these fishermen combine their catches, their sums must match the given values. For four friends, we can determine six unique pair sums: (a+b, a+c, a+d, b+c, b+d, c+d). We're given that these sums are 7, 9, 14, 14, 19, 21. 2. Let's list the sums and their combinations: [ a+b = 7, quad a+c = 9, quad a+d = 14, quad b+c = 14, quad b+d = 19, quad c+d = 21. ] 3. We next find the total sum of all individual catches by adding all the given pair sums together: [ (a+b) + (a+c) + (a+d) + (b+c) + (b+d) + (c+d) = 7 + 9 + 14 + 14 + 19 + 21. ] 4. Calculating the sum of the pair sums: [ 7 + 9 + 14 + 14 + 19 + 21 = 84. ] 5. Since each individual catch is included in three different sums, the above total is three times the combined sum of all individual catches: [ 3(a + b + c + d) = 84 quad Rightarrow quad a + b + c + d = frac{84}{3} = 28. ] 6. Now, we use the total sum a + b + c + d = 28 to determine individual catches. By setting equations using the sums provided, we can solve for each pair: [ a + b = 7, quad a + c = 9, quad a + d = 14, quad b + c = 14, quad b + d = 19, quad c + d = 21. ] 7. Given the total sum of 28: [ 28 - a = b + c + d quad text{(from the total sum equation).} ] 8. Substitute b + c + d back into the set equations: - From a + d = 14, we get d = 14 - a. - From a + b = 7, we get b = 7 - a. - From a + c = 9, we get c = 9 - a. 9. Utilizing b + c + d = 28 - a: [ (7 - a) + (9 - a) + (14 - a) = 28 - a. ] 10. Simplify and solve for a: [ 30 - 3a = 28 - a quad Rightarrow quad 2 = 2a quad Rightarrow quad a = 1. ] 11. Now, compute b, c, and d using the values obtained for a: - b = 7 - a = 7 - 1 = 6. - c = 9 - a = 9 - 1 = 8. - d = 14 - a = 14 - 1 = 13. 12. Therefore, the individual catches of the friends are: [ a = 1, quad b = 6, quad c = 8, quad d = 13. ] **Conclusion:** [ boxed{(1, 6, 8, 13)} ]

answer:Let's calculate the total amount Liam earned from selling his oranges first. Liam sold his oranges at 2.50 for 2, so each orange was sold for 2.50 / 2 = 1.25. Liam picked 40 oranges, so he earned 40 * 1.25 = 50 from selling his oranges. Now, let's find out how much Claire earned. We know that the total amount they want to save is 86, and Liam has already earned 50, so Claire must have earned 86 - 50 = 36 from selling her oranges. Claire sold her oranges at 1.20 each, so the number of oranges she sold is 36 / 1.20 per orange. Let's calculate that: 36 / 1.20 = 30 oranges. Therefore, Claire picked boxed{30} oranges.