answer:Emily offers to help with a chemistry project on bird feathers and suggests a trip to Hawk Mountain Sanctuary this weekend.

answer:When randomly picking 3 cards from a standard deck of 52 cards, we want to determine the probability that the 3 cards form a sequence of consecutive ranks, such as 3-4-5 or 8-9-10. To start, let's consider the number of possible sequences of 3 consecutive ranks in a deck of 52 cards. Since there are 13 ranks (Ace through King) and a sequence of 3 consecutive ranks can start from any of the first 11 ranks (Ace through Jack, since we cannot form a sequence starting from Queen or King), we have 11 possible starting points. Furthermore, each sequence can occur in any of the 4 suits (hearts, diamonds, clubs, or spades), giving us a total of 11 possible starting ranks and 4 possible suit combinations for each starting rank. For each starting rank, we can calculate the number of ways to choose the three cards. We can think of this as choosing 3 cards from a group of 4 cards (one from each suit) for the first rank, 4 cards for the second rank, and 4 cards for the third rank. This results in 4 choose 1 * 4 choose 1 * 4 choose 1 = 64 possible combinations for each of the 11 starting ranks. Therefore, the total number of consecutive sequences is 11 * 64 = 704. However, we can simplify the problem further by considering the total number of ways to choose 3 cards from 52. This can be calculated using combinations, resulting in 52 choose 3, which equals 52! / (3!(52-3)!) = 52! / (3! * 49!) = 22,100. Now, we can determine the probability by dividing the number of favorable outcomes (the number of consecutive sequences) by the total number of possible outcomes: probability = 704 / 22,100 = 176 / 5525. So, the probability that 3 randomly picked cards from a standard deck of 52 cards form a sequence of consecutive ranks is 176/5525, approximately 0.03185 or 3.185%.

answer:This problem is related to the concept of pairings or matchings in combinatorics. To solve it, we can use a recursive approach or simply consider the choices we have for forming pairs. To start, we have 10 numbers to pair. We can think of the first pair as being formed by choosing 2 numbers out of the 10 available. This can be done in 10 choose 2 ways, which is 10! / (2! * (10-2)!) = 45 ways. However, we have overcounted, as the order in which we choose the pairs doesn't matter (i.e., (1, 2) is the same as (2, 1)). After choosing the first pair, we have 8 numbers left to pair. We can think of the second pair as being formed by choosing 2 numbers out of the 8 remaining, which can be done in 8 choose 2 ways. Similarly, we overcount and the actual number of choices for the second pair is (8 choose 2) / 2. However, it's more efficient to approach this problem by using the concept of permutations and accounting for overcounting. For the first pair, we can choose any of the 10 numbers for the first element and then any of the remaining 9 numbers for the second element, giving us 10 * 9 possible combinations, but again we overcounted, as (1, 2) is the same as (2, 1). So, we actually have (10 * 9) / 2 possible combinations for the first pair. For the second pair, we can choose from the remaining 8 numbers for the first element and then any of the remaining 7 numbers for the second element, resulting in (8 * 7) / 2 possible combinations. We can continue this process for each of the 5 pairs, obtaining a total number of possible pairings as: [(10 * 9) / 2] * [(8 * 7) / 2] * [(6 * 5) / 2] * [(4 * 3) / 2] * [(2 * 1) / 2]. However, even with this, we still overcount the possible pairings. Each unique pairing can be obtained in 5! (5 factorial) ways, since there are 5! ways to reorder the pairs. To correct for this, we need to divide the above product by 5!: [(10 * 9) / 2] * [(8 * 7) / 2] * [(6 * 5) / 2] * [(4 * 3) / 2] * [(2 * 1) / 2] / 5!. Simplifying this expression and canceling out common factors, we get: (10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1) / (2^5 * 5!). This can be rewritten as 10! / (2^5 * 5!), where 10! is the total number of permutations of the 10 numbers, 2^5 is the number of permutations within each pair, and 5! is the number of permutations of the 5 pairs. Simplifying this expression gives us: (10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1) / (2 * 2 * 2 * 2 * 2 * 5 * 4 * 3 * 2 * 1) = 9 * 7 * 5 * 3 * 1 = 945. However, we can simplify this problem even further by noting that the above calculation can be rewritten as: 10! / [(2!)^5 * 5!]. This gives us the same answer but in a more concise and clear manner. The number of possible pairings is 10! / [(2!)^5 * 5!] = 945. Therefore, there are 945 different pairings possible.

answer:To find the total number of squares with side lengths that are multiples of 3, we need to consider the possible side lengths and the number of squares for each. For a 3x3 square, the top-left corner can be placed in 6 * 6 positions (since the square can't extend beyond the edges of the 8x8 checkerboard), resulting in 36 possible 3x3 squares. For a 6x6 square, the top-left corner can be placed in 3 * 3 positions, resulting in 9 possible 6x6 squares. There are no other square sizes with side lengths that are multiples of 3 that can fit within an 8x8 checkerboard. So, the total number of squares with side lengths that are multiples of 3 is the sum of these counts: 36 + 9 = 45. Therefore, there are 45 squares with side lengths that are multiples of 3.