answer：Okay, so I have this problem here: I need to find the number of four-digit even numbers that can be formed without repeating any digits, using the digits 1, 2, 3, 4, and 5. The options given are A: 8, B: 24, C: 48, D: 120. Hmm, let me try to figure this out step by step.First, I remember that an even number must end with an even digit. Looking at the digits I have—1, 2, 3, 4, 5—the even digits are 2 and 4. So, the last digit of my four-digit number has to be either 2 or 4. That gives me two possibilities for the last digit.Now, since I can't repeat any digits, once I choose the last digit, I have to use the remaining digits for the first three positions. Let me break this down:1. **Choosing the last digit:** As I thought, it can be either 2 or 4. So, there are 2 choices here.2. **Choosing the first digit:** Since it's a four-digit number, the first digit can't be zero, but in this case, all digits are from 1 to 5, so that's fine. After choosing the last digit, I have 4 remaining digits to choose from for the first position. So, 4 choices here.3. **Choosing the second digit:** After choosing the first and last digits, I have 3 remaining digits left. So, 3 choices for the second digit.4. **Choosing the third digit:** Now, I have 2 remaining digits left, so 2 choices for the third digit.Putting this all together, for each choice of the last digit (2 or 4), the number of possible four-digit numbers is 4 (first digit) × 3 (second digit) × 2 (third digit) = 24.Since there are 2 choices for the last digit, the total number of four-digit even numbers is 24 × 2 = 48.Wait, let me double-check that. So, if the last digit is 2, the first three digits can be any permutation of the remaining four digits (1, 3, 4, 5). The number of permutations is 4P3, which is 4 × 3 × 2 = 24. Similarly, if the last digit is 4, the first three digits can be any permutation of the remaining four digits (1, 2, 3, 5), which is also 24. So, 24 + 24 = 48. Yeah, that seems right.I think I might have confused myself earlier if I considered the last digit differently, but no, this seems consistent. So, the total number of four-digit even numbers without repeating digits from 1, 2, 3, 4, 5 is 48. Looking back at the options, that's option C.Just to make sure I didn't miss anything, let me think if there's another way to approach this problem. Maybe using combinations instead of permutations? Hmm, but since the order matters in forming numbers, permutations are the right way to go. Alternatively, I could calculate the total number of four-digit numbers without repetition and then see what fraction of them are even. The total number of four-digit numbers without repetition from 5 digits is 5P4 = 5 × 4 × 3 × 2 = 120. Since half of the numbers ending with even digits would be even, but wait, actually, in this case, only two out of the five digits are even, so the fraction is 2/5. So, 120 × (2/5) = 48. Yep, that also gives me 48. So, that confirms my earlier result.Alright, I feel confident now that the answer is 48, which is option C.

answer：First, I need to determine the mode of the data set. The mode is the number that appears most frequently. Looking at the scores: 90, 89, 90, 95, 92, 94, 93, 90, I notice that the score 90 appears three times, which is more than any other score. Therefore, the mode is 90.Next, I need to find the median of the data set. The median is the middle value when the numbers are arranged in order. Since there are eight scores, which is an even number, the median will be the average of the two middle numbers. First, I'll arrange the scores in ascending order: 89, 90, 90, 90, 92, 93, 94, 95. The two middle numbers are the fourth and fifth scores, which are 90 and 92. To find the median, I'll calculate the average of these two numbers: (90 + 92) / 2 = 91.Therefore, the mode is 90 and the median is 91.

answer：First, I need to determine the number of gallons in the first tank. Let's denote the number of gallons in the second tank as ( x ). Since the first tank is twice the size of the second tank, it contains ( 2x ) gallons.Next, I'll calculate the number of fish in each tank based on the given rules. In the second tank, each two-inch fish requires 2 gallons of water, so the number of two-inch fish is ( frac{x}{2} ).In the first tank, each three-inch fish requires 3 gallons of water, so the number of three-inch fish is ( frac{2x}{3} ).According to the problem, if one fish in the first tank eats another, the first tank will have 3 more fish than the second tank. This means that before any fish are eaten, the first tank has 4 more fish than the second tank.Setting up the equation:[frac{2x}{3} - 1 = frac{x}{2} + 3]To eliminate the fractions, I'll multiply the entire equation by 6:[4x - 6 = 3x + 18]Solving for ( x ):[x = 24]Therefore, the first tank, which is twice the size of the second tank, contains ( 2 times 24 = 48 ) gallons of water.

answer：To find the integer ( n ) such that ( 1 leq n leq 10 ) and ( n equiv 123456 pmod{11} ), I can use the divisibility rule for 11. This rule involves alternatingly adding and subtracting the digits of the number.Starting with the number 123456, I'll assign alternating signs to each digit:[1 - 2 + 3 - 4 + 5 - 6]Calculating this:[1 - 2 = -1][-1 + 3 = 2][2 - 4 = -2][-2 + 5 = 3][3 - 6 = -3]The result is (-3). To find the remainder when divided by 11, I'll add 11 to (-3):[-3 + 11 = 8]Therefore, ( 123456 equiv 8 pmod{11} ), and the integer ( n ) is 8.

answer：First, I need to add the binary numbers (10110_2), (110_2), (1_2), and (101_2). To do this, I'll align them properly by their least significant bits.Starting from the rightmost bit, I'll add each column, keeping track of any carryovers. In the first column, adding the bits gives me a sum that requires a carryover. I'll continue this process for each subsequent column, ensuring that I account for any carryovers from the previous addition.After adding all the columns and handling the carryovers, I'll compile the results to form the final binary sum.

answer：First, I recognize that the sum of the exterior angles of any polygon is always 360 degrees. Given that each interior angle is 150 degrees, I can find the measure of each exterior angle by subtracting the interior angle from 180 degrees: 180° - 150° = 30°.Next, to determine the number of sides (n) of the polygon, I divide the total sum of exterior angles by the measure of one exterior angle: 360° / 30° = 12. This tells me the polygon has 12 sides.Finally, to find the number of diagonals that can be drawn from one vertex, I use the formula n - 3, where n is the number of sides. So, 12 - 3 = 9. Therefore, there are 9 diagonals that can be drawn from one vertex of this polygon.