💡First, I need to understand the data provided. Tyler went on ten outings and caught the following number of fish: 1, 1, 2, 2, 2, 3, 3, 3, 4, 4. I should calculate the mean, median, and mode to compare them.To find the mean, I'll add up all the numbers and divide by the total number of outings. So, 1 + 1 + 2 + 2 + 2 + 3 + 3 + 3 + 4 + 4 equals 25. Dividing 25 by 10 gives a mean of 2.5.Next, the median is the middle value when the numbers are arranged in order. Since there are ten numbers, the median will be the average of the fifth and sixth numbers. Both the fifth and sixth numbers are 2 and 3, so the median is (2 + 3) / 2, which is also 2.5.Finally, the mode is the number that appears most frequently. Looking at the data, both 2 and 3 appear three times each, so there are two modes: 2 and 3.Now, comparing these values: the mean and median are both 2.5, and the mode is 2 or 3. This means the mode is less than the mean and median. Therefore, the correct statement is that the mode is less than the mean, which is equal to the median.

💡Okay, so I have this problem here: "What is the sum of the distinct prime integer divisors of 1260?" And there are multiple-choice answers: A) 9, B) 12, C) 16, D) 17, E) 32. Hmm, I need to figure this out step by step. Let me think about how to approach it.First, I remember that to find the prime divisors of a number, I need to factorize it into its prime components. Prime factorization, right? So, I should start by dividing 1260 by the smallest prime numbers and see how it breaks down.Let me start with 2. Is 1260 divisible by 2? Yes, because it's an even number. So, 1260 divided by 2 is 630. Okay, so 2 is a prime divisor. Now, I have 630 left. Is 630 divisible by 2? Yes, again, it's even. So, 630 divided by 2 is 315. Wait, but 315 isn't even, so I can't divide by 2 anymore. So, I've got 2 as a prime factor, and now I need to move on to the next prime number.The next prime after 2 is 3. Let's check if 315 is divisible by 3. To test divisibility by 3, I can add up the digits: 3 + 1 + 5 equals 9, and 9 is divisible by 3. So, yes, 315 divided by 3 is 105. Okay, so 3 is another prime factor. Now, I have 105 left. Let me check if 105 is divisible by 3 again. Adding the digits: 1 + 0 + 5 equals 6, which is divisible by 3. So, 105 divided by 3 is 35. Alright, so I have another 3 as a prime factor, but since we're looking for distinct prime factors, I only count 3 once.Moving on, the next prime number is 5. Let's see if 35 is divisible by 5. Yes, because it ends with a 5. So, 35 divided by 5 is 7. So, 5 is another prime factor. Now, I'm left with 7. Well, 7 is a prime number itself, so that's the last prime factor.So, putting it all together, the prime factors of 1260 are 2, 3, 5, and 7. Now, the question asks for the sum of the distinct prime integer divisors. That means I just need to add these prime factors together: 2 + 3 + 5 + 7.Let me do that addition step by step to make sure I don't make a mistake. 2 plus 3 is 5. 5 plus 5 is 10. 10 plus 7 is 17. So, the sum is 17.Looking back at the options, D is 17. So, that must be the correct answer. But wait, let me double-check my prime factorization to make sure I didn't miss anything.Starting with 1260, divided by 2 is 630, divided by 2 again is 315. Then, divided by 3 is 105, divided by 3 again is 35. Divided by 5 is 7, and 7 is prime. So, yes, the prime factors are indeed 2, 3, 5, and 7. Adding them up gives 17. I think I did it right.I guess another way to verify is to list all the divisors of 1260 and then pick out the prime ones and add them. But that might take longer. Let me see if I can do that quickly.The prime factors are 2, 3, 5, and 7. Any other prime factors? Let's see, 1260 divided by 2 is 630, which we've already handled. Divided by 3 is 420, which is also covered. Divided by 5 is 252, and divided by 7 is 180. So, I don't think there are any other prime factors beyond 2, 3, 5, and 7.Therefore, I'm confident that the sum is 17, which is option D.

💡First, I need to find the least common multiple (LCM) of the numbers 3, 4, and 9. To do this, I'll break down each number into its prime factors:- 3 is a prime number, so its prime factorization is simply 3.- 4 can be factored into 2 squared, which is (2^2).- 9 can be factored into 3 squared, which is (3^2).Next, I'll identify the highest power of each prime number that appears in the factorizations:- For the prime number 2, the highest power is (2^2) from the number 4.- For the prime number 3, the highest power is (3^2) from the number 9.Now, I'll calculate the LCM by multiplying these highest powers together:[text{LCM} = 2^2 times 3^2 = 4 times 9 = 36]With the LCM determined, I need to find the smallest three-digit number that is a multiple of 36. Since 36 is less than 100, I'll multiply it by successive integers until I reach the smallest three-digit multiple:[36 times 3 = 108]Therefore, the least three-digit positive integer that has 3, 4, and 9 as factors is 108.

💡Okay, so I need to figure out how many four-digit numbers N have the property that when you remove the leftmost digit, the resulting three-digit number is one eighth of N. Hmm, let me break this down step by step.First, let's denote the four-digit number N. Since it's a four-digit number, it can be represented as N = 1000a + x, where 'a' is the leftmost digit (which must be between 1 and 9, because it's a four-digit number) and 'x' is the remaining three-digit number. So, for example, if N is 1234, then a is 1 and x is 234.Now, according to the problem, the three-digit number x is one eighth of N. So, mathematically, that means x = N / 8. But since N is 1000a + x, we can substitute that into the equation:x = (1000a + x) / 8Let me write that down:x = (1000a + x) / 8To solve for x, I'll multiply both sides by 8 to get rid of the denominator:8x = 1000a + xNow, subtract x from both sides:7x = 1000aSo, x = (1000a) / 7Hmm, okay. So x is equal to 1000a divided by 7. But x has to be a three-digit integer because it's the number you get after removing the first digit of N. So, (1000a) must be divisible by 7, and the result must be a three-digit number.Let me think about the possible values of 'a'. Since 'a' is the first digit of a four-digit number, it can be any integer from 1 to 9. So, let's consider each possible value of 'a' and see if (1000a)/7 is an integer and a three-digit number.Starting with a = 1:x = 1000*1 / 7 = 1000 / 7 ≈ 142.857Hmm, that's not an integer. So, a = 1 doesn't work.Next, a = 2:x = 1000*2 / 7 = 2000 / 7 ≈ 285.714Still not an integer. Moving on.a = 3:x = 1000*3 / 7 = 3000 / 7 ≈ 428.571Not an integer either.a = 4:x = 1000*4 / 7 = 4000 / 7 ≈ 571.428Still not an integer.a = 5:x = 1000*5 / 7 = 5000 / 7 ≈ 714.285Not an integer.a = 6:x = 1000*6 / 7 = 6000 / 7 ≈ 857.142Still not an integer.a = 7:x = 1000*7 / 7 = 7000 / 7 = 1000Wait, 1000 is a four-digit number, but x needs to be a three-digit number. So, a = 7 gives x = 1000, which is invalid because x must be three digits.a = 8:x = 1000*8 / 7 = 8000 / 7 ≈ 1142.857That's way beyond three digits, so invalid.a = 9:x = 1000*9 / 7 = 9000 / 7 ≈ 1285.714Again, way too big.So, from a = 1 to a = 9, none of them give x as an integer three-digit number. Therefore, there are no four-digit numbers N that satisfy the given condition.Wait, but let me double-check to make sure I didn't make a mistake. Maybe I should verify if 1000a is divisible by 7 for any a from 1 to 6.Let's see:For a = 1: 1000 / 7 ≈ 142.857, not integer.a = 2: 2000 / 7 ≈ 285.714, not integer.a = 3: 3000 / 7 ≈ 428.571, not integer.a = 4: 4000 / 7 ≈ 571.428, not integer.a = 5: 5000 / 7 ≈ 714.285, not integer.a = 6: 6000 / 7 ≈ 857.142, not integer.So, none of these are integers. Therefore, my conclusion is correct that there are no such four-digit numbers N.I guess that's it. So, the answer is zero.**Final Answer**The number of such four-digit numbers is boxed{0}.

💡Okay, so I have this problem here where I need to find the domain of the function y = f(x + 3) + f(x²). The function f(x) has a domain of [0, 4]. Hmm, let me think about how to approach this.First, I remember that the domain of a function is all the possible input values (x-values) for which the function is defined. Since y is a combination of two functions, f(x + 3) and f(x²), both of these need to be defined for y to be defined. That means I need to find the values of x where both f(x + 3) and f(x²) are within the domain of f, which is [0, 4].Alright, let's break it down step by step.Starting with f(x + 3): For this to be defined, the expression inside f, which is (x + 3), must lie within [0, 4]. So, I can set up the inequality:0 ≤ x + 3 ≤ 4To solve for x, I'll subtract 3 from all parts of the inequality:0 - 3 ≤ x + 3 - 3 ≤ 4 - 3Which simplifies to:-3 ≤ x ≤ 1Okay, so x has to be between -3 and 1 for f(x + 3) to be defined.Now, moving on to f(x²): Similarly, the expression inside f here is x², so x² must also lie within [0, 4]. Let's write that inequality:0 ≤ x² ≤ 4To solve this, I'll consider the square root of both sides. Remember, when dealing with inequalities and square roots, I have to consider both the positive and negative roots.First, x² ≥ 0 is always true since any real number squared is non-negative. So, the lower bound is automatically satisfied.Now, for x² ≤ 4, taking the square root of both sides gives:|x| ≤ 2Which means:-2 ≤ x ≤ 2So, x has to be between -2 and 2 for f(x²) to be defined.Now, to find the domain of y, I need the values of x that satisfy both conditions simultaneously. That is, x has to be in the intersection of the two intervals we found.The first interval is [-3, 1], and the second interval is [-2, 2]. The intersection of these two intervals is where they overlap.Let me visualize this on a number line.- The interval [-3, 1] starts at -3 and goes up to 1.- The interval [-2, 2] starts at -2 and goes up to 2.The overlapping part is from -2 to 1 because that's where both intervals are present.So, the domain of y is [-2, 1].Looking at the options given:A: [-2,1]B: [1,2]C: [-2,-1]D: [-1,2]The correct answer should be A: [-2,1].Wait, let me double-check my reasoning to make sure I didn't make a mistake.For f(x + 3), x + 3 must be between 0 and 4, so x must be between -3 and 1. That seems right.For f(x²), x² must be between 0 and 4, so x must be between -2 and 2. That also makes sense because squaring any number outside of -2 to 2 would result in a number greater than 4, which is outside the domain of f.Therefore, the overlap is indeed from -2 to 1. So, I think I did it correctly.I don't see any errors in my calculations or reasoning. So, I'm confident that the domain is [-2,1], which is option A.**Final Answer**boxed{A}

💡First, I need to determine how many candies Barbara has initially. According to the problem, she starts with 18.0 candies.Next, the problem states that Barbara uses 9.0 candies. To find out how many candies she has left, I will subtract the number of candies she used from her initial amount.So, I will perform the subtraction: 18.0 minus 9.0.After calculating, I find that Barbara has 9.0 candies remaining.