💡First, let's denote Erica's wages as ( E ).Robin earns 30% more than Erica, so Robin's wages (( R )) can be calculated as:[R = E + 0.30E = 1.30E]Charles earns 30.76923076923077% more than Robin. To find Charles's wages (( C )), we use:[C = R + 0.3076923076923077R = 1.3076923076923077R]Substituting ( R = 1.30E ) into the equation for ( C ):[C = 1.3076923076923077 times 1.30E = 1.6999999999999997E]This means Charles earns approximately 70% more than Erica. To express this as a percentage:[text{Percent more} = 1.6999999999999997 times 100 approx 170%]

💡Okay, so I have this problem about purchasing energy-saving devices for a company. There are three parts to it, and I need to figure out each step by step. Let me start by understanding what each part is asking.**Part (1): Finding the Prices of A and B Type Devices**Alright, the company wants to buy 10 devices, and there are two models: A and B. They give me two specific cost comparisons:1. Buying 3 A-type devices costs 16,000 more than buying 2 B-type devices.2. Buying 2 A-type devices costs 6,000 less than buying 3 B-type devices.I need to find the individual prices of A and B. Hmm, this sounds like a system of equations problem. Let me denote the price of one A-type device as ( x ) dollars and the price of one B-type device as ( y ) dollars.From the first statement: 3A = 2B + 16,000. So, in equation form, that's:[ 3x = 2y + 16,000 ]Or, rearranged:[ 3x - 2y = 16,000 ]From the second statement: 2A = 3B - 6,000. So, the equation is:[ 2x = 3y - 6,000 ]Or, rearranged:[ 2x - 3y = -6,000 ]Now I have two equations:1. ( 3x - 2y = 16,000 )2. ( 2x - 3y = -6,000 )I need to solve this system of equations. Let me use the elimination method. Maybe I can eliminate one variable by making the coefficients equal.Let me multiply the first equation by 3 and the second equation by 2 to make the coefficients of ( y ) opposites.First equation multiplied by 3:[ 9x - 6y = 48,000 ]Second equation multiplied by 2:[ 4x - 6y = -12,000 ]Now, subtract the second new equation from the first new equation to eliminate ( y ):[ (9x - 6y) - (4x - 6y) = 48,000 - (-12,000) ]Simplify:[ 5x = 60,000 ]So, ( x = 12,000 ).Now plug ( x = 12,000 ) back into one of the original equations to find ( y ). Let's use the first equation:[ 3(12,000) - 2y = 16,000 ][ 36,000 - 2y = 16,000 ]Subtract 36,000 from both sides:[ -2y = -20,000 ]Divide by -2:[ y = 10,000 ]So, the price of an A-type device is 12,000 and a B-type device is 10,000.**Part (2): Possible Purchasing Plans Within the Budget**The company has a budget of 110,000 and wants to buy 10 devices. Let me denote the number of A-type devices as ( m ) and the number of B-type devices as ( 10 - m ).The total cost equation is:[ 12,000m + 10,000(10 - m) leq 110,000 ]Let me simplify this:[ 12,000m + 100,000 - 10,000m leq 110,000 ]Combine like terms:[ 2,000m + 100,000 leq 110,000 ]Subtract 100,000 from both sides:[ 2,000m leq 10,000 ]Divide by 2,000:[ m leq 5 ]Since ( m ) must be a non-negative integer (you can't buy a fraction of a device), the possible values for ( m ) are 0, 1, 2, 3, 4, 5. Therefore, the company can buy anywhere from 0 to 5 A-type devices and the rest will be B-type devices.**Part (3): Most Cost-Effective Purchasing Plan Considering Production Capacity**Now, there's an additional constraint: the total production capacity should not be less than 2040 tons per month. The production capacities are given as 240 tons per month for A-type and 180 tons per month for B-type.Let me set up the inequality for production capacity:[ 240m + 180(10 - m) geq 2040 ]Simplify:[ 240m + 1800 - 180m geq 2040 ]Combine like terms:[ 60m + 1800 geq 2040 ]Subtract 1800 from both sides:[ 60m geq 240 ]Divide by 60:[ m geq 4 ]So, ( m ) must be at least 4. From part (2), ( m ) can be up to 5. Therefore, the possible values for ( m ) are 4 and 5.Now, let's calculate the total cost for each case to find the most cost-effective plan.- For ( m = 4 ): [ 12,000(4) + 10,000(6) = 48,000 + 60,000 = 108,000 ] - For ( m = 5 ): [ 12,000(5) + 10,000(5) = 60,000 + 50,000 = 110,000 ]Comparing the two, buying 4 A-type and 6 B-type devices costs 108,000, which is cheaper than buying 5 A-type and 5 B-type devices at 110,000. Therefore, the most cost-effective plan is to buy 4 A-type devices and 6 B-type devices.**Final Answer**1. The price of an A-type device is boxed{12000} dollars and the price of a B-type device is boxed{10000} dollars.2. The possible purchasing plans are buying boxed{0}, boxed{1}, boxed{2}, boxed{3}, boxed{4}, or boxed{5} A-type devices with the rest being B-type devices.3. The most cost-effective purchasing plan is to buy boxed{4} A-type devices and boxed{6} B-type devices, costing a total of boxed{108000} dollars.

💡First, identify the possible outcomes for matching colors. Abe and Bob can show either both green or both red jelly beans.Next, calculate the probability of both showing a green jelly bean:- Abe has 2 green jelly beans out of 5 total, so the probability that Abe shows a green jelly bean is (frac{2}{5}).- Bob has 2 green jelly beans out of 7 total, so the probability that Bob shows a green jelly bean is (frac{2}{7}).- The events are independent; thus, the probability that both show green is: [ frac{2}{5} times frac{2}{7} = frac{4}{35} ]Then, calculate the probability of both showing a red jelly bean:- Abe has 3 red jelly beans out of 5 total, so the probability that Abe shows a red jelly bean is (frac{3}{5}).- Bob has 3 red jelly beans out of 7 total, so the probability that Bob shows a red jelly bean is (frac{3}{7}).- The probability that both show red is: [ frac{3}{5} times frac{3}{7} = frac{9}{35} ]Finally, sum the probabilities of the favorable outcomes to find the total probability of matching colors:[frac{4}{35} + frac{9}{35} = frac{13}{35}]The probability that Abe and Bob show jelly beans of the same color is (frac{13}{35}).

💡Alright, so I've got this probability problem here, and I need to figure it out step by step. Let me read it again:"Xiao Ming throws a die with uniform density three times and observes the number of points on the upper face each time. It is known that the numbers of points in the three throws are all different. The probability that the sum of the three numbers of points does not exceed 8 is ( )."The options are A: 19/20, B: 1/20, C: 4/5, D: 1/5.Okay, so first, let's break down the problem. We have a die, which I assume is a standard six-sided die with faces numbered 1 through 6. Xiao Ming throws it three times, and each time, the number of points is different. So, we're dealing with three distinct numbers from 1 to 6.We need to find the probability that the sum of these three distinct numbers is less than or equal to 8.Alright, so probability is generally the number of favorable outcomes divided by the total number of possible outcomes. So, I need to figure out two things:1. The total number of ways Xiao Ming can throw the die three times with all different numbers.2. The number of those outcomes where the sum of the three numbers is ≤ 8.Then, the probability will be the second number divided by the first.Let's start with the total number of possible outcomes where all three numbers are different. Since each die has 6 faces, the first throw can be any number from 1 to 6. The second throw has to be different from the first, so there are 5 possibilities. The third throw has to be different from both the first and the second, so there are 4 possibilities.So, the total number of possible outcomes is 6 * 5 * 4. Let me calculate that:6 * 5 = 30, and 30 * 4 = 120.So, there are 120 possible outcomes where all three numbers are different.Now, onto the number of favorable outcomes where the sum is ≤ 8. Hmm, this seems a bit trickier. I need to list all possible sets of three distinct numbers from 1 to 6 that add up to 8 or less.Let me think about how to approach this systematically. Since the numbers are distinct and range from 1 to 6, the smallest possible sum is 1 + 2 + 3 = 6. The next possible sums would be 7 and 8.Wait, so the minimum sum is 6, and we need sums up to 8. So, the possible sums are 6, 7, and 8.Let me list all possible triplets (sets of three distinct numbers) that add up to 6, 7, or 8.Starting with the smallest sum, which is 6:- The only triplet that adds up to 6 is {1, 2, 3}.Now, moving on to sum = 7:- What triplets add up to 7? Let's see:1 + 2 + 4 = 71 + 3 + 3 = 7, but this has duplicates, so it's invalid since all numbers must be different.2 + 2 + 3 = 7, again duplicates, so invalid.So, the only valid triplet for sum = 7 is {1, 2, 4}.Next, sum = 8:- Let's find all triplets that add up to 8.1 + 2 + 5 = 81 + 3 + 4 = 82 + 2 + 4 = 8, duplicates, invalid.2 + 3 + 3 = 8, duplicates, invalid.So, the valid triplets for sum = 8 are {1, 2, 5} and {1, 3, 4}.Wait, is there another triplet? Let me check:1 + 4 + 3 is the same as {1, 3, 4}, which we already have.2 + 3 + 3 is invalid.3 + 3 + 2 is invalid.So, no, only two triplets for sum = 8.So, in total, the triplets that meet the condition are:- {1, 2, 3} (sum = 6)- {1, 2, 4} (sum = 7)- {1, 2, 5} (sum = 8)- {1, 3, 4} (sum = 8)So, that's four triplets.But wait, each triplet can be arranged in different orders, right? Because the die is thrown three times, and the order matters. So, each set of three numbers can be permuted in 3! = 6 ways.So, for each of these four triplets, there are 6 possible sequences.Therefore, the total number of favorable outcomes is 4 * 6 = 24.Wait, let me confirm that. If I have the triplet {1, 2, 3}, the possible sequences are all the permutations of these three numbers, which are 6 in total. Similarly for {1, 2, 4}, {1, 2, 5}, and {1, 3, 4}.Yes, so 4 triplets * 6 permutations each = 24 favorable outcomes.So, the number of favorable outcomes is 24, and the total number of possible outcomes is 120.Therefore, the probability is 24 / 120.Simplifying that, 24 divided by 120 is equal to 1/5.So, the probability is 1/5, which corresponds to option D.Wait, just to make sure I didn't miss any triplets. Let me double-check.For sum = 6: only {1, 2, 3}.Sum = 7: {1, 2, 4}.Sum = 8: {1, 2, 5} and {1, 3, 4}.Is there another triplet for sum = 8? Let's see:What about {2, 3, 3}? No, duplicates.{2, 2, 4}? Duplicates.{1, 4, 3} is same as {1, 3, 4}.{1, 5, 2} is same as {1, 2, 5}.So, no, I think that's all.Therefore, 4 triplets, each with 6 permutations, 24 total.24 / 120 = 1/5.So, the answer should be D: 1/5.**Final Answer**boxed{D}

💡Okay, so I need to figure out what tan(405 degrees) is. Hmm, I remember that the tangent function has something to do with the unit circle and angles. Let me think... I think it's related to the ratio of sine over cosine, right? So tan(theta) is sin(theta)/cos(theta). But I'm not sure how that helps me with 405 degrees.Wait, 405 degrees is more than 360 degrees. I remember that angles in the unit circle repeat every 360 degrees because a full circle is 360 degrees. So maybe I can subtract 360 degrees from 405 degrees to find an equivalent angle that's between 0 and 360 degrees. Let me try that: 405 minus 360 is 45 degrees. So tan(405 degrees) should be the same as tan(45 degrees). That makes sense because after a full rotation of 360 degrees, the angle repeats.Now, what's tan(45 degrees)? I think that's one of those special angles where the sine and cosine are equal. Let me recall... Yes, in a 45-45-90 triangle, both legs are equal, so the sides are in the ratio 1:1:sqrt(2). So sin(45 degrees) is sqrt(2)/2 and cos(45 degrees) is also sqrt(2)/2. Therefore, tan(45 degrees) is sin(45)/cos(45), which is (sqrt(2)/2) divided by (sqrt(2)/2). The sqrt(2)/2 cancels out, leaving 1. So tan(45 degrees) is 1.Since tan(405 degrees) is the same as tan(45 degrees), that means tan(405 degrees) is also 1. I think that's it. But just to make sure, let me visualize the unit circle. Starting at 0 degrees, going all the way around 360 degrees brings me back to the starting point, which is the same as 0 degrees. So adding another 45 degrees would place me at 45 degrees, which is in the first quadrant where both sine and cosine are positive. Therefore, their ratio, which is tangent, is positive as well. So yes, tan(405 degrees) is indeed 1.I don't think I made any mistakes here. I reduced the angle correctly, used the special triangle ratios, and double-checked using the unit circle. Everything seems consistent. So I'm confident that the answer is 1.

💡First, I recognize that Cade gave away 8 marbles and was left with 79 marbles.To find out how many marbles Cade initially had, I need to add the number of marbles he gave away to the number of marbles he had left.So, I add 8 marbles to 79 marbles.This calculation gives me a total of 87 marbles.Therefore, Cade initially had 87 marbles.