🧠:<think>Alright, so I've got these two probability problems to solve, both related to Dr. Smith's fossil analysis. Let me take them one at a time and think through each step carefully.Starting with the first problem about Type A fossils. It says that the depth distribution is modeled by a Gaussian function, which is another name for the normal distribution. The formula given is:[f(x) = frac{1}{sigma sqrt{2pi}} e^{-frac{(x-mu)^2}{2sigma^2}}]They've provided the mean ((mu)) as 10 meters and the standard deviation ((sigma)) as 2 meters. The question is asking for the probability that a randomly selected Type A fossil is found between 8 meters and 12 meters deep.Okay, so since this is a normal distribution, I remember that probabilities correspond to the area under the curve between those two points. To find this, I can use the cumulative distribution function (CDF) for the normal distribution. The CDF gives the probability that a random variable is less than or equal to a certain value.So, the probability we want is:[P(8 leq X leq 12) = P(X leq 12) - P(X leq 8)]To compute this, I need to standardize the values 8 and 12. That is, convert them into z-scores because standard normal tables or functions use z-scores.The z-score formula is:[z = frac{x - mu}{sigma}]So, for x = 8:[z_1 = frac{8 - 10}{2} = frac{-2}{2} = -1]And for x = 12:[z_2 = frac{12 - 10}{2} = frac{2}{2} = 1]So now, I need to find the probability that Z is between -1 and 1. In terms of the standard normal distribution, this is:[P(-1 leq Z leq 1) = P(Z leq 1) - P(Z leq -1)]I remember that the standard normal distribution is symmetric around 0, so ( P(Z leq -1) = 1 - P(Z leq 1) ). Alternatively, I can look up the values in a z-table or use a calculator.Looking up z = 1 in the standard normal table, the cumulative probability is approximately 0.8413. For z = -1, it's 1 - 0.8413 = 0.1587.So, subtracting these:[0.8413 - 0.1587 = 0.6826]Therefore, the probability that a Type A fossil is found between 8 and 12 meters is approximately 68.26%.Wait, let me double-check. Since the mean is 10, and we're looking at one standard deviation below and above, I recall that about 68% of the data lies within one standard deviation in a normal distribution. So, that seems to align. So, that makes sense.Moving on to the second problem about Type B fossils. It says that their presence follows a Poisson distribution with an average ((lambda)) of 3 fossils per square meter. The question is asking for the probability that in a 2 square meter area, at least 5 fossils are found.Hmm, okay. So, first, I need to adjust the lambda for the 2 square meter area. Since lambda is per square meter, for 2 square meters, lambda becomes:[lambda_{text{total}} = 3 times 2 = 6]So, now, we're dealing with a Poisson distribution with (lambda = 6). We need the probability that the number of fossils, X, is at least 5. That is:[P(X geq 5) = 1 - P(X leq 4)]So, I need to calculate the cumulative probability from 0 to 4 and subtract that from 1.The Poisson probability mass function is:[P(X = k) = frac{e^{-lambda} lambda^k}{k!}]So, I need to compute this for k = 0,1,2,3,4 and sum them up.Let me write down each term:For k=0:[P(X=0) = frac{e^{-6} times 6^0}{0!} = frac{e^{-6} times 1}{1} = e^{-6} approx 0.002478752]k=1:[P(X=1) = frac{e^{-6} times 6^1}{1!} = frac{e^{-6} times 6}{1} approx 0.014872513]k=2:[P(X=2) = frac{e^{-6} times 6^2}{2!} = frac{e^{-6} times 36}{2} approx 0.044617538]k=3:[P(X=3) = frac{e^{-6} times 6^3}{3!} = frac{e^{-6} times 216}{6} approx 0.089235076]k=4:[P(X=4) = frac{e^{-6} times 6^4}{4!} = frac{e^{-6} times 1296}{24} approx 0.133852614]Now, let's sum these up:0.002478752 + 0.014872513 = 0.0173512650.017351265 + 0.044617538 = 0.0619688030.061968803 + 0.089235076 = 0.1512038790.151203879 + 0.133852614 = 0.285056493So, the cumulative probability P(X ≤ 4) is approximately 0.285056493.Therefore, P(X ≥ 5) = 1 - 0.285056493 ≈ 0.714943507.So, approximately 71.49% chance of finding at least 5 fossils in a 2 square meter area.Wait, let me verify the calculations step by step to make sure I didn't make an arithmetic error.Calculating each term:For k=0: e^{-6} ≈ 0.002478752. That's correct.k=1: 0.002478752 * 6 = 0.014872512. Correct.k=2: 0.014872512 * 6 / 2 = 0.044617536. Correct.k=3: 0.044617536 * 6 / 3 = 0.089235072. Correct.k=4: 0.089235072 * 6 / 4 = 0.133852608. Correct.Adding them up:0.002478752 + 0.014872512 = 0.0173512640.017351264 + 0.044617536 = 0.06196880.0619688 + 0.089235072 = 0.1512038720.151203872 + 0.133852608 = 0.28505648Yes, so cumulative is approximately 0.28505648.Subtracting from 1: 1 - 0.28505648 ≈ 0.71494352.So, approximately 71.49%.Alternatively, if I use a calculator or Poisson table, but I think my manual calculation is correct.Alternatively, I can use the formula for the Poisson CDF:[P(X leq k) = e^{-lambda} sum_{i=0}^{k} frac{lambda^i}{i!}]Which is exactly what I did.So, summarizing:1. For Type A fossils, the probability is approximately 68.26%.2. For Type B fossils, the probability is approximately 71.49%.I think that's all. I don't see any mistakes in my reasoning, but let me just recap.First problem: Normal distribution, mean 10, standard deviation 2. Calculated z-scores for 8 and 12, found the area between them, which is about 68.26%. That seems right.Second problem: Poisson distribution, lambda 3 per square meter, so 6 for 2 square meters. Calculated P(X >=5) by subtracting P(X<=4) from 1. Calculated each term, summed them, got approximately 0.285, so 1 - 0.285 ≈ 0.715 or 71.5%. That seems correct.Yeah, I think that's solid.Final Answer1. The probability for Type A fossils is boxed{0.6826}.2. The probability for Type B fossils is boxed{0.7149}.</think>

🧠:<think>Okay, so I have this problem about a street artist and a photographer. Let me try to figure it out step by step. First, the problem says that the time until a mural is cleaned up follows a normal distribution with an average of 7 days and a standard deviation of 2 days. The photographer can only take photos if the mural stays up for at least 3 days and at most 10 days. So, part 1 is asking for the probability that the photographer can successfully capture a mural. That means I need to find the probability that the time until cleanup is between 3 and 10 days. Since it's a normal distribution, I remember that I can use z-scores to standardize the values and then use the standard normal distribution table or a calculator to find the probabilities. Let me recall the formula for z-score: z = (X - μ) / σ, where X is the value, μ is the mean, and σ is the standard deviation.So, for X = 3 days:z1 = (3 - 7) / 2 = (-4)/2 = -2And for X = 10 days:z2 = (10 - 7) / 2 = 3/2 = 1.5Now, I need to find the area under the standard normal curve between z = -2 and z = 1.5. I think I can use the standard normal distribution table for this. Let me remember how to use it. The table gives the probability that Z is less than a certain value. So, I can find P(Z < 1.5) and P(Z < -2), and then subtract the latter from the former to get the probability between -2 and 1.5.Looking up z = 1.5 in the table: I think that's about 0.9332. And for z = -2: I remember that the table might not have negative z-scores, but I know that the distribution is symmetric, so P(Z < -2) is the same as 1 - P(Z < 2). Looking up z = 2: that's approximately 0.9772, so P(Z < -2) = 1 - 0.9772 = 0.0228.So, the probability between -2 and 1.5 is 0.9332 - 0.0228 = 0.9104.Wait, let me double-check that. So, P(-2 < Z < 1.5) = P(Z < 1.5) - P(Z < -2) = 0.9332 - 0.0228 = 0.9104. Yeah, that seems right.So, the probability is approximately 0.9104, or 91.04%.Hmm, that seems high. Let me think again. The mean is 7 days, so 3 days is two standard deviations below the mean, and 10 days is 1.5 standard deviations above. Since the normal distribution is bell-shaped, most of the data is within the mean plus or minus a couple standard deviations. So, the area between -2 and 1.5 should cover a large portion. Maybe 91% is correct.Okay, moving on to part 2. The photographer wants to document as many murals as possible over a year, which is 365 days. The street artist paints a new mural every 12 days on average. So, how many murals will the photographer document in a year, given the probability from part 1?First, let me figure out how many murals are painted in a year. If a new mural is painted every 12 days, then in 365 days, the number of murals is 365 / 12. Let me calculate that.365 divided by 12 is approximately 30.4167. So, about 30 murals in a year. But since you can't have a fraction of a mural, maybe it's 30 murals.But wait, actually, the artist paints a new mural every 12 days, so starting from day 0, the next mural is on day 12, then day 24, etc. So, in 365 days, how many intervals of 12 days are there?Let me compute 365 / 12. 12*30=360, so 365-360=5. So, 30 full intervals, meaning 30 murals. But actually, starting from day 0, the first mural is day 0, then day 12, ..., up to day 360, which is the 31st mural? Wait, no, because 0 to 12 is the first interval, so 365 days would include 30 intervals, resulting in 31 murals? Hmm, I need to clarify.Wait, if the artist paints a new mural every 12 days, starting from day 0, then the murals are painted on day 0, 12, 24, ..., up to the last day before or equal to 365.So, the number of murals is floor(365 / 12) + 1. Let's compute 365 / 12 ≈ 30.4167, so floor is 30, plus 1 is 31. So, 31 murals in a year.But wait, actually, the last mural would be on day 360, which is 30*12=360, and the next would be on day 372, which is beyond 365. So, in 365 days, there are 31 murals: day 0, 12, 24, ..., 360.So, 31 murals in total.But wait, the photographer can only photograph a mural if it's up for at least 3 days and at most 10 days. So, each mural has a probability of 0.9104 of being photographed.Therefore, the expected number of murals photographed is 31 * 0.9104.Let me compute that. 31 * 0.9104 ≈ 31 * 0.91 = approximately 28.21.But let me do it more accurately. 31 * 0.9104:31 * 0.9 = 27.931 * 0.0104 = 0.3224So, total is 27.9 + 0.3224 ≈ 28.2224.So, approximately 28.22 murals. Since you can't photograph a fraction of a mural, we might round it to 28 murals.But wait, actually, in probability terms, the expected value can be a fractional number, so maybe we should present it as approximately 28.22, but since the question says "how many murals will the photographer successfully document," it might expect an integer. So, 28 or 28.22.But let me think again. The number of murals is 31, each with a success probability of ~0.9104, so the expectation is 31 * 0.9104 ≈ 28.22. So, about 28 murals.Alternatively, maybe the number of opportunities is 365 / 12 ≈ 30.4167, so approximately 30 murals, each with 0.9104 chance, so 30 * 0.9104 ≈ 27.31, which is about 27 murals.Wait, now I'm confused because earlier I thought it was 31 murals, but maybe it's 30.Let me clarify: If the artist paints a new mural every 12 days, starting from day 0, then the number of murals in 365 days is the number of times 12 fits into 365, plus 1 if starting from day 0.So, 365 / 12 ≈ 30.4167, so 30 full intervals, meaning 31 murals (including day 0). But if the artist starts on day 0, then the next is day 12, ..., last one on day 360, which is 30*12=360. So, 31 murals in total.But wait, the photographer can only photograph a mural if it's up for at least 3 days and at most 10 days. So, does the photographer have the chance to photograph each mural? Or is there a time constraint?Wait, the problem says the photographer can only photograph a mural if it remains for at least 3 days and at most 10 days. So, for each mural, the time it's up is a random variable, and if that time is between 3 and 10 days, the photographer can capture it.So, each mural has an independent probability of ~0.9104 of being captured. So, over 31 murals, the expected number is 31 * 0.9104 ≈ 28.22.But the problem says "over the course of a year (365 days)", so maybe we need to consider the number of murals that can be painted in 365 days, which is 365 / 12 ≈ 30.4167, so 30 murals, because you can't have a fraction. So, 30 murals, each with probability ~0.9104, so 30 * 0.9104 ≈ 27.31, which is approximately 27 murals.Wait, but if the first mural is on day 0, then the next on day 12, ..., the last one on day 360, which is within 365 days. So, that's 31 murals. So, 31 is correct.Alternatively, maybe the artist paints a new mural every 12 days, meaning the time between murals is 12 days, so in 365 days, the number of murals is floor(365 / 12) + 1 = 30 + 1 = 31.Yes, that makes sense. So, 31 murals.Therefore, the expected number is 31 * 0.9104 ≈ 28.22, so approximately 28 murals.But let me check if the photographer can photograph multiple murals in a year. Since each mural is independent, the expectation is additive, so 31 * 0.9104 ≈ 28.22.So, rounding to the nearest whole number, it's 28 murals.Alternatively, maybe the problem expects us to use the exact value, so 28.22, but since it's a count, 28 is the integer part.Wait, but in probability, the expectation can be a non-integer, so maybe we should present it as approximately 28.22, but the question says "how many murals will the photographer successfully document in a year," which probably expects an integer. So, 28.But let me think again. If the photographer has a 91.04% chance per mural, and there are 31 murals, the expected number is 28.22. So, approximately 28.Alternatively, maybe the problem expects us to use 365 / 12 ≈ 30.4167, so 30 murals, leading to 30 * 0.9104 ≈ 27.31, so 27 murals.Wait, I'm getting conflicting numbers here. Let me clarify:If the artist paints a new mural every 12 days, starting from day 0, then the murals are on days 0, 12, 24, ..., 360. So, how many is that?From day 0 to day 360, that's 360 / 12 = 30 intervals, so 31 murals.Yes, that's correct. So, 31 murals.Therefore, the expected number is 31 * 0.9104 ≈ 28.22, so approximately 28 murals.But let me check if the photographer can photograph a mural only once, and each mural is independent. So, yes, expectation is linear, so 31 * 0.9104 ≈ 28.22.So, the answer is approximately 28 murals.Wait, but let me think about the timeline. Each mural is up for a certain number of days, and the photographer can photograph it if it's up for between 3 and 10 days. So, for each mural, the photographer has a certain probability of capturing it, independent of others.Therefore, the expected number is indeed the number of murals multiplied by the probability.So, 31 * 0.9104 ≈ 28.22, which is approximately 28.Alternatively, if we use more precise calculations:0.9104 * 31:31 * 0.9 = 27.931 * 0.0104 = 0.3224Total: 27.9 + 0.3224 = 28.2224So, 28.2224, which is approximately 28.22.But since the question asks for how many murals, it's better to round to the nearest whole number, so 28.Wait, but sometimes in such problems, they might expect you to use the floor function, but I think expectation can be a decimal, but since the question is about the number of murals, which must be an integer, it's better to round to the nearest whole number.So, 28.Alternatively, maybe the problem expects the exact value, so 28.22, but I think 28 is acceptable.Wait, but let me check the initial probability calculation again.I had z1 = -2, which gives P(Z < -2) = 0.0228z2 = 1.5, which gives P(Z < 1.5) = 0.9332So, the area between them is 0.9332 - 0.0228 = 0.9104Yes, that's correct.So, the probability is 0.9104.Therefore, the expected number is 31 * 0.9104 ≈ 28.22, so approximately 28.Alternatively, if the problem considers the number of murals as 30, then 30 * 0.9104 ≈ 27.31, which is approximately 27.But I think 31 is correct because starting from day 0, the first mural is on day 0, then day 12, ..., up to day 360, which is 31 murals.So, I think 28 is the correct answer.Wait, but let me think about the timeline again. If the artist paints a new mural every 12 days, then in 365 days, the number of murals is floor(365 / 12) + 1 = 30 + 1 = 31.Yes, that's correct.So, 31 murals, each with a 91.04% chance of being photographed, so 31 * 0.9104 ≈ 28.22, which is approximately 28.Therefore, the answers are:1. Approximately 91.04% probability.2. Approximately 28 murals.But let me write the exact values.For part 1, the probability is P(3 < X < 10) = P(-2 < Z < 1.5) = Φ(1.5) - Φ(-2) = 0.9332 - 0.0228 = 0.9104.So, 0.9104 or 91.04%.For part 2, number of murals is 31, so expected number is 31 * 0.9104 = 28.2224, which is approximately 28.Alternatively, if we use more precise z-table values, maybe the probability is slightly different.Wait, let me check the exact z-values.For z = 1.5, the exact value is 0.933196, which is approximately 0.9332.For z = -2, the exact value is 0.02275, which is approximately 0.0228.So, the difference is 0.9332 - 0.0228 = 0.9104.So, that's correct.Therefore, the answers are:1. 0.9104 or 91.04%2. Approximately 28 murals.But let me think again about part 2. The problem says the artist paints a new mural every 12 days on average. So, does that mean the time between murals is 12 days, or the number of days per mural is 12? Wait, no, the time until cleanup is 7 days on average, but the artist paints a new mural every 12 days. So, the time between murals is 12 days, independent of how long each mural stays up.So, in 365 days, the number of murals is 365 / 12 ≈ 30.4167, so 30 full murals, but since the first mural is on day 0, the last one is on day 360, which is 30*12=360, so 31 murals.Yes, that's correct.So, the photographer can attempt to photograph each of these 31 murals, each with a probability of 0.9104. So, the expected number is 31 * 0.9104 ≈ 28.22, which is approximately 28.Therefore, the final answers are:1. The probability is approximately 91.04%.2. The photographer will successfully document approximately 28 murals in a year.But let me write the exact decimal for part 1, which is 0.9104, and for part 2, 28.2224, but since it's a count, 28.Alternatively, if we use more precise z-values, maybe the probability is slightly different, but I think 0.9104 is accurate enough.So, to summarize:1. The probability is approximately 0.9104 or 91.04%.2. The expected number of murals documented is approximately 28.I think that's it.</think>

🧠:<think>Okay, so I have this problem where a business owner hired a retired golf professional, and they want to figure out the financial impact over five years. There are two parts: calculating the total revenue from new players and calculating the total salary paid to the professional. Let me try to tackle each part step by step.Starting with part 1: The number of new players per year is given by the function ( P(t) = 50e^{0.05t} ), where ( t ) is the number of years since the professional was hired. The average revenue per player per year is 500. I need to write an integral to represent the total revenue from these new players over the first five years and then evaluate it.Hmm, okay, so revenue is typically calculated as the number of customers multiplied by the revenue per customer. In this case, since the number of players is changing over time, it's a bit more involved. Instead of a simple multiplication, I think I need to integrate the revenue over the time period because the number of players is a continuous function.So, for each small time interval ( dt ), the number of new players is ( P(t) ), and each contributes 500 in revenue. Therefore, the revenue from new players in that small interval would be ( 500 times P(t) times dt ). To get the total revenue over five years, I need to sum all these small revenues from ( t = 0 ) to ( t = 5 ).Putting that into an integral, it should be:[text{Total Revenue} = int_{0}^{5} 500 times P(t) , dt]Substituting ( P(t) ) into the integral:[text{Total Revenue} = int_{0}^{5} 500 times 50e^{0.05t} , dt]Simplifying that, 500 multiplied by 50 is 25,000, so:[text{Total Revenue} = 25,000 int_{0}^{5} e^{0.05t} , dt]Now, I need to evaluate this integral. The integral of ( e^{kt} ) with respect to ( t ) is ( frac{1}{k}e^{kt} ). So here, ( k = 0.05 ), so the integral becomes:[25,000 left[ frac{1}{0.05} e^{0.05t} right]_0^5]Calculating ( frac{1}{0.05} ) is 20, so:[25,000 times 20 left[ e^{0.05 times 5} - e^{0} right]]Simplify the exponents:( 0.05 times 5 = 0.25 ), so:[500,000 left[ e^{0.25} - 1 right]]Now, I need to compute ( e^{0.25} ). I remember that ( e^{0.25} ) is approximately 1.2840254. Let me double-check that with a calculator. Yes, ( e^{0.25} approx 1.2840254 ).So substituting that in:[500,000 times (1.2840254 - 1) = 500,000 times 0.2840254]Calculating that:First, 500,000 times 0.2 is 100,000.Then, 500,000 times 0.08 is 40,000.500,000 times 0.0040254 is approximately 500,000 * 0.004 = 2,000, and 500,000 * 0.0000254 ≈ 12.7.Adding those together: 100,000 + 40,000 = 140,000; 140,000 + 2,000 = 142,000; 142,000 + 12.7 ≈ 142,012.7.Wait, that seems a bit off. Let me do it more accurately.0.2840254 multiplied by 500,000:0.2840254 * 500,000 = (0.2 + 0.08 + 0.0040254) * 500,0000.2 * 500,000 = 100,0000.08 * 500,000 = 40,0000.0040254 * 500,000 = 2,012.7Adding those together: 100,000 + 40,000 = 140,000; 140,000 + 2,012.7 = 142,012.7So, approximately 142,012.70.Wait, but let me verify with another method. 0.2840254 * 500,000.Alternatively, 500,000 * 0.2840254 = 500,000 * (284,025.4 / 1,000,000) = (500,000 / 1,000,000) * 284,025.4 = 0.5 * 284,025.4 = 142,012.7Yes, that's correct. So, the total revenue is approximately 142,012.70.Wait, but let me check if I did the integral correctly. The integral of ( e^{0.05t} ) is ( frac{1}{0.05}e^{0.05t} ), which is 20e^{0.05t}. Then evaluating from 0 to 5:20(e^{0.25} - 1). Then multiplied by 25,000.Wait, no, hold on. Wait, the integral was 25,000 times the integral of e^{0.05t} dt, which is 25,000 * (20(e^{0.25} - 1)) = 500,000(e^{0.25} - 1). So that part is correct.And e^{0.25} is approximately 1.2840254, so 1.2840254 - 1 = 0.2840254. Multiply by 500,000 gives 142,012.7.So, the total revenue is approximately 142,012.70 over five years.Wait, but let me think again. Is this the correct approach? Because the number of players is given per year, and the revenue per player is per year. So, if P(t) is the number of new players per year, then integrating P(t) over five years gives the total number of new players over five years. Then, multiplying by 500 would give the total revenue.Alternatively, is P(t) the number of new players per year, so each year t, you have P(t) new players, each contributing 500 per year. So, the revenue each year would be 500 * P(t), and integrating that over five years would give the total revenue.Wait, yes, that's exactly what I did. So, the integral of 500 * P(t) dt from 0 to 5 is correct.Alternatively, if I think about it as the number of players each year is P(t), and each contributes 500 per year, so the annual revenue is 500 * P(t). Therefore, integrating over five years gives the total revenue.So, that seems correct.So, moving on to part 2: The retired professional is paid a yearly salary that increases geometrically. The initial salary in the first year is 80,000, and it increases by 4% each year. I need to write and evaluate the sum that represents the total salary paid over the first five years.Okay, so this is a geometric series where each term is 4% higher than the previous one. The first term is 80,000, and the common ratio is 1.04.The formula for the sum of a geometric series is:[S_n = a frac{r^n - 1}{r - 1}]Where ( a ) is the first term, ( r ) is the common ratio, and ( n ) is the number of terms.In this case, ( a = 80,000 ), ( r = 1.04 ), and ( n = 5 ).So, plugging in the values:[S_5 = 80,000 times frac{1.04^5 - 1}{1.04 - 1}]First, compute ( 1.04^5 ). Let me calculate that.1.04^1 = 1.041.04^2 = 1.08161.04^3 = 1.0816 * 1.04 ≈ 1.1248641.04^4 ≈ 1.124864 * 1.04 ≈ 1.169858561.04^5 ≈ 1.16985856 * 1.04 ≈ 1.2166529024So, approximately 1.2166529024.Therefore, ( 1.04^5 - 1 ≈ 0.2166529024 ).The denominator is ( 1.04 - 1 = 0.04 ).So, the sum becomes:[S_5 = 80,000 times frac{0.2166529024}{0.04}]Calculating ( frac{0.2166529024}{0.04} ):0.2166529024 / 0.04 = 5.41632256So, ( S_5 = 80,000 times 5.41632256 )Calculating that:80,000 * 5 = 400,00080,000 * 0.41632256 ≈ 80,000 * 0.4 = 32,000; 80,000 * 0.01632256 ≈ 1,305.8048So, adding those together: 32,000 + 1,305.8048 ≈ 33,305.8048Therefore, total sum is 400,000 + 33,305.8048 ≈ 433,305.8048So, approximately 433,305.80.Wait, let me do that multiplication more accurately.80,000 * 5.41632256First, 80,000 * 5 = 400,00080,000 * 0.41632256:Let me compute 0.41632256 * 80,000:0.4 * 80,000 = 32,0000.01632256 * 80,000 = 1,305.8048So, adding together: 32,000 + 1,305.8048 = 33,305.8048Therefore, total is 400,000 + 33,305.8048 = 433,305.8048So, approximately 433,305.80.Alternatively, using another method:5.41632256 * 80,000 = (5 + 0.41632256) * 80,000 = 5*80,000 + 0.41632256*80,000 = 400,000 + 33,305.8048 = 433,305.8048Yes, that's correct.So, the total salary over five years is approximately 433,305.80.Wait, but let me confirm the formula again. The sum of a geometric series is ( S_n = a frac{r^n - 1}{r - 1} ). Here, a = 80,000, r = 1.04, n = 5. So, plugging in:( S_5 = 80,000 times frac{1.04^5 - 1}{0.04} ). We calculated 1.04^5 ≈ 1.2166529024, so 1.2166529024 - 1 = 0.2166529024. Divided by 0.04 is 5.41632256. Multiply by 80,000 gives 433,305.8048.Yes, that seems correct.Alternatively, I can compute each year's salary and sum them up to verify.Year 1: 80,000Year 2: 80,000 * 1.04 = 83,200Year 3: 83,200 * 1.04 = 86,528Year 4: 86,528 * 1.04 ≈ 86,528 + 3,461.12 = 89,989.12Year 5: 89,989.12 * 1.04 ≈ 89,989.12 + 3,599.5648 ≈ 93,588.6848Now, summing these:80,000 + 83,200 = 163,200163,200 + 86,528 = 249,728249,728 + 89,989.12 = 339,717.12339,717.12 + 93,588.6848 ≈ 433,305.8048Yes, that's the same result as before. So, the total salary is approximately 433,305.80.So, summarizing:1. The total revenue from new players over five years is approximately 142,012.70.2. The total salary paid to the professional over five years is approximately 433,305.80.Wait, but let me check if I interpreted the salary correctly. The problem says the salary increases geometrically, starting at 80,000 with a 4% increase each year. So, that means each year's salary is 4% higher than the previous year. So, the first year is 80,000, the second is 80,000*1.04, third is 80,000*(1.04)^2, and so on up to the fifth year, which is 80,000*(1.04)^4.Wait, actually, when n = 5, the exponents go from 0 to 4, because the first term is year 1: 80,000*(1.04)^0 = 80,000.So, the sum is from k=0 to k=4 of 80,000*(1.04)^k.Which is indeed a geometric series with n=5 terms, first term a=80,000, ratio r=1.04.So, the formula is correct.Therefore, the calculations are accurate.So, I think I've got both parts figured out.Final Answer1. The total revenue from new players over the first five years is boxed{142012.70} dollars.2. The total salary paid to the professional over the first five years is boxed{433305.80} dollars.</think>

🧠:<think>First, I need to calculate the Compound Annual Growth Rate (CAGR) for Hiroshima's population from 1945 to 2023. The population grew from 340,000 to 1,200,000 over 78 years. Using the CAGR formula, I'll set up the equation and solve for the growth rate.Next, I'll address the exponential decay of Cesium-137. Given its half-life of 30 years, I'll use the exponential decay formula to determine how much of the initial 100 grams remains after 78 years. This involves calculating the decay constant and applying it to the decay equation.</think>