🧠:<think>Alright, so I have these two statistical problems to solve. Let me take them one by one and think through each step carefully. I'm a bit nervous because I don't want to make any mistakes, but I'll try my best.Starting with the first problem. It's about a job applicant who wants to analyze their interview performance using statistics. They collected data from 5 interviewers, each giving scores in four categories: Technical Skills (T), Communication (C), Problem-Solving (P), and Cultural Fit (F). The total scores are believed to follow a multivariate normal distribution with a specific mean vector and covariance matrix.The question is asking for the probability that the applicant's total score in Technical Skills (T) and Communication (C) combined exceeds 16. Hmm, okay. So, I need to find P(T + C > 16). First, let me recall what a multivariate normal distribution is. It's a generalization of the univariate normal distribution to higher dimensions. Each variable has a normal distribution, and their joint distribution is also normal. The key parameters are the mean vector and the covariance matrix.Given that, I know that any linear combination of the variables in a multivariate normal distribution is also normally distributed. So, T + C should be a normal random variable. Therefore, I can find its mean and variance and then compute the probability.Let me write down the mean vector μ and the covariance matrix Σ:μ = [7, 6, 5, 4]^TΣ is given as:4 2 1 02 3 0.5 0.21 0.5 2 0.10 0.2 0.1 1So, the variables are T, C, P, F in that order. So, the first element is T, the second is C, etc.Since we're interested in T + C, let's denote this as a new variable, say S = T + C.To find the distribution of S, we need to compute its mean and variance.The mean of S is just the sum of the means of T and C. So, E[S] = E[T] + E[C] = 7 + 6 = 13.Next, the variance of S. Since S is a linear combination of T and C, the variance is Var(T) + Var(C) + 2*Cov(T, C). Looking at the covariance matrix Σ, the covariance between T and C is the element in the first row and second column, which is 2. So, Cov(T, C) = 2.Therefore, Var(S) = Var(T) + Var(C) + 2*Cov(T, C) = 4 + 3 + 2*2 = 4 + 3 + 4 = 11.Wait, hold on, is that correct? Let me double-check. Var(T) is the variance of T, which is the (1,1) element of Σ, so 4. Var(C) is the (2,2) element, which is 3. Cov(T, C) is the (1,2) element, which is 2. So, yes, Var(S) = 4 + 3 + 2*2 = 11. That seems right.So, S ~ N(13, 11). Now, we need to find P(S > 16). To compute this probability, we can standardize S. Let me recall that for a normal variable X ~ N(μ, σ²), the standardized variable Z = (X - μ)/σ follows a standard normal distribution N(0,1).So, let's compute Z for S = 16.Z = (16 - 13)/sqrt(11) = 3 / sqrt(11) ≈ 3 / 3.3166 ≈ 0.904.So, we need to find P(Z > 0.904). Looking at standard normal tables or using a calculator, P(Z > 0.904) is equal to 1 - Φ(0.904), where Φ is the standard normal CDF.Looking up Φ(0.904), I can use a Z-table or approximate it. Alternatively, I remember that Φ(0.9) is approximately 0.8159 and Φ(0.91) is approximately 0.8186. Since 0.904 is between 0.9 and 0.91, let's interpolate.The difference between 0.9 and 0.91 is 0.01 in Z, and the corresponding Φ values increase by about 0.8186 - 0.8159 = 0.0027.0.904 is 0.004 above 0.9, so the increase would be (0.004 / 0.01) * 0.0027 ≈ 0.00108.So, Φ(0.904) ≈ 0.8159 + 0.00108 ≈ 0.81698.Therefore, P(Z > 0.904) ≈ 1 - 0.81698 ≈ 0.18302.So, approximately 18.3% chance that T + C exceeds 16.Wait, but let me confirm this calculation because sometimes I might make a mistake in the interpolation or the Z value.Alternatively, using a calculator, if I compute 3 / sqrt(11):sqrt(11) ≈ 3.3166, so 3 / 3.3166 ≈ 0.9045.Looking up 0.9045 in a standard normal table, or using a calculator, the cumulative probability is approximately 0.8169, so 1 - 0.8169 ≈ 0.1831.Yes, so about 18.31%. So, approximately 18.3%.Alternatively, if I use more precise methods, maybe using a calculator or software, but since I don't have that here, I think 18.3% is a reasonable approximation.So, the probability is approximately 18.3%.Wait, but let me think again. The question says "the applicant's total score in Technical Skills (T) and Communication (C) combined exceeds 16." So, that's exactly what I computed. So, I think that's correct.So, moving on to the second problem.The second problem is about hypothesis testing. The applicant wants to test whether the mean score in Cultural Fit (F) for all rejected applicants is 4 against the alternative that it is not 4. They collected a sample of 30 rejected applicants, with a sample mean of 3.7 and a sample standard deviation of 0.9. They want to conduct a hypothesis test at the 5% significance level.Alright, so this is a two-tailed test because the alternative is that the mean is not equal to 4.Given that, the null hypothesis H0: μ = 4, and the alternative hypothesis H1: μ ≠ 4.The sample size is 30, which is moderately large, and the sample data is assumed to follow a normal distribution. So, we can use a t-test here because the population standard deviation is unknown, and we're using the sample standard deviation.Wait, but with n = 30, sometimes people use the z-test as an approximation, but since the population standard deviation is unknown, the t-test is more appropriate. However, with n = 30, the t-distribution is quite close to the normal distribution, so either could be used, but I think the t-test is more precise here.So, let's proceed with a t-test.The test statistic is calculated as:t = (sample mean - hypothesized mean) / (sample standard deviation / sqrt(n))Plugging in the numbers:t = (3.7 - 4) / (0.9 / sqrt(30)) = (-0.3) / (0.9 / 5.4772) ≈ (-0.3) / (0.1643) ≈ -1.826.So, the t-statistic is approximately -1.826.Now, we need to find the critical value or the p-value to make a decision.Since it's a two-tailed test at 5% significance level, the total alpha is 0.05, so each tail has 0.025.With n = 30, the degrees of freedom (df) is 29.Looking up the critical value for a t-distribution with df = 29 and alpha = 0.025, the critical value is approximately ±2.045.So, the critical region is t < -2.045 or t > 2.045.Our calculated t-statistic is -1.826, which is greater than -2.045. Therefore, it does not fall into the rejection region.Alternatively, we can compute the p-value. The p-value is the probability of observing a t-statistic as extreme as -1.826 or more, under the null hypothesis.Since it's a two-tailed test, the p-value is 2 * P(t < -1.826).Looking up the t-distribution table for df = 29, the value -1.826 is between the critical values for 0.05 and 0.10. Wait, let me think.Wait, actually, the critical value for 0.05 (two-tailed) is ±2.045, and for 0.10 it's ±1.699.Wait, no, actually, for df = 29, the critical values are:- For 0.05 two-tailed: ±2.045- For 0.10 two-tailed: ±1.699Wait, but our t-statistic is -1.826, which is between -2.045 and -1.699. So, the p-value is between 0.05 and 0.10.Wait, actually, let me think again. The t-statistic is -1.826, so in absolute terms, it's 1.826.Looking at the t-table for df = 29:- The value 1.699 corresponds to 0.10 two-tailed.- The value 1.826 is higher than 1.699, so it's more extreme.Wait, actually, the t-table might not have 1.826, but let's see.Alternatively, perhaps using a calculator or software, but since I don't have that, I can estimate.Alternatively, I can use the fact that for df = 29, the t-value of 1.826 is approximately corresponding to a p-value around 0.075.Wait, because 1.699 is 0.10, and 2.045 is 0.05. So, 1.826 is closer to 1.699 than to 2.045.Wait, let me compute the difference.From 1.699 to 2.045 is an increase of 0.346.1.826 is 1.826 - 1.699 = 0.127 above 1.699.So, 0.127 / 0.346 ≈ 0.367, so about 36.7% of the way from 1.699 to 2.045.Since the p-value decreases from 0.10 to 0.05 as the t-value increases from 1.699 to 2.045, the p-value for 1.826 would be approximately 0.10 - (0.10 - 0.05) * (0.127 / 0.346) ≈ 0.10 - 0.05 * 0.367 ≈ 0.10 - 0.0183 ≈ 0.0817.So, approximately 0.0817, which is about 8.17%. Since it's a two-tailed test, the p-value is approximately 0.0817.Wait, but actually, the p-value is the area in both tails beyond |t|, so if the t-statistic is -1.826, the p-value is 2 * P(t < -1.826). Since the t-distribution is symmetric, this is equal to 2 * P(t > 1.826).But in the t-table, we usually look up the upper tail probability. So, for t = 1.826, df = 29, the upper tail probability is approximately 0.04 (since 1.699 is 0.10, 1.826 is higher, so the upper tail is less than 0.05 but more than 0.025). Wait, no, 2.045 is 0.025.Wait, maybe I should think differently. Let me recall that for df = 29, the critical value for 0.05 is 2.045, so a t-value of 1.826 is less than that, so the upper tail probability is greater than 0.05.Wait, no, actually, the critical value for 0.05 is 2.045, so a t-value of 1.826 is less than that, meaning the upper tail probability is greater than 0.05.Wait, no, actually, no. Wait, the critical value is the value beyond which the probability is 0.025 (for two-tailed). So, if the t-value is 1.826, which is less than 2.045, the upper tail probability is greater than 0.025.Wait, I'm getting confused. Let me think again.The t-table gives critical values for certain alpha levels. For example, for df = 29, the critical value for alpha = 0.025 is 2.045. So, if our t-statistic is 1.826, which is less than 2.045, the p-value is greater than 0.05 (since 1.826 is less extreme than 2.045).Wait, no, actually, the p-value is the probability of getting a t-statistic as extreme or more extreme than the observed one. So, if our t-statistic is 1.826, which is less than 2.045, the p-value is greater than 0.05.Wait, but actually, for a two-tailed test, the p-value is 2 * P(t > |1.826|). Since 1.826 is less than 2.045, P(t > 1.826) is greater than 0.025. Therefore, the p-value is greater than 0.05.Wait, but how much greater? Let me try to estimate.Looking at the t-table, for df = 29, the critical values are:- 1.699 for 0.10- 2.045 for 0.05So, 1.826 is between 1.699 and 2.045.The difference between 1.699 and 2.045 is 0.346.1.826 - 1.699 = 0.127.So, 0.127 / 0.346 ≈ 0.367, so about 36.7% of the way from 1.699 to 2.045.The corresponding p-value for 1.699 is 0.10, and for 2.045 is 0.05. So, the p-value decreases from 0.10 to 0.05 as t increases from 1.699 to 2.045.Therefore, the p-value for 1.826 would be approximately 0.10 - (0.10 - 0.05) * (0.127 / 0.346) ≈ 0.10 - 0.05 * 0.367 ≈ 0.10 - 0.0183 ≈ 0.0817.So, approximately 0.0817, which is about 8.17%. Since it's a two-tailed test, the p-value is approximately 0.0817.Wait, but actually, the p-value is the area in both tails beyond |t|. Since the t-statistic is -1.826, the p-value is 2 * P(t < -1.826). Since the t-distribution is symmetric, this is equal to 2 * P(t > 1.826).But in the t-table, we usually look up the upper tail probability. So, for t = 1.826, df = 29, the upper tail probability is approximately 0.04 (since 1.699 is 0.10, 1.826 is higher, so the upper tail is less than 0.05 but more than 0.025). Wait, no, 2.045 is 0.025.Wait, perhaps I should use a more accurate method. Alternatively, I can use the fact that for large degrees of freedom, the t-distribution approximates the normal distribution.But with df = 29, it's still somewhat close to the normal, but not extremely so.Alternatively, using a calculator or software, but since I don't have that, I can use a linear approximation.Alternatively, perhaps I can use the formula for the t-distribution's cumulative distribution function, but that's complicated without a calculator.Alternatively, I can recall that for df = 30, the critical value for 0.05 is approximately 2.042, which is very close to 2.045 for df = 29, so the p-value for t = 1.826 would be roughly similar.But perhaps I can use the following approach: for a t-distribution with 29 degrees of freedom, the p-value for t = 1.826 can be approximated using the normal distribution.The t-value of 1.826 is approximately 1.826 standard errors away from the mean.In the standard normal distribution, the upper tail probability for Z = 1.826 is approximately 0.034 (since Z = 1.82 corresponds to 0.0344, and Z = 1.83 corresponds to 0.0336). So, approximately 0.034.But since it's a t-distribution with 29 degrees of freedom, the p-value is slightly higher than the normal approximation because the t-distribution has fatter tails.Wait, actually, the t-distribution with higher degrees of freedom is closer to the normal distribution, so the p-value would be slightly higher than the normal approximation.Wait, but in our case, the t-value is 1.826, which is less than the critical value of 2.045 for 0.05, so the p-value is greater than 0.05.Wait, but earlier, I thought it was approximately 0.0817, which is about 8.17%.Alternatively, perhaps I can use the following formula for the p-value:p-value ≈ 2 * (1 - Φ(|t| * sqrt((df)/(df + t²))))But I'm not sure if that's correct. Alternatively, perhaps I can use the approximation:For a t-distribution with ν degrees of freedom, the p-value can be approximated using the normal distribution with a continuity correction, but I'm not sure.Alternatively, perhaps I can use the following formula:p-value ≈ 2 * [1 - Φ(t)] for t > 0.But since we're dealing with a two-tailed test, it's 2 times the one-tailed p-value.Wait, but without a calculator, it's hard to get an exact value. So, perhaps I can just state that the p-value is approximately between 0.05 and 0.10, and since it's greater than 0.05, we fail to reject the null hypothesis.Alternatively, using the critical value approach, since the calculated t-statistic of -1.826 is greater than -2.045, it does not fall into the rejection region. Therefore, we fail to reject the null hypothesis.So, in conclusion, at the 5% significance level, there is not enough evidence to reject the null hypothesis that the mean score in Cultural Fit (F) is 4.Wait, but let me double-check the calculations.Sample mean = 3.7Hypothesized mean = 4Sample standard deviation = 0.9n = 30t = (3.7 - 4) / (0.9 / sqrt(30)) = (-0.3) / (0.9 / 5.4772) ≈ (-0.3) / 0.1643 ≈ -1.826.Yes, that's correct.Degrees of freedom = 30 - 1 = 29.Critical value for two-tailed test at 5% is ±2.045.Since -1.826 is greater than -2.045, we fail to reject H0.Therefore, the conclusion is that there is not enough evidence to reject the null hypothesis that the mean score in Cultural Fit is 4.So, summarizing both problems:1. The probability that T + C exceeds 16 is approximately 18.3%.2. The hypothesis test fails to reject the null hypothesis at the 5% significance level, so there's not enough evidence to conclude that the mean Cultural Fit score differs from 4.I think that's it. I hope I didn't make any mistakes in the calculations.</think>

🧠:<think>Alright, so I have this problem about a graduate student who took out a loan and is trying to figure out how much they'll owe after a grace period and then what their monthly payments will be. Let me try to break this down step by step.First, the student took out a 100,000 loan with an annual interest rate of 6%, compounded monthly. They're in a 2-year Master's program, and after that, there's a 1-year grace period where they don't have to make any payments, but interest still accrues. Then, they plan to repay the loan over 10 years. So, Sub-problem 1 is to calculate the total amount owed at the end of the 1-year grace period. Hmm, okay. So, I need to find out how much the loan will grow to after 1 year with monthly compounding.I remember the formula for compound interest is A = P(1 + r/n)^(nt), where:- A is the amount of money accumulated after n years, including interest.- P is the principal amount (100,000 in this case).- r is the annual interest rate (decimal form, so 6% would be 0.06).- n is the number of times that interest is compounded per year (monthly, so 12).- t is the time the money is invested for in years (1 year here).So plugging in the numbers:A = 100,000 * (1 + 0.06/12)^(12*1)Let me compute that. First, 0.06 divided by 12 is 0.005. Then, 1 + 0.005 is 1.005. Now, 12*1 is 12, so we're raising 1.005 to the 12th power.I can calculate 1.005^12. I remember that (1 + r)^n can be calculated using logarithms or just multiplying step by step. Maybe I can approximate it or use the rule of 72? Wait, no, the rule of 72 is for doubling time. Maybe I should just compute it step by step.Alternatively, I know that (1 + 0.005)^12 is approximately e^(0.06) because for small r, (1 + r/n)^(nt) ≈ e^(rt). But e^0.06 is approximately 1.0618365. But wait, is that accurate enough? Maybe I should calculate it more precisely.Alternatively, I can use the formula for monthly compounding. Let me compute 1.005^12.1.005^1 = 1.0051.005^2 = 1.005 * 1.005 = 1.0100251.005^3 = 1.010025 * 1.005 ≈ 1.0150751251.005^4 ≈ 1.015075125 * 1.005 ≈ 1.020201561.005^5 ≈ 1.02020156 * 1.005 ≈ 1.025250751.005^6 ≈ 1.02525075 * 1.005 ≈ 1.030377881.005^7 ≈ 1.03037788 * 1.005 ≈ 1.035555021.005^8 ≈ 1.03555502 * 1.005 ≈ 1.040772551.005^9 ≈ 1.04077255 * 1.005 ≈ 1.046035731.005^10 ≈ 1.04603573 * 1.005 ≈ 1.051357411.005^11 ≈ 1.05135741 * 1.005 ≈ 1.056727201.005^12 ≈ 1.05672720 * 1.005 ≈ 1.06214204So, approximately 1.06214204. Therefore, A ≈ 100,000 * 1.06214204 ≈ 106,214.20.Wait, let me check that multiplication: 100,000 * 1.06214204 is indeed 106,214.20.Alternatively, using a calculator, 1.005^12 is approximately 1.0616778, so 100,000 * 1.0616778 ≈ 106,167.78.Hmm, so my manual calculation was a bit off, but close. Maybe I made a rounding error somewhere. Let me check with a calculator function.Alternatively, I can use the formula for compound interest:A = P(1 + r/n)^(nt)So, plugging in the numbers:A = 100,000 * (1 + 0.06/12)^(12*1)A = 100,000 * (1.005)^12A ≈ 100,000 * 1.0616778A ≈ 106,167.78So, approximately 106,167.78.Wait, so my initial manual calculation was 1.06214204, which is about 106,214.20, but the calculator gives 1.0616778, which is 106,167.78. So, the exact amount is approximately 106,167.78.So, the total amount owed at the end of the 1-year grace period is approximately 106,167.78.Okay, so that's Sub-problem 1.Now, moving on to Sub-problem 2. The student gets a job immediately after the grace period with a starting salary of 60,000 per year, paid monthly. They receive an annual raise of 3%. We need to determine the monthly payment required to fully repay the loan over 10 years, and also calculate what percentage of the student's total income over the 10-year period will go towards repaying the loan.Alright, so first, we need to calculate the monthly payment. For that, we can use the loan payment formula, which is:M = P * [i(1 + i)^n] / [(1 + i)^n - 1]Where:- M is the monthly payment- P is the principal loan amount (which, after the grace period, is 106,167.78)- i is the monthly interest rate (annual rate divided by 12)- n is the number of payments (10 years * 12 months = 120 payments)So, let's compute each part.First, the principal P is 106,167.78.The annual interest rate is 6%, so the monthly interest rate i is 0.06 / 12 = 0.005.The number of payments n is 10 * 12 = 120.Plugging into the formula:M = 106,167.78 * [0.005*(1 + 0.005)^120] / [(1 + 0.005)^120 - 1]First, let's compute (1 + 0.005)^120. That's the same as 1.005^120.I can compute that using logarithms or recognize that 1.005^120 is approximately e^(0.005*120) = e^0.6 ≈ 1.8221188. But actually, 1.005^120 is a bit more precise.Alternatively, I can use the formula for compound interest again. Let me compute it step by step.But perhaps it's faster to use the approximation or a calculator.Alternatively, I can note that (1 + 0.005)^120 is the same as (1.005)^120.I know that (1.005)^12 ≈ 1.0616778, as we computed earlier. So, (1.005)^120 is (1.0616778)^10.Compute (1.0616778)^10.Let me compute that:First, compute ln(1.0616778) ≈ 0.0597.Then, 10 * 0.0597 ≈ 0.597.Then, e^0.597 ≈ 1.815.Alternatively, compute it step by step:1.0616778^2 ≈ 1.0616778 * 1.0616778 ≈ 1.1274861.0616778^4 ≈ (1.127486)^2 ≈ 1.2712491.0616778^8 ≈ (1.271249)^2 ≈ 1.6161531.0616778^10 ≈ 1.616153 * 1.127486 ≈ 1.8167So, approximately 1.8167.Therefore, (1.005)^120 ≈ 1.8167.So, going back to the formula:M = 106,167.78 * [0.005 * 1.8167] / [1.8167 - 1]Compute numerator: 0.005 * 1.8167 ≈ 0.0090835Denominator: 1.8167 - 1 = 0.8167So, M ≈ 106,167.78 * (0.0090835 / 0.8167)Compute 0.0090835 / 0.8167 ≈ 0.01112So, M ≈ 106,167.78 * 0.01112 ≈ ?Compute 106,167.78 * 0.01 = 1,061.6778106,167.78 * 0.00112 ≈ 106,167.78 * 0.001 = 106.16778; 106,167.78 * 0.00012 ≈ 12.74013So, total ≈ 106.16778 + 12.74013 ≈ 118.9079So, total M ≈ 1,061.6778 + 118.9079 ≈ 1,180.5857So, approximately 1,180.59 per month.Wait, but let me check this with a more precise calculation.Alternatively, using the formula:M = P * [i(1 + i)^n] / [(1 + i)^n - 1]We have:i = 0.005n = 120(1 + i)^n = 1.005^120 ≈ 1.8166967So, numerator: 0.005 * 1.8166967 ≈ 0.00908348Denominator: 1.8166967 - 1 = 0.8166967So, M = 106,167.78 * (0.00908348 / 0.8166967)Compute 0.00908348 / 0.8166967 ≈ 0.01112So, M ≈ 106,167.78 * 0.01112 ≈ 1,180.59So, approximately 1,180.59 per month.Alternatively, using a financial calculator or Excel's PMT function, which is PMT(rate, nper, pv, [fv], [type]).In Excel, it would be PMT(0.005, 120, -106167.78) ≈ ?Let me compute that:PMT(0.005, 120, -106167.78) ≈ ?Using the formula:M = -PMT(0.005, 120, 106167.78) ≈ ?Wait, actually, in Excel, the formula is PMT(rate, nper, pv, [fv], [type]). So, if we input PMT(0.005, 120, -106167.78), it should give us the monthly payment.But since I don't have Excel here, I can use the formula again.Alternatively, I can use the approximation we did earlier, which is about 1,180.59.But let me check with another method.We can also use the present value of an annuity formula.The present value P is equal to M * [1 - (1 + i)^-n] / iSo, rearranged, M = P * i / [1 - (1 + i)^-n]So, M = 106,167.78 * 0.005 / [1 - (1.005)^-120]Compute (1.005)^-120 = 1 / (1.005)^120 ≈ 1 / 1.8166967 ≈ 0.5503So, 1 - 0.5503 ≈ 0.4497So, M ≈ 106,167.78 * 0.005 / 0.4497 ≈ 530.8389 / 0.4497 ≈ 1,180.59Yes, same result. So, the monthly payment is approximately 1,180.59.Okay, so that's the monthly payment required.Now, the second part is to calculate what percentage of the student's total income over the 10-year repayment period will go towards repaying the loan.So, first, we need to calculate the student's total income over 10 years, considering the annual raise of 3%.The student starts with a salary of 60,000 per year, paid monthly. So, the monthly salary is 60,000 / 12 = 5,000 per month.Each year, the salary increases by 3%. So, the salary in year 1 is 60,000, year 2 is 60,000 * 1.03, year 3 is 60,000 * (1.03)^2, and so on, up to year 10.We need to compute the total income over 10 years, which is the sum of the annual salaries from year 1 to year 10.Alternatively, since the salary increases each year, it's a geometric series.The total income T can be calculated as:T = S * [ (1 - (1 + g)^n ) / (1 - (1 + g)) ] where S is the starting salary, g is the growth rate, and n is the number of years.Wait, actually, the formula for the sum of a geometric series is S * [ (1 - (1 + g)^n ) / (1 - (1 + g)) ] but that might not be correct. Let me think.Actually, the sum of a geometric series is a * (r^n - 1) / (r - 1), where a is the first term, r is the common ratio.In this case, the first term a is 60,000, the common ratio r is 1.03, and the number of terms n is 10.So, T = 60,000 * [ (1.03^10 - 1) / (1.03 - 1) ]Compute 1.03^10 first.1.03^10 ≈ 1.343916379So, 1.343916379 - 1 = 0.343916379Divide by (1.03 - 1) = 0.03So, 0.343916379 / 0.03 ≈ 11.4638793Therefore, T ≈ 60,000 * 11.4638793 ≈ 60,000 * 11.4638793 ≈ 687,832.76So, total income over 10 years is approximately 687,832.76.Alternatively, let me compute it step by step to verify.Year 1: 60,000Year 2: 60,000 * 1.03 = 61,800Year 3: 61,800 * 1.03 = 63,654Year 4: 63,654 * 1.03 ≈ 65,557.62Year 5: 65,557.62 * 1.03 ≈ 67,524.35Year 6: 67,524.35 * 1.03 ≈ 69,549.88Year 7: 69,549.88 * 1.03 ≈ 71,636.48Year 8: 71,636.48 * 1.03 ≈ 73,785.57Year 9: 73,785.57 * 1.03 ≈ 75,998.13Year 10: 75,998.13 * 1.03 ≈ 78,278.07Now, summing these up:Year 1: 60,000Year 2: 61,800 → Total: 121,800Year 3: 63,654 → Total: 185,454Year 4: 65,557.62 → Total: 251,011.62Year 5: 67,524.35 → Total: 318,535.97Year 6: 69,549.88 → Total: 388,085.85Year 7: 71,636.48 → Total: 459,722.33Year 8: 73,785.57 → Total: 533,507.90Year 9: 75,998.13 → Total: 609,506.03Year 10: 78,278.07 → Total: 687,784.10So, approximately 687,784.10, which is very close to our earlier calculation of 687,832.76. The slight difference is due to rounding errors in each year's salary.So, total income ≈ 687,784.10.Now, the total amount paid towards the loan is the monthly payment multiplied by the number of months.Monthly payment is approximately 1,180.59, over 120 months.Total loan repayment: 1,180.59 * 120 ≈ ?Compute 1,180.59 * 100 = 118,0591,180.59 * 20 = 23,611.80Total: 118,059 + 23,611.80 ≈ 141,670.80So, total repayment is approximately 141,670.80.Now, to find the percentage of total income that goes towards the loan:Percentage = (Total repayment / Total income) * 100So, Percentage ≈ (141,670.80 / 687,784.10) * 100 ≈ ?Compute 141,670.80 / 687,784.10 ≈ 0.2059Multiply by 100: ≈ 20.59%So, approximately 20.59% of the student's total income over the 10-year period will go towards repaying the loan.Wait, let me double-check the total repayment. The monthly payment is 1,180.59, over 120 months, so 1,180.59 * 120.Let me compute 1,180.59 * 120:1,180.59 * 100 = 118,0591,180.59 * 20 = 23,611.80Total: 118,059 + 23,611.80 = 141,670.80Yes, that's correct.And total income is approximately 687,784.10.So, 141,670.80 / 687,784.10 ≈ 0.2059 or 20.59%.So, approximately 20.59%.Alternatively, if we use more precise numbers, maybe it's slightly different, but it's roughly 20.6%.So, summarizing:Sub-problem 1: The total amount owed after the grace period is approximately 106,167.78.Sub-problem 2: The monthly payment is approximately 1,180.59, and the percentage of total income going towards the loan is approximately 20.59%.I think that's it. Let me just make sure I didn't make any calculation errors.Wait, for Sub-problem 1, I used the compound interest formula correctly, right? Yes, P = 100,000, r = 0.06, n = 12, t = 1. So, A = 100,000*(1 + 0.06/12)^(12*1) ≈ 106,167.78. That seems correct.For Sub-problem 2, the monthly payment calculation using the loan formula gave us approximately 1,180.59, and the total repayment over 10 years is about 141,670.80. The total income over 10 years is approximately 687,784.10, so the percentage is roughly 20.59%. That seems reasonable.I think that's all.</think>

🧠:<think>Okay, so I need to solve these two sub-problems for the Mongolian filmmaker. Let me start with Sub-problem 1.First, the filmmaker is planning to shoot in two locations: the Gobi Desert and the Altai Mountains. The distances are 700 km and 1200 km respectively. The travel cost function is given as C(d) = 50 + 0.5d, where d is the distance in kilometers. The filmmaker wants to minimize travel costs and maximize shooting time under optimal lighting.Wait, but the problem says the filmmaker is considering travel times, costs, and lighting. So I need to calculate the total travel cost and total time spent in optimal lighting.Let me break it down.First, the travel cost. The filmmaker needs to go from the starting point to Gobi Desert, then to Altai Mountains, and then back? Or is it a round trip for each location? Hmm, the problem doesn't specify the exact travel route, just the distances from the starting point. So maybe it's a round trip for each location? Or perhaps a one-way trip? Hmm, the problem says "travel costs" for shooting in each location. So maybe the filmmaker goes to each location, spends 3 days shooting, and returns? Or is it a round trip for each location?Wait, the problem says the filmmaker plans to spend 3 days shooting in each location. So perhaps the filmmaker goes to Gobi, spends 3 days, then goes to Altai, spends 3 days, and then returns? Or is it a round trip for each location? Hmm, the problem isn't entirely clear, but let me assume that the filmmaker is making a round trip for each location. So, going to Gobi and back, then going to Altai and back. So total distance would be 2*700 + 2*1200.Wait, but that might not be the case. Alternatively, the filmmaker could go from starting point to Gobi, then from Gobi to Altai, then from Altai back to starting point. That would be a more efficient route, but the problem doesn't specify the exact path, so maybe it's safer to assume that each location is a separate round trip.But actually, in real terms, if you have two locations, it's more efficient to go to one, then to the other, and back. So the total distance would be 700 (to Gobi) + (1200 - 700) (from Gobi to Altai) + 1200 (back from Altai to starting point). Wait, is Gobi and Altai in the same direction from the starting point? Probably not, since Gobi is in the south and Altai is in the west or north. So maybe the distance from Gobi to Altai isn't simply 1200 - 700. Hmm, this is getting complicated.Wait, the problem might be simplifying things, so perhaps it's just a round trip to each location. So total distance is 2*700 + 2*1200. Let me check the problem statement again.It says: "the Gobi Desert is 700 km away from the starting point, and the Altai Mountains are 1200 km away." So if the filmmaker is starting from point A, goes to Gobi (700 km), then goes to Altai (1200 km from A), but the distance from Gobi to Altai isn't given. So unless we assume that Gobi and Altai are in opposite directions from A, the total distance would be 700 + 1200 + 1200 + 700? Wait, that doesn't make sense.Wait, perhaps the filmmaker goes from A to Gobi (700 km), spends 3 days, then goes from Gobi to Altai (unknown distance), spends 3 days, then returns from Altai to A (1200 km). But since the distance from Gobi to Altai isn't given, maybe we can't calculate that. So perhaps the problem is assuming that each location is a separate round trip, meaning the filmmaker goes to Gobi, comes back, then goes to Altai, comes back. So total distance would be 2*700 + 2*1200.Let me proceed with that assumption.So total distance is 2*700 + 2*1200 = 1400 + 2400 = 3800 km.Then, the total travel cost is C(d) = 50 + 0.5d. So for 3800 km, it's 50 + 0.5*3800.Calculating that: 0.5*3800 = 1900, plus 50 is 1950 USD.Wait, but is that correct? Because the cost function is per trip? Or per kilometer? Wait, the function is given as C(d) = 50 + 0.5d, where d is the distance traveled in kilometers. So it's a linear function, so total cost is 50 + 0.5*d, where d is total distance.So if total distance is 3800 km, then total cost is 50 + 0.5*3800 = 50 + 1900 = 1950 USD.Okay, that seems straightforward.Now, the total time spent in optimal lighting conditions. The Gobi has 6 hours per day, and Altai has 4 hours per day. The filmmaker spends 3 days in each location.So in Gobi: 3 days * 6 hours/day = 18 hours.In Altai: 3 days * 4 hours/day = 12 hours.Total optimal lighting time: 18 + 12 = 30 hours.So that's Sub-problem 1 done.Now, Sub-problem 2: The filmmaker wants to use the Golden Ratio for composition. The Golden Ratio is approximately 1.618, so phi ≈ 1.618.Given that the height of a scene's frame is h, the width should be w = phi * h.The maximum height available is 8 meters. So, maximum width w = 1.618 * 8.Calculating that: 1.618 * 8 = let's see, 1.6 * 8 = 12.8, 0.018*8=0.144, so total is 12.8 + 0.144 = 12.944 meters.So maximum width is approximately 12.944 meters.Then, the filmmaker wants to stitch together 5 such frames side by side for a panoramic sequence. So total horizontal length would be 5 * width.So 5 * 12.944 = let's calculate that.12.944 * 5: 10*5=50, 2.944*5=14.72, so total is 50 + 14.72 = 64.72 meters.So total horizontal length is 64.72 meters.Wait, but let me double-check the calculations.First, w = phi * h = 1.618 * 8.1.618 * 8:1 * 8 = 80.618 * 8 = 4.944So total is 8 + 4.944 = 12.944 meters. Correct.Then, 5 frames side by side: 5 * 12.944.12.944 * 5:12 * 5 = 600.944 * 5 = 4.72Total: 60 + 4.72 = 64.72 meters. Correct.So, summarizing:Sub-problem 1:Total travel cost: 1950 USDTotal optimal lighting time: 30 hoursSub-problem 2:Maximum width per frame: 12.944 metersTotal panoramic length: 64.72 metersI think that's all. Let me just make sure I didn't miss anything.Wait, in Sub-problem 1, is the travel cost per trip or per kilometer? The function is C(d) = 50 + 0.5d, so per kilometer, with a fixed cost of 50. So total distance is 3800 km, so total cost is 50 + 0.5*3800 = 1950. That seems right.And the optimal lighting time: 3 days at each location, 6 and 4 hours per day respectively. So 18 + 12 = 30 hours. Correct.Sub-problem 2: Golden Ratio, width = phi * height, height is 8, so width is 12.944. Then 5 frames side by side: 5*12.944=64.72. Correct.I think that's all.Final AnswerSub-problem 1: The total travel cost is boxed{1950} USD and the total time spent in optimal lighting conditions is boxed{30} hours.Sub-problem 2: The maximum width of the frame is boxed{12.944} meters and the total horizontal length of the panoramic sequence is boxed{64.72} meters.</think>

🧠:<think>Okay, so I have two sub-problems to solve here. Let me tackle them one by one.Starting with Sub-problem 1: I need to find the exact month(s) x where the effectiveness of the exhibits in Museum A and Museum B are the same. The effectiveness functions are given as E_A(x) = 3x² - 4x + 5 and E_B(x) = -2x³ + 5x² + 2x + 1. So, I need to set these two equations equal to each other and solve for x.Let me write that out:3x² - 4x + 5 = -2x³ + 5x² + 2x + 1Hmm, okay, so I need to bring all terms to one side to set the equation to zero. Let me subtract the right side from both sides:3x² - 4x + 5 - (-2x³ + 5x² + 2x + 1) = 0Simplify that:3x² - 4x + 5 + 2x³ - 5x² - 2x - 1 = 0Now, combine like terms:2x³ + (3x² - 5x²) + (-4x - 2x) + (5 - 1) = 0Calculating each term:2x³ - 2x² - 6x + 4 = 0So, the equation simplifies to:2x³ - 2x² - 6x + 4 = 0Hmm, this is a cubic equation. Solving cubic equations can be tricky, but maybe I can factor this. Let me try to factor out a common term first. I notice each term is even, so I can factor out a 2:2(x³ - x² - 3x + 2) = 0So, dividing both sides by 2:x³ - x² - 3x + 2 = 0Now, let's see if I can factor this cubic. I'll try rational root theorem. The possible rational roots are factors of the constant term over factors of the leading coefficient. So, possible roots are ±1, ±2.Let me test x=1:1 - 1 - 3 + 2 = (1 -1) + (-3 + 2) = 0 -1 = -1 ≠ 0Not a root.x= -1:-1 - 1 + 3 + 2 = (-1 -1) + (3 + 2) = -2 + 5 = 3 ≠ 0Not a root.x=2:8 - 4 - 6 + 2 = (8 -4) + (-6 + 2) = 4 -4 = 0Yes! x=2 is a root.So, (x - 2) is a factor. Let's perform polynomial division or use synthetic division to factor it out.Using synthetic division with x=2:Coefficients: 1 | -1 | -3 | 2Bring down the 1.Multiply 1 by 2: 2. Add to next coefficient: -1 + 2 = 1.Multiply 1 by 2: 2. Add to next coefficient: -3 + 2 = -1.Multiply -1 by 2: -2. Add to last coefficient: 2 + (-2) = 0. Perfect.So, the cubic factors as (x - 2)(x² + x - 1) = 0So, now set each factor equal to zero:x - 2 = 0 => x = 2x² + x - 1 = 0Using quadratic formula for the second factor:x = [-b ± √(b² - 4ac)] / 2aHere, a=1, b=1, c=-1So,x = [-1 ± √(1 + 4)] / 2 = [-1 ± √5]/2So, the roots are x = (-1 + √5)/2 and x = (-1 - √5)/2But since x represents the number of months, it must be a positive real number. So, let's evaluate:(-1 + √5)/2 ≈ (-1 + 2.236)/2 ≈ 1.236/2 ≈ 0.618, which is positive.(-1 - √5)/2 ≈ (-1 - 2.236)/2 ≈ -3.236/2 ≈ -1.618, which is negative. So, we can disregard this root.Therefore, the solutions are x = 2 and x ≈ 0.618 months.Wait, but x is the number of months since the exhibit opened, so it can be a fractional number of months, but in the context of the problem, are we to consider only integer months? The problem says "exact month(s)", so perhaps fractional months are acceptable as exact solutions.So, the exact solutions are x = 2 and x = (-1 + √5)/2.Let me write that as x = (sqrt(5) - 1)/2 and x = 2.So, that's Sub-problem 1 done.Now, moving on to Sub-problem 2: Comparing the total number of visitors each museum attracts by the end of the first 12 months.Museum A's visitor trend is V_A(t) = 100e^{0.05t}, and Museum B's is V_B(t) = 150 ln(t + 1) + 50.We need to calculate the total number of visitors each museum will have attracted by the end of 12 months. So, I think this means we need to compute the integral of the visitor function from t=0 to t=12, because the total visitors would be the area under the curve over that period.So, for Museum A, total visitors T_A = ∫₀¹² 100e^{0.05t} dtFor Museum B, total visitors T_B = ∫₀¹² [150 ln(t + 1) + 50] dtLet me compute each integral separately.Starting with Museum A:T_A = ∫₀¹² 100e^{0.05t} dtThe integral of e^{kt} dt is (1/k)e^{kt} + C, so:T_A = 100 * [ (1/0.05) e^{0.05t} ] from 0 to 12Simplify:100 / 0.05 = 2000So,T_A = 2000 [ e^{0.05*12} - e^{0} ] = 2000 [ e^{0.6} - 1 ]That's the exact expression. Let me compute e^{0.6}:But since the problem says to provide exact analytical solutions without approximations, I can leave it as e^{0.6}.So, T_A = 2000(e^{0.6} - 1)Now, Museum B:T_B = ∫₀¹² [150 ln(t + 1) + 50] dtWe can split this into two integrals:T_B = 150 ∫₀¹² ln(t + 1) dt + 50 ∫₀¹² dtCompute each integral separately.First, ∫ ln(t + 1) dt. Let me use substitution.Let u = t + 1, so du = dt. When t=0, u=1; when t=12, u=13.So, ∫ ln(u) du from 1 to 13.The integral of ln(u) du is u ln(u) - u + C.So,∫₀¹² ln(t + 1) dt = [ (t + 1) ln(t + 1) - (t + 1) ] from 0 to 12Evaluate at 12:(13 ln13 - 13)Evaluate at 0:(1 ln1 - 1) = (0 - 1) = -1So, the definite integral is (13 ln13 - 13) - (-1) = 13 ln13 - 13 + 1 = 13 ln13 - 12So, the first part is 150*(13 ln13 - 12)Now, the second integral: ∫₀¹² 50 dt = 50*(12 - 0) = 600So, T_B = 150*(13 ln13 - 12) + 600Let me compute that:150*13 ln13 = 1950 ln13150*(-12) = -1800So, T_B = 1950 ln13 - 1800 + 600 = 1950 ln13 - 1200So, T_B = 1950 ln13 - 1200Therefore, the total visitors for Museum A is 2000(e^{0.6} - 1) and for Museum B is 1950 ln13 - 1200.To compare them, we can leave them in exact form, but perhaps we can compute their approximate numerical values to see which is larger, but since the problem says to provide exact solutions, I think we just need to present these expressions.But let me just make sure I didn't make a mistake in the integrals.For Museum A: Integral of 100e^{0.05t} from 0 to 12.Yes, that's 100*(1/0.05)(e^{0.6} - 1) = 2000(e^{0.6} - 1). Correct.For Museum B: Integral of 150 ln(t + 1) + 50.Yes, split into two integrals. The integral of ln(t + 1) is (t + 1) ln(t + 1) - (t + 1). Evaluated from 0 to 12: (13 ln13 -13) - (1 ln1 -1) = 13 ln13 -13 - (-1) = 13 ln13 -12. Multiply by 150: 1950 ln13 - 1800. Then add the integral of 50, which is 600. So, 1950 ln13 -1200. Correct.So, I think that's correct.Therefore, the exact total visitors for Museum A is 2000(e^{0.6} - 1) and for Museum B is 1950 ln13 - 1200.I can also compute these to compare, but since the problem says to provide exact solutions, I think that's sufficient.Wait, but maybe the problem expects the total number of visitors, which might be the sum of visitors each month? Or is it the integral? Hmm, the functions are given as V_A(t) and V_B(t), which are visitor trends. So, if t is the number of months since the start of the year, then V_A(t) is the number of visitors at month t. So, to get the total visitors over 12 months, we need to sum V_A(t) from t=0 to t=11, or integrate over t from 0 to 12? Hmm, the problem says "attracts visitors according to the function V_A(t)", so it's a continuous model, so integrating over 12 months makes sense.Yes, so the total visitors would be the integral from 0 to 12 of V_A(t) dt and similarly for V_B(t). So, my approach is correct.Therefore, the exact total visitors are as I found.Final AnswerSub-problem 1: The effectiveness is the same at months ( boxed{2} ) and ( boxed{dfrac{sqrt{5} - 1}{2}} ).Sub-problem 2: Museum A attracts ( boxed{2000(e^{0.6} - 1)} ) visitors, and Museum B attracts ( boxed{1950 ln 13 - 1200} ) visitors by the end of the first 12 months.</think>