🧠:<think>Okay, so I have this problem where I need to help the principal allocate resources to three STEM programs: Robotics, Computer Science, and Environmental Science. The goal is to maximize the benefit function given by ( f(x, y, z) = 3x^2 + 2y^2 + 4z^2 - 2xy - yz + 3xz ), with the constraint that the total resources ( x + y + z = 10 ). Hmm, sounds like an optimization problem with a constraint. I think I need to use Lagrange multipliers for this.First, let me recall how Lagrange multipliers work. If I have a function to maximize, ( f(x, y, z) ), subject to a constraint ( g(x, y, z) = 0 ), then I can set up the equations ( nabla f = lambda nabla g ), where ( lambda ) is the Lagrange multiplier. So, in this case, my constraint is ( x + y + z - 10 = 0 ), so ( g(x, y, z) = x + y + z - 10 ).Let me compute the gradients. The gradient of ( f ) is:( nabla f = left( frac{partial f}{partial x}, frac{partial f}{partial y}, frac{partial f}{partial z} right) )Calculating each partial derivative:- ( frac{partial f}{partial x} = 6x - 2y + 3z )- ( frac{partial f}{partial y} = 4y - 2x - z )- ( frac{partial f}{partial z} = 8z - y + 3x )And the gradient of ( g ) is:( nabla g = (1, 1, 1) )So, according to the method, I need to set up the system of equations:1. ( 6x - 2y + 3z = lambda )2. ( 4y - 2x - z = lambda )3. ( 8z - y + 3x = lambda )4. ( x + y + z = 10 )So, now I have four equations with four variables: x, y, z, and λ. I need to solve this system.Let me write equations 1, 2, and 3 without λ:1. ( 6x - 2y + 3z = lambda )2. ( -2x + 4y - z = lambda )3. ( 3x - y + 8z = lambda )So, since all of them equal λ, I can set them equal to each other.First, set equation 1 equal to equation 2:( 6x - 2y + 3z = -2x + 4y - z )Bring all terms to the left:( 6x + 2x - 2y - 4y + 3z + z = 0 )Simplify:( 8x - 6y + 4z = 0 )Divide by 2:( 4x - 3y + 2z = 0 ) --- Let's call this equation A.Now, set equation 2 equal to equation 3:( -2x + 4y - z = 3x - y + 8z )Bring all terms to the left:( -2x - 3x + 4y + y - z - 8z = 0 )Simplify:( -5x + 5y - 9z = 0 )Divide by... Hmm, maybe leave it as is for now. Let's call this equation B: ( -5x + 5y - 9z = 0 )Now, set equation 1 equal to equation 3:Wait, actually, I already set 1=2 and 2=3, so maybe I can use equations A and B along with the constraint equation.So, equations:A: ( 4x - 3y + 2z = 0 )B: ( -5x + 5y - 9z = 0 )And the constraint: ( x + y + z = 10 )So, now we have three equations:1. ( 4x - 3y + 2z = 0 )2. ( -5x + 5y - 9z = 0 )3. ( x + y + z = 10 )Let me try to solve this system.First, maybe express one variable in terms of others from equation A or B.Looking at equation A: ( 4x - 3y + 2z = 0 )Let me solve for z:( 2z = -4x + 3y )( z = -2x + (3/2)y ) --- Equation A1Now, plug this into equation B:( -5x + 5y - 9z = 0 )Substitute z:( -5x + 5y - 9(-2x + (3/2)y) = 0 )Compute:( -5x + 5y + 18x - (27/2)y = 0 )Combine like terms:(-5x + 18x) + (5y - 27/2 y) = 013x + (10/2 y - 27/2 y) = 013x - (17/2)y = 0Multiply both sides by 2 to eliminate fraction:26x - 17y = 0 --- Equation C: 26x = 17y => y = (26/17)xNow, from equation A1, z = -2x + (3/2)ySubstitute y:z = -2x + (3/2)(26/17 x) = -2x + (78/34)x = -2x + (39/17)xConvert -2x to -34/17 x:z = (-34/17 + 39/17)x = (5/17)xSo, now we have y and z in terms of x:y = (26/17)xz = (5/17)xNow, plug these into the constraint equation: x + y + z = 10Substitute y and z:x + (26/17)x + (5/17)x = 10Combine terms:x + (26x + 5x)/17 = 10x + (31x)/17 = 10Convert x to 17/17 x:(17x + 31x)/17 = 1048x /17 = 10Multiply both sides by 17:48x = 170x = 170 /48Simplify:Divide numerator and denominator by 2:85 /24 ≈ 3.5417So, x = 85/24Now, compute y:y = (26/17)x = (26/17)*(85/24)Simplify:26 and 24 have a common factor of 2: 26/24 = 13/1285 and 17: 85 ÷17=5So, (13/12)*(5) = 65/12 ≈5.4167Similarly, z = (5/17)x = (5/17)*(85/24)Simplify:85 ÷17=5So, (5)*(5)/24 =25/24≈1.0417Let me check if x + y + z =10:x=85/24≈3.5417y=65/12≈5.4167z=25/24≈1.0417Adding them up:3.5417 +5.4167≈8.9584 +1.0417≈10. So, correct.So, the resource allocation is:x=85/24 ≈3.5417y=65/12≈5.4167z=25/24≈1.0417But since the resources are in hundreds of units, and total is 10, which is 1000 units. So, x≈354.17 units, y≈541.67 units, z≈104.17 units.Wait, but let me confirm if these are correct by plugging back into the original equations.First, compute gradients:Compute ∇f:At (85/24, 65/12,25/24):Compute each partial derivative:df/dx =6x -2y +3z6*(85/24)= (6*85)/24=510/24=21.25-2y= -2*(65/12)= -130/12≈-10.83333z=3*(25/24)=75/24≈3.125Total:21.25 -10.8333 +3.125≈13.5417Similarly, df/dy=4y -2x -z4*(65/12)=260/12≈21.6667-2x= -2*(85/24)= -170/24≈-7.0833-z= -25/24≈-1.0417Total≈21.6667 -7.0833 -1.0417≈13.5417df/dz=8z - y +3x8*(25/24)=200/24≈8.3333-y= -65/12≈-5.41673x=3*(85/24)=255/24≈10.625Total≈8.3333 -5.4167 +10.625≈13.5417So, all partial derivatives equal≈13.5417, which should be equal to λ.And gradient of g is (1,1,1), so λ=13.5417.So, that seems consistent.Therefore, the allocation is x=85/24, y=65/12, z=25/24.But let me write them as fractions:x=85/24, y=65/12=130/24, z=25/24So, in terms of hundreds of units, x≈3.5417, y≈5.4167, z≈1.0417.So, that's part 1 done.Now, moving on to part 2: The principal wants to ensure balanced student engagement over the academic year. The engagement levels are represented by vector e, subject to transformation matrix A. The initial engagement levels are e0 = [3;2;1], and we need to find e such that A*e = e0.So, given matrix A:A = [1 1 -1;-1 2 1;2 1 1]And e0 = [3;2;1]We need to solve A*e = e0 for e.So, set up the system:1. e1 + e2 - e3 = 32. -e1 + 2e2 + e3 = 23. 2e1 + e2 + e3 = 1So, three equations:1. e1 + e2 - e3 = 32. -e1 + 2e2 + e3 = 23. 2e1 + e2 + e3 = 1Let me write this in augmented matrix form:[1 1 -1 | 3-1 2 1 | 22 1 1 |1]Let me perform row operations to solve this.First, let's label the equations:Equation 1: e1 + e2 - e3 = 3Equation 2: -e1 + 2e2 + e3 = 2Equation 3: 2e1 + e2 + e3 = 1Let me try to eliminate e1 from equations 2 and 3.First, add equation 1 and equation 2:(1 + (-1))e1 + (1 + 2)e2 + (-1 +1)e3 = 3 + 2So, 0e1 + 3e2 + 0e3 =5 => 3e2=5 => e2=5/3≈1.6667So, e2=5/3.Now, plug e2=5/3 into equation 1 and equation 3.From equation 1: e1 + (5/3) - e3 =3 => e1 - e3=3 -5/3=4/3So, e1 - e3=4/3 --- Equation 4From equation 3: 2e1 + (5/3) + e3=1 => 2e1 + e3=1 -5/3= -2/3So, 2e1 + e3= -2/3 --- Equation 5Now, we have equations 4 and 5:4. e1 - e3=4/35. 2e1 + e3= -2/3Let me add equations 4 and 5:(e1 + 2e1) + (-e3 + e3)=4/3 -2/33e1=2/3 => e1=2/9≈0.2222Now, from equation 4: e1 - e3=4/3 => e3= e1 -4/3= (2/9) -4/3= (2/9 -12/9)= -10/9≈-1.1111So, e3= -10/9So, the solution is:e1=2/9, e2=5/3, e3= -10/9Let me verify by plugging back into the original equations.Equation 1: e1 + e2 - e3= 2/9 +5/3 - (-10/9)=2/9 +15/9 +10/9=27/9=3 ✔️Equation 2: -e1 +2e2 +e3= -2/9 +10/3 + (-10/9)= (-2/9 -10/9) +10/3= (-12/9) +30/9=18/9=2 ✔️Equation 3:2e1 +e2 +e3=4/9 +5/3 + (-10/9)= (4/9 -10/9) +15/9= (-6/9) +15/9=9/9=1 ✔️All equations are satisfied. So, the transformed engagement levels are e1=2/9, e2=5/3, e3= -10/9.But wait, engagement levels being negative? That seems odd. Maybe I made a mistake.Wait, let me check the calculations again.From equation 1: e1 + e2 - e3 =3Equation 2: -e1 +2e2 +e3=2Equation 3:2e1 +e2 +e3=1Adding equation1 and equation2:(e1 -e1) + (e2 +2e2) + (-e3 +e3)=3 +2 => 3e2=5 => e2=5/3. Correct.Then equation1: e1 +5/3 -e3=3 => e1 -e3=4/3. Correct.Equation3:2e1 +5/3 +e3=1 =>2e1 +e3= -2/3. Correct.Then adding equations 4 and5: e1 -e3 +2e1 +e3=4/3 -2/3 =>3e1=2/3 =>e1=2/9. Correct.Then e3= e1 -4/3=2/9 -12/9= -10/9. Correct.So, the solution is correct, even though e3 is negative. Maybe in the context, negative engagement could mean disengagement or something. So, perhaps it's acceptable.So, the transformed engagement levels are e1=2/9, e2=5/3, e3= -10/9.So, summarizing:1. Resource allocation: x=85/24≈3.5417, y=65/12≈5.4167, z=25/24≈1.0417 (in hundreds of units)2. Engagement levels: e1=2/9≈0.2222, e2=5/3≈1.6667, e3= -10/9≈-1.1111I think that's it.Final Answer1. The optimal resource allocation is ( x = boxed{dfrac{85}{24}} ), ( y = boxed{dfrac{65}{12}} ), and ( z = boxed{dfrac{25}{24}} ).2. The transformed engagement levels are ( mathbf{e} = boxed{begin{bmatrix} dfrac{2}{9} dfrac{5}{3} -dfrac{10}{9} end{bmatrix}} ).</think>

🧠:<think>Alright, so I have this problem where a district superintendent is looking at whether a new curriculum has improved math test scores. There are two parts: part a is a hypothesis test, and part b is projecting future scores and probabilities. Let me tackle them one by one.Starting with part a. They want a hypothesis test at a 5% significance level. I need to define the null and alternative hypotheses first. The null hypothesis, H0, is that there's no significant change in the average test score after the curriculum change. The alternative hypothesis, Ha, is that there is a significant increase. So, H0: μ = 75 and Ha: μ > 75. That makes sense because we're testing if the new curriculum caused an improvement.Next, I need to calculate the test statistic. Since we have two large samples (1,000 students each), and the population standard deviation is known (10), I can use a z-test. The formula for the z-test statistic is:z = (x̄2 - x̄1) / (σ * sqrt(1/n1 + 1/n2))Wait, no, actually, since both samples are from the same population before and after, but actually, in this case, it's two independent samples. So, the formula should be:z = (x̄2 - x̄1) / (σ * sqrt(1/n1 + 1/n2))But wait, hold on. The standard deviation is the same for both years, so σ1 = σ2 = 10. And both sample sizes are 1,000. So plugging in the numbers:x̄1 = 75, x̄2 = 78, σ = 10, n1 = n2 = 1000.So, z = (78 - 75) / (10 * sqrt(1/1000 + 1/1000)).Calculating the denominator first: sqrt(2/1000) = sqrt(0.002) ≈ 0.04472.So, denominator is 10 * 0.04472 ≈ 0.4472.Numerator is 3.So, z ≈ 3 / 0.4472 ≈ 6.708.That's a pretty high z-score. The critical z-value for a one-tailed test at 5% significance is 1.645. Since 6.708 is way higher than 1.645, we can reject the null hypothesis. So, the increase is statistically significant.Wait, but hold on, another thought: since the samples are from the same population but different years, is it a paired test? But no, the students are different each year, so it's independent samples. So, the z-test for two independent means is appropriate here.Alternatively, sometimes people use a t-test, but with such large sample sizes, the z-test is fine.So, conclusion: the increase is statistically significant at the 5% level.Moving on to part b. The superintendent wants to project the impact over 5 years with a 2% annual growth in the student population. But wait, the test scores: the improvement was 3 points over one year. So, is the improvement additive each year, or is it multiplicative?The problem says "the same improvement pattern continues." So, I think it means the average increases by 3 points each year. So, after 5 years, the mean would be 75 + 5*3 = 90? Wait, no, wait. Wait, in the first year after the change, it was 78, which is 75 + 3. So, if the improvement continues, each subsequent year adds another 3 points? Or is the improvement 3 points in the first year, and then it's compounded?Wait, the problem says "the same improvement pattern continues." The improvement was 3 points over one year. So, if it's a linear improvement, each year adds 3 points. So, after 5 years, the mean would be 75 + 5*3 = 90. But wait, that seems too high because the standard deviation is 10, so 90 would be 1.5 standard deviations above the original mean. But wait, actually, the mean is increasing each year by 3 points.But wait, hold on, in the first year after the change, the mean was 78, which is 75 + 3. So, if the improvement continues, does that mean each year the mean increases by another 3 points? So, year 1: 75, year 2: 78, year 3: 81, year 4: 84, year 5: 87. So, after 5 years, the mean would be 87? Wait, but the problem says "the same improvement pattern continues." So, if the improvement is 3 points each year, then yes, after 5 years, it's 75 + 5*3 = 90. Wait, but that would be 90, but in the first year, it was 78, so that's 3 points. So, each year, the mean increases by 3 points. So, after 5 years, it's 75 + 3*5 = 90.But wait, hold on, actually, the first year after the change was 78, which is year 2. So, if we're projecting for the next 5 years, starting from year 2, which was 78, then year 3 would be 81, year 4: 84, year 5: 87, year 6: 90, year 7: 93. Wait, but the problem says "after 5 years," so starting from the year after the change, which was year 2, so 5 years later would be year 7, with a mean of 93? Hmm, maybe I'm overcomplicating.Wait, the problem says: "project the potential impact of this curriculum change on future cohorts. Assume the same improvement pattern continues and that the student population grows by 2% annually." So, the improvement pattern is an increase of 3 points per year? Or is it a 3-point increase in the first year, and then perhaps the same rate of improvement?Wait, the problem says "the same improvement pattern continues." So, in the first year after the change, the mean increased by 3 points. So, if the pattern continues, each subsequent year also has a 3-point increase. So, after 5 years, the mean would be 75 + 5*3 = 90. But wait, actually, the first year after the change was 78, so that's 75 + 3. So, if we're projecting 5 years after the change, that would be 75 + 5*3 = 90.But wait, the problem says "the same improvement pattern continues." So, maybe it's not additive, but multiplicative? Like, the improvement rate is 3 points per year, so it's a linear increase. So, yes, 3 points each year.But let me think again. The first year after the change, the mean was 78, which is 3 points higher. So, if the same improvement continues, each year the mean increases by another 3 points. So, after 5 years, the mean would be 75 + 5*3 = 90.But wait, the problem also mentions the student population grows by 2% annually. So, does that affect the mean test score? Or is the mean test score independent of the population size? Because the test scores are per student, so the population growth might affect the total number of students, but the mean is per student. So, unless the population growth affects the distribution of test scores, the mean should remain the same regardless of population size. So, the 2% growth might be a red herring for the mean projection, unless it's implying that the sample size increases, but in part b, it says "for a cohort of the same size (1,000 students)." So, even though the population grows, we're still considering a sample of 1,000 students each year. So, the mean test score after 5 years would be 75 + 5*3 = 90.Wait, but that seems too high because the standard deviation is 10, so 90 is 1.5 standard deviations above the original mean. But in reality, test scores can't go beyond certain limits, but the problem doesn't specify any upper bound, so we can assume it's possible.But wait, actually, in the first year, the mean was 78, which is 3 points higher. So, if the same improvement continues, each year it's another 3 points. So, after 5 years, it's 75 + 5*3 = 90.But wait, another thought: maybe the improvement is 3 points in the first year, and then it's compounded annually? Like, 3% improvement each year? But the problem says "the same improvement pattern continues," and the improvement was 3 points, not 3%. So, it's more likely an additive 3 points each year.So, expected mean after 5 years is 90.Now, the second part: determine the probability that the average test score will exceed 80 in the fifth year for a cohort of 1,000 students.Wait, but if the expected mean after 5 years is 90, then the probability that the average exceeds 80 is almost 1, because 80 is below the mean. But wait, maybe I misinterpreted the projection.Wait, hold on. Let me read part b again: "Calculate the expected mean test score after 5 years, and determine the probability that the average test score will exceed 80 in the fifth year for a cohort of the same size (1,000 students)."So, the expected mean after 5 years is 90. So, the average test score is 90, and we need the probability that the average exceeds 80. Since 80 is below the mean, and the distribution is normal, the probability is more than 0.5. But actually, since 80 is 10 points below 90, and the standard deviation is 10, we can calculate the z-score.Wait, but the standard deviation for the sample mean is σ / sqrt(n). Since the standard deviation of the population is 10, and the sample size is 1,000, the standard error is 10 / sqrt(1000) ≈ 0.3162.So, for the fifth year, the expected mean is 90, and we want P(average > 80). So, z = (80 - 90) / (10 / sqrt(1000)) = (-10) / 0.3162 ≈ -31.62.Wait, that's a very low z-score. The probability that Z is less than -31.62 is practically 0. So, the probability that the average exceeds 80 is 1 - 0 = 1. But that can't be right because 80 is much lower than 90. Wait, but actually, the z-score is (80 - 90) / (10 / sqrt(1000)) ≈ -31.62, which is way in the left tail. So, the probability that the average is less than 80 is practically 0, so the probability that it's greater than 80 is 1.But that seems counterintuitive because 80 is 10 points below 90, but with a standard error of ~0.316, it's 31.62 standard errors away. So, yes, the probability is effectively 1.Wait, but maybe I made a mistake in the expected mean. Let me double-check. If the improvement is 3 points each year, starting from 75, then after 1 year: 78, after 2 years: 81, 3:84, 4:87, 5:90. So, yes, 90 is correct.Alternatively, maybe the improvement is 3 points in the first year, and then it's compounded annually at 2%, but that doesn't make much sense because 2% growth is for the population, not the test scores.Wait, the problem says "the same improvement pattern continues and that the student population grows by 2% annually." So, the improvement pattern is the 3-point increase, and the population grows by 2%. But the test scores are per student, so the population growth doesn't affect the mean test score. So, the mean increases by 3 points each year regardless of population growth.So, the expected mean after 5 years is 90.Therefore, the probability that the average exceeds 80 is 1, because 80 is way below the mean, and the standard error is so small that the distribution is tightly clustered around 90.But wait, let me think again. Maybe the improvement is not additive but multiplicative. Like, each year the mean increases by 3 points, but that 3 points is a percentage of the previous year's mean. But the problem says "the same improvement pattern continues," and the improvement was 3 points, not a percentage. So, it's more likely additive.Alternatively, maybe the improvement is 3 points in the first year, and then it's compounded annually at the same rate. But 3 points is an absolute improvement, not a percentage, so compounding wouldn't apply.So, I think the expected mean after 5 years is 90, and the probability that the average exceeds 80 is 1.But wait, let me calculate it properly. The z-score is (80 - 90) / (10 / sqrt(1000)) ≈ -31.62. The probability that Z is less than -31.62 is effectively 0, so the probability that Z is greater than -31.62 is 1. Therefore, the probability that the average exceeds 80 is 1.But that seems too certain. Maybe I'm missing something. Let me check the calculations again.The standard error (SE) is σ / sqrt(n) = 10 / sqrt(1000) ≈ 0.3162.The expected mean after 5 years is 90.We want P(average > 80) = P(Z > (80 - 90)/0.3162) = P(Z > -31.62). Since the normal distribution is symmetric, P(Z > -31.62) = 1 - P(Z < -31.62). But P(Z < -31.62) is practically 0, so P(Z > -31.62) ≈ 1.Therefore, the probability is 1, or 100%.But that seems a bit too certain, but mathematically, it's correct because 80 is so far below the mean relative to the standard error.Alternatively, maybe the improvement is not 3 points each year, but the same percentage increase. But the problem states the improvement was 3 points, not a percentage. So, I think additive is correct.So, to summarize part b:Expected mean after 5 years: 90.Probability that average exceeds 80: 1.But wait, another thought: maybe the improvement is not 3 points each year, but the same relative improvement. The first year, the mean went from 75 to 78, which is an increase of 4%. So, maybe each year, the mean increases by 4%. So, after 5 years, it would be 75*(1.04)^5 ≈ 75*1.2167 ≈ 91.25. But the problem says "the same improvement pattern continues," and the improvement was 3 points, not 4%. So, I think additive is correct.Alternatively, if it's a 4% increase each year, but that's not what the problem states. The problem states the improvement was 3 points, so additive.Therefore, I think my initial conclusion is correct.So, final answers:a) The z-test statistic is approximately 6.708, which is greater than the critical value of 1.645, so we reject H0. The increase is statistically significant.b) The expected mean after 5 years is 90, and the probability that the average exceeds 80 is 1.But wait, in part b, the problem says "the same improvement pattern continues." So, if the improvement was 3 points in the first year, does that mean each subsequent year also has a 3-point increase? Or is it that the improvement rate is 3 points per year, so after 5 years, it's 75 + 5*3 = 90.Yes, that's correct.So, I think that's it.</think>

🧠:<think>Alright, so I have these two probability questions to solve about a Chinese soccer team's performance. Let me take them one by one.Starting with the first question: The team conceded an average of 1.2 goals per match, and we need to find the probability that they conceded exactly 2 goals in a randomly selected match. It says the number of goals conceded follows a Poisson distribution. Hmm, okay, I remember that the Poisson distribution is used for counting the number of times an event happens in a fixed interval of time or space. So, in this case, the number of goals conceded per match fits that description.The formula for the Poisson probability mass function is P(k) = (λ^k * e^(-λ)) / k!, where λ is the average rate (which is 1.2 here), k is the number of occurrences (which is 2 in this case), and e is the base of the natural logarithm, approximately 2.71828.So, plugging in the numbers: P(2) = (1.2^2 * e^(-1.2)) / 2!Let me compute each part step by step.First, 1.2 squared is 1.44.Next, e^(-1.2). I need to calculate that. I know that e^(-1) is about 0.3679, and e^(-0.2) is approximately 0.8187. So, multiplying those together: 0.3679 * 0.8187 ≈ 0.3012. Alternatively, I can use a calculator for more precision, but since I don't have one, I'll go with this approximation.Then, 2! is 2.So, putting it all together: (1.44 * 0.3012) / 2.First, multiply 1.44 by 0.3012. Let me do that:1.44 * 0.3 = 0.4321.44 * 0.0012 = 0.001728Adding those together: 0.432 + 0.001728 ≈ 0.433728Now, divide that by 2: 0.433728 / 2 ≈ 0.216864So, approximately 0.2169, or 21.69%.Wait, let me check if my approximation for e^(-1.2) was accurate enough. Maybe I should use a better method. I remember that e^x can be approximated by its Taylor series: e^x = 1 + x + x^2/2! + x^3/3! + ... So, e^(-1.2) would be 1 - 1.2 + (1.44)/2 - (1.728)/6 + (2.0736)/24 - ... Let's compute a few terms.First term: 1Second term: -1.2Third term: (1.44)/2 = 0.72Fourth term: -(1.728)/6 ≈ -0.288Fifth term: (2.0736)/24 ≈ 0.0864Sixth term: -(2.48832)/120 ≈ -0.020736Adding these up:1 - 1.2 = -0.2-0.2 + 0.72 = 0.520.52 - 0.288 = 0.2320.232 + 0.0864 = 0.31840.3184 - 0.020736 ≈ 0.297664So, after six terms, we have approximately 0.2977. That's closer to the actual value, which I think is around 0.3012, so my initial approximation was pretty close. So, maybe 0.2977 is a better estimate.So, let's recalculate with e^(-1.2) ≈ 0.2977.So, 1.2^2 is 1.44.Multiply that by 0.2977: 1.44 * 0.2977 ≈ Let's compute 1 * 0.2977 = 0.2977, 0.44 * 0.2977 ≈ 0.130988. Adding together: 0.2977 + 0.130988 ≈ 0.428688.Divide by 2! which is 2: 0.428688 / 2 ≈ 0.214344.So, approximately 0.2143, or 21.43%.Hmm, so depending on the precision of e^(-1.2), the probability is roughly 21.4% to 21.7%. That seems reasonable.Alternatively, if I use a calculator for e^(-1.2), it's approximately 0.301194.So, 1.44 * 0.301194 ≈ 1.44 * 0.3 = 0.432, 1.44 * 0.001194 ≈ 0.001717. So total is approximately 0.432 + 0.001717 ≈ 0.433717.Divide by 2: 0.433717 / 2 ≈ 0.216858.So, approximately 0.2169, or 21.69%. So, rounding to four decimal places, 0.2169.Therefore, the probability is approximately 21.69%.Okay, moving on to the second question: The team's total number of goals scored over the season followed a Gaussian (normal) distribution with a mean of 58 goals and a standard deviation of 6 goals. We need to find the probability that the team scored between 50 and 65 goals over the entire season.Alright, so for a normal distribution, we can standardize the values and use the Z-table or the standard normal distribution to find probabilities.First, let's recall that the Z-score is calculated as Z = (X - μ) / σ, where X is the value, μ is the mean, and σ is the standard deviation.So, we need to find P(50 < X < 65). To find this, we can compute the Z-scores for 50 and 65, then find the area under the standard normal curve between these two Z-scores.Let's compute Z1 for X = 50:Z1 = (50 - 58) / 6 = (-8) / 6 ≈ -1.3333And Z2 for X = 65:Z2 = (65 - 58) / 6 = 7 / 6 ≈ 1.1667So, we need to find P(-1.3333 < Z < 1.1667).To find this probability, we can use the standard normal distribution table or a calculator. The probability that Z is less than 1.1667 minus the probability that Z is less than -1.3333.Looking up Z = 1.1667 in the standard normal table. Let me recall that Z = 1.16 corresponds to approximately 0.8770, and Z = 1.17 corresponds to approximately 0.8790. Since 1.1667 is closer to 1.17, maybe around 0.8785.Similarly, for Z = -1.3333, which is the same as -1.33. Looking up Z = -1.33, the table gives the probability that Z is less than -1.33. From the table, Z = -1.33 corresponds to approximately 0.0918.So, the probability between -1.3333 and 1.1667 is P(Z < 1.1667) - P(Z < -1.3333) ≈ 0.8785 - 0.0918 ≈ 0.7867.So, approximately 78.67%.Alternatively, let me use more precise Z-values.For Z = 1.1667, let's interpolate between Z=1.16 and Z=1.17.Z=1.16: 0.8770Z=1.17: 0.8790Difference per 0.01 Z is 0.0020.So, 1.1667 is 0.0067 above 1.16.So, 0.0067 / 0.01 = 0.67 of the way from 1.16 to 1.17.So, the increase is 0.67 * 0.0020 ≈ 0.00134.Thus, P(Z < 1.1667) ≈ 0.8770 + 0.00134 ≈ 0.87834.Similarly, for Z = -1.3333, which is -1.33 - 0.0033.Looking up Z = -1.33: 0.0918Z = -1.34: 0.0901So, the difference between Z=-1.33 and Z=-1.34 is 0.0918 - 0.0901 = 0.0017 over 0.01 Z.So, for Z = -1.3333, which is 0.0033 below -1.33, the probability decreases by (0.0033 / 0.01) * 0.0017 ≈ 0.000561.So, P(Z < -1.3333) ≈ 0.0918 - 0.000561 ≈ 0.091239.Thus, the probability between -1.3333 and 1.1667 is approximately 0.87834 - 0.091239 ≈ 0.7871.So, approximately 78.71%.Alternatively, using a calculator or precise Z-table, but I think 78.7% is a good approximation.So, summarizing:1. The probability of conceding exactly 2 goals in a match is approximately 21.69%.2. The probability of scoring between 50 and 65 goals over the season is approximately 78.7%.I think that's it. Let me just double-check my calculations.For the first part, Poisson with λ=1.2, k=2:P(2) = (1.2^2 * e^-1.2) / 2! ≈ (1.44 * 0.3012) / 2 ≈ 0.4337 / 2 ≈ 0.21685, which is about 21.69%. That seems correct.For the second part, normal distribution with μ=58, σ=6, finding P(50 < X < 65):Z1 = (50 - 58)/6 ≈ -1.3333Z2 = (65 - 58)/6 ≈ 1.1667Looking up these Z-scores, the probability between them is roughly 78.7%. That seems reasonable because 50 is about 1.33σ below the mean, and 65 is about 1.17σ above. The area between them should be a large chunk, more than 70%, which 78.7% is.So, I think my answers are correct.Final Answer1. The probability is boxed{0.2169}.2. The probability is boxed{0.7867}.</think>

🧠:<think>Alright, so I have this problem about an employee transporting boxes. Let me try to understand it step by step.First, the employee has a truck that can carry up to 2000 cubic feet of merchandise. There are two types of boxes: Type A and Type B. Type A boxes are 2x3x4 feet, and Type B are 3x3x3 feet. The store requires a specific ratio of 3 Type A boxes for every 2 Type B boxes. Okay, so part 1 is asking for the maximum number of each type of box that can be transported in a single trip while maintaining that ratio and not exceeding the truck's capacity. Part 2 is about figuring out the minimum number of trips needed if the store requires 300 Type A and 200 Type B boxes, with the ratio maintained in each trip.Let me start with part 1.First, I need to figure out the volume of each box. For Type A, the volume is length x width x height, so 2*3*4. Let me calculate that: 2*3 is 6, 6*4 is 24. So each Type A box is 24 cubic feet.Type B boxes are 3x3x3, so that's 3*3*3. 3*3 is 9, 9*3 is 27. So each Type B box is 27 cubic feet.Now, the ratio required is 3 Type A to 2 Type B. So for every 3 Type A boxes, there should be 2 Type B boxes. That means if I let the number of Type A boxes be 3x and Type B boxes be 2x, where x is some integer, then I can express the total volume in terms of x.So total volume would be (3x * 24) + (2x * 27). Let me compute that:3x * 24 = 72x2x * 27 = 54xSo total volume is 72x + 54x = 126xThis total volume must be less than or equal to 2000 cubic feet.So 126x ≤ 2000To find x, I divide both sides by 126:x ≤ 2000 / 126Let me compute that. 2000 divided by 126.Well, 126*15 is 1890, because 126*10 is 1260, 126*5 is 630, so 1260+630=1890.2000 - 1890 = 110. So 110/126 is approximately 0.873.So x ≤ approximately 15.873. But x has to be an integer because you can't have a fraction of a box. So x can be at most 15.Therefore, the number of Type A boxes is 3x = 3*15 = 45Number of Type B boxes is 2x = 2*15 = 30Let me check the total volume: 45*24 + 30*27.45*24: 40*24=960, 5*24=120, so total 960+120=108030*27: 30*27=810Total volume: 1080 + 810 = 1890 cubic feet, which is under 2000.If I try x=16, then:3x=48, 2x=32Volume: 48*24 + 32*2748*24: 40*24=960, 8*24=192, total 960+192=115232*27: 30*27=810, 2*27=54, total 810+54=864Total volume: 1152 + 864 = 2016, which exceeds 2000. So x=16 is too much.Therefore, the maximum is x=15, giving 45 Type A and 30 Type B boxes.So that's part 1 done.Now, part 2: The store needs 300 Type A and 200 Type B boxes. The employee can make multiple trips, but each trip must maintain the 3:2 ratio. We need to find the minimum number of trips required.First, let's see what the ratio is. The store requires 300 Type A and 200 Type B. Let's check the ratio: 300:200 simplifies to 3:2, which is exactly the required ratio. So that's good.So each trip must carry boxes in a 3:2 ratio, and each trip can carry up to 2000 cubic feet.From part 1, we know that in one trip, the maximum number of boxes is 45 Type A and 30 Type B, totaling 75 boxes, with a volume of 1890 cubic feet.But maybe we can carry more in a trip? Wait, in part 1, we found the maximum number while maintaining the ratio. So 45 and 30 is the maximum per trip.But wait, actually, if the store's total requirement is 300 Type A and 200 Type B, which is exactly 10 times the 30 and 20 ratio. Wait, 300 is 10*30, 200 is 10*20? Wait, no, 300 Type A and 200 Type B is 3:2 ratio, same as 30:20, which is same as 3:2.Wait, but in part 1, each trip can carry 45 Type A and 30 Type B, which is 3:2 ratio as well.So, if each trip can carry 45 Type A and 30 Type B, how many trips would be needed to reach 300 Type A and 200 Type B?Let me compute how many trips are needed if each trip carries 45 Type A and 30 Type B.Number of trips needed for Type A: 300 / 45 = 6.666...Number of trips needed for Type B: 200 / 30 ≈ 6.666...So, since you can't do a fraction of a trip, you need to round up to 7 trips.But wait, 7 trips would carry 7*45=315 Type A and 7*30=210 Type B. But the store only needs 300 Type A and 200 Type B. So that would be 15 extra Type A and 10 extra Type B. Is that acceptable? The problem says the store has a total requirement, so maybe they can take extra, but it's better to see if we can do it in 7 trips without exceeding.Alternatively, maybe we can optimize the number of boxes per trip to minimize the number of trips, but still maintaining the 3:2 ratio each time.Wait, perhaps in each trip, we can carry a different number of boxes as long as the ratio is maintained, not necessarily the maximum per trip.But since the goal is to minimize the number of trips, it's better to carry as much as possible each trip.But in part 1, we found that the maximum per trip is 45 Type A and 30 Type B. So if we use that, we need 7 trips, but that would exceed the required number.Alternatively, maybe we can adjust the number of boxes per trip so that we don't exceed the required numbers.Wait, but the ratio has to be maintained in each trip, so each trip must have 3 Type A for every 2 Type B.So, let me think of it as each trip has 3k Type A and 2k Type B, for some integer k.We need total 300 Type A and 200 Type B.So, the total number of trips would be such that the sum over all trips of (3k_i) = 300 and sum over all trips of (2k_i) = 200.But since each trip can have a different k_i, but we need to minimize the number of trips.But actually, since the ratio is fixed, the total number of Type A and Type B boxes must be in the ratio 3:2, which they are (300:200=3:2). So, the total number of "units" is 300/3 = 100 units of Type A and 200/2 = 100 units of Type B. So, total 100 units.Each trip can carry a certain number of units, where each unit is 3 Type A and 2 Type B.But wait, in part 1, each trip can carry 15 units (since 45 Type A is 15*3, and 30 Type B is 15*2). So each trip can carry 15 units.But the total units needed are 100. So, 100 / 15 ≈ 6.666, so 7 trips.Wait, that seems consistent with earlier.But maybe we can carry more units per trip if we don't carry the maximum volume.Wait, but the volume per trip is limited to 2000 cubic feet.In part 1, 15 units (45 Type A and 30 Type B) take up 1890 cubic feet. Maybe we can carry more units if we adjust the number of boxes, but keeping the 3:2 ratio.Wait, but 15 units is already the maximum number of units per trip without exceeding the volume. Because if we try to carry more units, say 16 units, that would be 48 Type A and 32 Type B, which as calculated earlier, would take 2016 cubic feet, exceeding the truck's capacity.So, 15 units per trip is the maximum. Therefore, to carry 100 units, we need 7 trips, since 6 trips would carry 90 units, leaving 10 units, which would require a 7th trip.But let me verify:Each trip: 15 units (45A, 30B)6 trips: 6*15=90 units, which is 270A and 180BRemaining: 300-270=30A and 200-180=20BSo, remaining is 30A and 20B, which is exactly 10 units (since 30A is 10*3, 20B is 10*2). So, 10 units can be carried in one trip, but wait, 10 units would be 30A and 20B, which is 30*24 + 20*27 = 720 + 540 = 1260 cubic feet, which is less than 2000. So, actually, in the 7th trip, we can carry only 10 units, which is less than the maximum 15 units.But the question is asking for the minimum number of trips, so 7 trips is the answer.Alternatively, maybe we can combine the remaining 10 units with some other trips to reduce the number of trips, but since each trip must maintain the 3:2 ratio, we can't just carry extra in one trip without maintaining the ratio.Wait, perhaps if we adjust the number of units per trip, but keeping the ratio, we can have some trips carry more and some carry less, but overall, the total units would still require 7 trips.Wait, let me think differently. Maybe instead of always carrying 15 units per trip, we can have some trips carry more units if possible, but I don't think so because 15 units is already the maximum per trip without exceeding the volume.Alternatively, maybe we can carry different numbers of units per trip, but the total number of trips would still be 7 because 100 units divided by 15 units per trip is approximately 6.666, so 7 trips.Therefore, the minimum number of trips is 7.Wait, but let me check if 7 trips is indeed the minimum. Suppose we try to carry 16 units in one trip, but as we saw, that would exceed the volume. So, no, we can't carry more than 15 units per trip.Alternatively, maybe we can have some trips carry 15 units and some carry fewer, but the total number of trips would still be 7.Wait, let me calculate the total volume for 7 trips:6 trips carrying 15 units each: 6*1890=11340 cubic feet1 trip carrying 10 units: 1260 cubic feetTotal volume: 11340 + 1260 = 12600 cubic feetBut that's just the total volume, but the problem is about the number of trips, not the total volume. So, regardless of the volume, the number of trips is 7.Alternatively, maybe we can optimize the number of units per trip to minimize the number of trips, but I think 7 is the minimum because 6 trips can only carry 90 units, leaving 10 units, which needs another trip.So, the answer is 7 trips.Wait, but let me think again. Maybe if we adjust the number of units per trip, not all trips have to carry the same number of units, as long as each trip maintains the 3:2 ratio.So, perhaps in some trips, we can carry more units, and in others, fewer, but overall, the total units would still be 100.But since each trip can carry a maximum of 15 units, and 100 divided by 15 is 6 with a remainder of 10, so we need 7 trips.Yes, that seems correct.So, to summarize:Part 1: Maximum per trip is 45 Type A and 30 Type B.Part 2: Minimum number of trips is 7.Final Answer1. The maximum number of Type A boxes is boxed{45} and Type B boxes is boxed{30}.2. The minimum number of trips required is boxed{7}.</think>