💡Alright, so I've got this probability problem here, and I need to figure out which pair of events are mutually exclusive. Let me try to break it down step by step.First, let's understand the setup. There's a group of 4 male students and 2 female students, making a total of 6 students. We're selecting 2 students randomly for a speech contest. The question is about identifying which pair of events cannot happen at the same time, meaning they are mutually exclusive.The options are:- A: "Exactly 1 male student" and "Exactly 1 male student"- B: "At least 1 male student" and "All male students"- C: "At least 1 male student" and "At least 1 female student"- D: "At least 1 male student" and "All female students"Okay, let's tackle each option one by one.**Option A: "Exactly 1 male student" and "Exactly 1 male student"**Hmm, this seems a bit odd because both events are the same. If we're selecting 2 students, "Exactly 1 male student" means one male and one female. So, both events are identical. Since they're the same event, they can certainly happen together. Therefore, they are not mutually exclusive. In fact, they're the same event, so they're not even different events to begin with. So, A is not the answer.**Option B: "At least 1 male student" and "All male students"**Alright, "At least 1 male student" means that in the selected 2 students, there is 1 or 2 male students. "All male students" specifically means both selected students are male. So, if both are male, then it certainly satisfies "At least 1 male student." In other words, "All male students" is a subset of "At least 1 male student." Therefore, these two events can happen together. They are not mutually exclusive. So, B is not the answer.**Option C: "At least 1 male student" and "At least 1 female student"**Okay, "At least 1 male student" means 1 or 2 males, and "At least 1 female student" means 1 or 2 females. If we select 2 students, it's possible to have 1 male and 1 female. In that case, both events occur simultaneously. Therefore, these two events can happen together, so they are not mutually exclusive. So, C is not the answer.**Option D: "At least 1 male student" and "All female students"**Now, "At least 1 male student" means 1 or 2 males, and "All female students" means both selected students are female. If we have "All female students," that means there are 0 male students selected. Therefore, "At least 1 male student" cannot happen if "All female students" has occurred, and vice versa. These two events cannot happen at the same time. Hence, they are mutually exclusive.So, after analyzing all the options, the correct answer is D.**Final Answer**boxed{D}

💡Okay, so I have this function f(x) defined piecewise. It's log base a of (x + a - 1) when x is greater than 1, and it's a linear function (2a - 1)x - a when x is less than or equal to 1. The problem says that the function satisfies (f(x₁) - f(x₂))/(x₁ - x₂) > 0 for any real numbers x₁ ≠ x₂. I need to find the range of the real number a.Hmm, so the condition (f(x₁) - f(x₂))/(x₁ - x₂) > 0 means that the function is strictly increasing. Because if the difference quotient is positive, it means that as x increases, f(x) also increases. So, f(x) is a strictly increasing function over its entire domain.Since f(x) is piecewise, I need to ensure that both pieces are increasing and that there's no decrease at the point where the function changes from one piece to the other, which is at x = 1.First, let's look at the linear part: (2a - 1)x - a for x ≤ 1. For this to be increasing, the slope must be positive. The slope here is (2a - 1). So, 2a - 1 > 0. Solving that gives 2a > 1, so a > 1/2.But wait, the logarithmic function is log base a of (x + a - 1). For the logarithmic function to be defined, the argument x + a - 1 must be positive. Since x > 1, the smallest value x can take is just above 1. So, plugging x = 1 into the argument gives 1 + a - 1 = a. Therefore, a must be positive. But since the logarithmic function is also part of the increasing function, its base a must be greater than 1 because logarithmic functions with base greater than 1 are increasing. If the base were between 0 and 1, the logarithmic function would be decreasing, which would contradict the overall function being increasing.So, from the logarithmic part, we have a > 1.Now, combining the two conditions from the linear part and the logarithmic part: a > 1/2 and a > 1. The stricter condition is a > 1, so that's our starting point.Next, we need to ensure that the function is continuous at x = 1, or at least that there's no jump discontinuity that would cause the function to decrease. Wait, actually, the function doesn't necessarily have to be continuous, but it does have to be increasing everywhere. So, even if there's a jump, the right-hand limit at x = 1 must be greater than or equal to the left-hand limit.Let me compute the left-hand limit as x approaches 1 from below. That's f(1) from the linear part: (2a - 1)(1) - a = 2a - 1 - a = a - 1.Now, the right-hand limit as x approaches 1 from above is f(1) from the logarithmic part: log_a(1 + a - 1) = log_a(a) = 1.So, for the function to be increasing, the value of the function at x = 1 from the right must be greater than or equal to the value from the left. That is, 1 ≥ a - 1. Solving that gives a ≤ 2.So, combining all the conditions: a > 1 and a ≤ 2. Therefore, the range of a is (1, 2].Wait, but let me double-check. The function is strictly increasing, so at x = 1, the right-hand value must be greater than the left-hand value. So, 1 > a - 1, which would give a < 2. But if a = 2, then 1 = a - 1, which is 1 = 1. So, at a = 2, the function is still increasing because the right-hand limit equals the left-hand limit, meaning there's no decrease. So, a can be equal to 2.Therefore, the range is 1 < a ≤ 2, which is (1, 2].Let me check the options: A is (1, ∞), B is (1, 2), C is (1, 2], D is (2, ∞). So, the correct answer is C.I think that's it. I considered both pieces needing to be increasing, ensured the transition at x = 1 doesn't cause a decrease, and solved the inequalities accordingly.

💡First, we need to determine the total fraction of the pizza that Alex, Beth, Cyril, and Eve ate together. - Alex ate (frac{1}{7}) of the pizza.- Beth ate (frac{1}{5}) of the pizza.- Cyril ate (frac{1}{6}) of the pizza.- Eve ate (frac{1}{9}) of the pizza.To add these fractions together, we need a common denominator. The least common multiple (LCM) of 7, 5, 6, and 9 is 630. Converting each fraction to have a denominator of 630:- Alex: (frac{1}{7} = frac{90}{630})- Beth: (frac{1}{5} = frac{126}{630})- Cyril: (frac{1}{6} = frac{105}{630})- Eve: (frac{1}{9} = frac{70}{630})Adding these together:[frac{90}{630} + frac{126}{630} + frac{105}{630} + frac{70}{630} = frac{391}{630}]This means the total fraction of the pizza eaten by Alex, Beth, Cyril, and Eve is (frac{391}{630}). Therefore, the fraction of the pizza that Dan ate is:[1 - frac{391}{630} = frac{239}{630}]Now, we compare the fractions of the pizza each sibling ate:- Dan: (frac{239}{630})- Beth: (frac{126}{630})- Cyril: (frac{105}{630})- Alex: (frac{90}{630})- Eve: (frac{70}{630})Arranging these in decreasing order:[text{Dan, Beth, Cyril, Alex, Eve}]

💡First, identify the total number of employees and the number of employees to be selected. There are 4 employees, and 2 need to be chosen.Next, calculate the total number of possible ways to select 2 employees out of 4 using the combination formula ( C(n, k) ), which is ( C(4, 2) = 6 ).Determine the number of ways to select 2 employees without choosing either A or B. Since only C and D remain, there is only 1 way to choose both of them.Calculate the probability of not selecting A or B, which is ( frac{1}{6} ).Finally, find the probability of selecting at least one of A or B by subtracting the probability of not selecting them from 1. This gives ( 1 - frac{1}{6} = frac{5}{6} ).

💡Okay, so I have this problem here: If the line y = 2x + b is a tangent to the curve y = e^x + x, then what is the value of the real number b? Hmm, let me try to figure this out step by step.First, I remember that for a line to be tangent to a curve, two conditions must be satisfied. The first condition is that the line must touch the curve at exactly one point. The second condition is that the slope of the line must be equal to the derivative of the curve at that point of contact. So, I need to find a point (x, y) where both the curve and the line meet, and the slope of the line matches the derivative of the curve at that point.Alright, let's start by finding the derivative of the curve y = e^x + x. The derivative, which gives the slope of the tangent line at any point x, is y' = e^x + 1. So, the slope of the tangent line at any point x on the curve is e^x + 1.Now, the given line is y = 2x + b. The slope of this line is 2. Since the line is tangent to the curve, their slopes must be equal at the point of contact. That means e^x + 1 must equal 2. Let me write that down:e^x + 1 = 2Subtracting 1 from both sides gives:e^x = 1To solve for x, I can take the natural logarithm of both sides:ln(e^x) = ln(1)Which simplifies to:x = 0So, the point of contact is at x = 0. Now, I need to find the corresponding y-coordinate on the curve. Plugging x = 0 into the curve equation y = e^x + x:y = e^0 + 0 = 1 + 0 = 1So, the point of contact is (0, 1).Now, since this point also lies on the tangent line y = 2x + b, I can substitute x = 0 and y = 1 into the line equation to find b:1 = 2(0) + bSimplifying:1 = 0 + bSo, b = 1Wait, that seems straightforward. Let me double-check to make sure I didn't make any mistakes.First, I found the derivative correctly: y' = e^x + 1. Then, I set that equal to the slope of the line, which is 2, and solved for x. That gave me x = 0. Plugging x = 0 back into the original curve equation gave me y = 1. Then, using the point (0, 1) in the line equation, I solved for b and got 1. Is there another way to verify this? Maybe by plugging the value of b back into the line equation and checking if it's indeed tangent to the curve.So, if b = 1, the line is y = 2x + 1. Let's see if this line touches the curve y = e^x + x at exactly one point. We already know that at x = 0, both the curve and the line equal 1. Let's check another point. For example, at x = 1:Curve: y = e^1 + 1 ≈ 2.718 + 1 ≈ 3.718Line: y = 2(1) + 1 = 3So, at x = 1, the curve is higher than the line. What about at x = -1?Curve: y = e^{-1} + (-1) ≈ 0.368 - 1 ≈ -0.632Line: y = 2(-1) + 1 = -2 + 1 = -1Here, the curve is higher than the line again. It seems like the line is tangent at x = 0 and doesn't intersect the curve anywhere else, which makes sense because the curve is exponential and grows faster than the linear line as x increases, and it also decreases but not as rapidly as the line as x decreases.Therefore, my answer seems correct. The value of b is 1.**Final Answer**The value of the real number ( b ) is boxed{1}.

💡To determine the number of days the contractor was absent, let's define the variables:- Let ( x ) be the number of days the contractor worked.- Let ( y ) be the number of days the contractor was absent.We know that the total number of days is 30, so:[ x + y = 30 ]The contractor earns Rs. 25 for each day worked and is fined Rs. 7.50 for each day absent. The total amount received is Rs. 425:[ 25x - 7.5y = 425 ]First, solve the first equation for ( x ):[ x = 30 - y ]Substitute ( x ) into the second equation:[ 25(30 - y) - 7.5y = 425 ]Expand and simplify:[ 750 - 25y - 7.5y = 425 ][ 750 - 32.5y = 425 ]Isolate ( y ):[ -32.5y = 425 - 750 ][ -32.5y = -325 ][ y = frac{-325}{-32.5} ][ y = 10 ]Therefore, the contractor was absent for 10 days.