💡Okay, so I have this problem where I need to find the cumulative distribution functions (CDFs) for two independent random variables X and Y, their joint probability density function (PDF), and the joint CDF. Let me try to break this down step by step.First, let's look at the given densities. For X, the density function p₁(x) is 0.5 between -1 and 1, and 0 otherwise. Similarly, for Y, the density function p₂(y) is 0.5 between 0 and 2, and 0 otherwise. Both X and Y are independent, which is important because independence will help in finding the joint PDF and CDF.Starting with the first part: finding the CDFs F₁(x) and F₂(y). I remember that the CDF is the integral of the PDF from negative infinity up to a point x. So for F₁(x), I need to integrate p₁(t) from -∞ to x.Let me think about the intervals for X. If x is less than or equal to -1, the integral will be zero because p₁(t) is zero in that region. If x is between -1 and 1, the integral will be the area under p₁(t) from -1 to x, which is a rectangle with height 0.5 and width (x - (-1)) = x + 1. So the integral should be 0.5*(x + 1). If x is greater than 1, the integral will cover the entire area under p₁(t), which is 1 because the total area under a PDF must be 1.Similarly, for F₂(y), the CDF for Y. If y is less than or equal to 0, the integral is zero. If y is between 0 and 2, the integral is the area under p₂(t) from 0 to y, which is 0.5*y. If y is greater than 2, the integral covers the entire area, so it's 1.Okay, that seems straightforward. Now, moving on to the joint PDF. Since X and Y are independent, their joint PDF is just the product of their individual PDFs. So p(x, y) = p₁(x)*p₂(y). That means wherever both p₁(x) and p₂(y) are non-zero, the joint PDF will be 0.5*0.5 = 0.25. Otherwise, it's zero.So the joint PDF will be 0.25 in the region where -1 < x ≤ 1 and 0 < y ≤ 2, and zero elsewhere.Finally, the joint CDF F(x, y). I recall that for independent variables, the joint CDF is the product of the marginal CDFs. So F(x, y) = F₁(x)*F₂(y). Let me verify that.If x ≤ -1 or y ≤ 0, then F₁(x) or F₂(y) is zero, so the joint CDF is zero. If x is between -1 and 1 and y is between 0 and 2, then F(x, y) = 0.5(x + 1)*0.5y = 0.25(x + 1)y. If x > 1 and y ≤ 2, then F₁(x) is 1, so F(x, y) = 1*0.5y = 0.5y. Similarly, if y > 2 and x ≤ 1, then F₂(y) is 1, so F(x, y) = 0.5(x + 1)*1 = 0.5(x + 1). If both x > 1 and y > 2, then F(x, y) = 1*1 = 1.Wait, let me make sure I didn't mix up any cases. For example, if x is between -1 and 1 and y is greater than 2, F₂(y) is 1, so F(x, y) should be F₁(x)*1 = 0.5(x + 1). Similarly, if x is greater than 1 and y is between 0 and 2, it's 1*F₂(y) = 0.5y. And if both are beyond their limits, it's 1.I think that covers all the cases. Let me just recap:1. F₁(x) is 0 for x ≤ -1, 0.5(x + 1) for -1 < x ≤ 1, and 1 for x > 1.2. F₂(y) is 0 for y ≤ 0, 0.5y for 0 < y ≤ 2, and 1 for y > 2.3. The joint PDF p(x, y) is 0.25 in the rectangle -1 < x ≤ 1 and 0 < y ≤ 2, else 0.4. The joint CDF F(x, y) is the product of F₁(x) and F₂(y), considering all possible combinations of x and y.I think that's all. Let me just visualize it to make sure. For F₁(x), it's a linear increase from 0 to 1 as x goes from -1 to 1. For F₂(y), it's a linear increase from 0 to 1 as y goes from 0 to 2. The joint PDF is a uniform distribution over the rectangle defined by x between -1 and 1 and y between 0 and 2. The joint CDF will be a combination of these, increasing in both x and y directions, with different expressions depending on which region we're in.I don't see any mistakes in my reasoning, so I think I'm ready to write down the final answers.

💡Okay, so I have this problem here where I need to determine which of the given inequalities is true for any φ in the interval (0, π/2). The options are:(A) sin(sin φ) < cos φ < cos(cos φ)(B) sin(sin φ) > cos φ > cos(cos φ)(C) sin(cos φ) > cos φ > cos(sin φ)(D) sin(cos φ) < cos φ < cos(sin φ)Hmm, let me try to break this down step by step. First, I should recall some basic properties of sine and cosine functions, especially within the interval (0, π/2). I remember that both sine and cosine functions are positive in this interval. Also, sine is increasing and cosine is decreasing in this interval. That might be useful. Let me consider each option one by one.Starting with option (A): sin(sin φ) < cos φ < cos(cos φ). Hmm, okay. So, let's analyze sin(sin φ) and cos(cos φ). Since φ is between 0 and π/2, sin φ is between 0 and 1. Similarly, cos φ is also between 0 and 1. Now, sin(sin φ) would be the sine of a number between 0 and 1. Since sine is increasing on (0, π/2), and 1 is less than π/2 (which is approximately 1.5708), sin(sin φ) would be less than sin(1). Similarly, cos(cos φ) would be the cosine of a number between 0 and 1. Since cosine is decreasing on (0, π/2), cos(cos φ) would be greater than cos(1).So, sin(sin φ) < sin(1) and cos(cos φ) > cos(1). But what about cos φ? Since φ is between 0 and π/2, cos φ is between 0 and 1, but decreasing as φ increases. Wait, so is sin(sin φ) less than cos φ? Let's test with a specific value. Let's take φ = π/4, which is approximately 0.7854.Compute sin(sin(π/4)): sin(√2/2) ≈ sin(0.7071) ≈ 0.6496.Compute cos(π/4): √2/2 ≈ 0.7071.So, sin(sin(π/4)) ≈ 0.6496 < 0.7071 ≈ cos(π/4). So, that holds.Now, is cos φ < cos(cos φ)? Let's check with φ = π/4.cos(π/4) ≈ 0.7071.cos(cos(π/4)) = cos(√2/2) ≈ cos(0.7071) ≈ 0.7616.So, 0.7071 < 0.7616, which is true. So, for φ = π/4, option (A) holds.But wait, let me test another value to be sure. Let's take φ approaching 0. Let φ = 0.1 radians.Compute sin(sin(0.1)): sin(0.0998) ≈ 0.0998.cos(0.1) ≈ 0.9950.cos(cos(0.1)) = cos(0.9950) ≈ 0.5403.So, sin(sin(0.1)) ≈ 0.0998 < cos(0.1) ≈ 0.9950, which is true.But cos(0.1) ≈ 0.9950 > cos(cos(0.1)) ≈ 0.5403. Wait, that contradicts the second inequality in option (A). So, cos φ is not less than cos(cos φ) in this case. Therefore, option (A) is not always true.Hmm, so option (A) fails when φ approaches 0. So, maybe (A) is not the correct answer.Moving on to option (B): sin(sin φ) > cos φ > cos(cos φ).Let me test this with φ = π/4 again.sin(sin(π/4)) ≈ 0.6496, cos(π/4) ≈ 0.7071, and cos(cos(π/4)) ≈ 0.7616.So, 0.6496 < 0.7071, which contradicts the first inequality in (B). So, sin(sin φ) is not greater than cos φ here. So, option (B) is false.Option (C): sin(cos φ) > cos φ > cos(sin φ).Let me test this with φ = π/4.sin(cos(π/4)) = sin(√2/2) ≈ sin(0.7071) ≈ 0.6496.cos(π/4) ≈ 0.7071.cos(sin(π/4)) = cos(√2/2) ≈ cos(0.7071) ≈ 0.7616.So, sin(cos φ) ≈ 0.6496 < cos φ ≈ 0.7071, which contradicts the first inequality in (C). So, option (C) is false.Finally, option (D): sin(cos φ) < cos φ < cos(sin φ).Again, testing with φ = π/4.sin(cos(π/4)) ≈ 0.6496 < cos(π/4) ≈ 0.7071 < cos(sin(π/4)) ≈ 0.7616.So, this holds for φ = π/4.Let me test another value, say φ approaching 0, φ = 0.1 radians.sin(cos(0.1)) = sin(0.9950) ≈ 0.8415.cos(0.1) ≈ 0.9950.cos(sin(0.1)) = cos(0.0998) ≈ 0.9950.Wait, so sin(cos(0.1)) ≈ 0.8415 < cos(0.1) ≈ 0.9950, which is true.But cos(0.1) ≈ 0.9950 and cos(sin(0.1)) ≈ 0.9950. Wait, are they equal? Let me compute more accurately.cos(0.1) ≈ 0.995004165.cos(sin(0.1)) = cos(0.099833417) ≈ 0.995004165.Wait, so they are approximately equal? That's interesting. So, for φ approaching 0, cos φ and cos(sin φ) are almost equal.But let me check φ = 0.5 radians.Compute sin(cos(0.5)): sin(cos(0.5)) ≈ sin(0.87758) ≈ 0.7692.cos(0.5) ≈ 0.87758.cos(sin(0.5)) = cos(0.4794) ≈ 0.8872.So, sin(cos(0.5)) ≈ 0.7692 < cos(0.5) ≈ 0.87758 < cos(sin(0.5)) ≈ 0.8872.So, this holds as well.Wait, but when φ approaches 0, cos φ and cos(sin φ) are almost equal. So, is cos φ < cos(sin φ) always true?Wait, cos is a decreasing function on (0, π/2). So, if sin φ < φ, which is true because for φ in (0, π/2), sin φ < φ.Therefore, since sin φ < φ, and cos is decreasing, cos(sin φ) > cos(φ).Yes, that makes sense. So, cos(sin φ) > cos φ.Similarly, sin(cos φ) < cos φ because cos φ is in (0,1), and sin x < x for x in (0, π/2). Since cos φ is in (0,1), which is a subset of (0, π/2), sin(cos φ) < cos φ.Therefore, combining these two results, we have sin(cos φ) < cos φ < cos(sin φ), which is option (D).So, after testing specific values and reasoning through the properties of sine and cosine functions, I conclude that option (D) is correct.

💡Okay, so I need to find the equation of a line that passes through the intersection point of two given lines and is perpendicular to another line. Hmm, let me break this down step by step.First, the problem gives me two lines: 2x - y - 3 = 0 and 4x - 3y - 5 = 0. I need to find where these two lines intersect because that point will be on the line I'm trying to find. To find the intersection, I can solve these two equations simultaneously.Let me write them down again:1. 2x - y - 3 = 02. 4x - 3y - 5 = 0I can solve this system of equations using substitution or elimination. Maybe elimination is easier here. If I multiply the first equation by 3, it might help eliminate y.Multiplying the first equation by 3:3*(2x - y - 3) = 0 => 6x - 3y - 9 = 0Now I have:1. 6x - 3y - 9 = 02. 4x - 3y - 5 = 0If I subtract the second equation from the first, I can eliminate y:(6x - 3y - 9) - (4x - 3y - 5) = 0 - 0Simplify:6x - 3y - 9 - 4x + 3y + 5 = 0Combine like terms:(6x - 4x) + (-3y + 3y) + (-9 + 5) = 0So, 2x + 0y - 4 = 0 => 2x - 4 = 0 => 2x = 4 => x = 2Now that I have x = 2, I can plug this back into one of the original equations to find y. Let's use the first equation:2x - y - 3 = 0 => 2*(2) - y - 3 = 0 => 4 - y - 3 = 0 => 1 - y = 0 => y = 1So, the intersection point is (2, 1). Got that down.Next, the line I'm looking for has to be perpendicular to the line 2x + 3y + 5 = 0. I remember that the slopes of perpendicular lines are negative reciprocals of each other. So, first, I need to find the slope of the given line.Let me rewrite 2x + 3y + 5 = 0 in slope-intercept form (y = mx + b) to find its slope.Starting with 2x + 3y + 5 = 0:3y = -2x - 5Divide both sides by 3:y = (-2/3)x - 5/3So, the slope (m) of this line is -2/3.Since the line I'm looking for is perpendicular, its slope should be the negative reciprocal of -2/3. The reciprocal of 2/3 is 3/2, and the negative of that is 3/2. Wait, no, hold on. The reciprocal of -2/3 is -3/2, and the negative of that would be 3/2. So, yes, the slope of the line we want is 3/2.Alright, so now I have a point (2, 1) and a slope (3/2). I can use the point-slope form of a line to write the equation.Point-slope form is: y - y1 = m(x - x1)Plugging in the values:y - 1 = (3/2)(x - 2)Let me simplify this equation to get it into standard form (Ax + By + C = 0).First, distribute the 3/2 on the right side:y - 1 = (3/2)x - 3Now, multiply every term by 2 to eliminate the fraction:2*(y - 1) = 2*((3/2)x - 3)Simplify:2y - 2 = 3x - 6Now, bring all terms to one side:-3x + 2y - 2 + 6 = 0Simplify:-3x + 2y + 4 = 0But usually, we like the coefficient of x to be positive. So, multiply the entire equation by -1:3x - 2y - 4 = 0So, the equation of the line is 3x - 2y - 4 = 0.Let me just double-check my steps to make sure I didn't make any mistakes.1. Found intersection point by solving the two equations: (2,1). That seems correct.2. Found slope of the given line: -2/3. Then took the negative reciprocal to get 3/2. That makes sense because perpendicular slopes are negative reciprocals.3. Used point-slope form with (2,1) and slope 3/2: y - 1 = (3/2)(x - 2). That looks right.4. Simplified to standard form: 3x - 2y - 4 = 0. Let me check the algebra:Starting from y - 1 = (3/2)(x - 2)Multiply both sides by 2: 2y - 2 = 3x - 6Bring all terms to left: -3x + 2y + 4 = 0Multiply by -1: 3x - 2y - 4 = 0. Yep, that's correct.So, I think I did everything right. The equation of the line is 3x - 2y - 4 = 0.**Final Answer**The equation of the line is boxed{3x - 2y - 4 = 0}.

💡First, I need to determine the stock price at the end of the first year after an 80% increase. Starting with the initial price of 150, an 80% increase means the stock price will rise by 0.80 multiplied by 150, which equals 120. Adding this increase to the initial price gives a new stock price of 270 at the end of the first year.Next, I need to calculate the stock price at the end of the second year after a 30% decrease from the price at the end of the first year. A 30% decrease means the stock price will drop by 0.30 multiplied by 270, which equals 81. Subtracting this decrease from the price at the end of the first year gives a final stock price of 189 at the end of the second year.

💡First, I need to determine the interval within ([-2, 3]) where the inequality ((x + 1)(x - 3) leq 0) holds true. To do this, I'll find the critical points by setting each factor equal to zero:1. (x + 1 = 0) leads to (x = -1).2. (x - 3 = 0) leads to (x = 3).These critical points divide the interval ([-2, 3]) into three sub-intervals: ([-2, -1)), ([-1, 3)), and (3). I'll test each sub-interval to see where the inequality holds:- For (x) in ([-2, -1)), let's choose (x = -1.5): [ (-1.5 + 1)(-1.5 - 3) = (-0.5)(-4.5) = 2.25 > 0 ] The inequality does not hold in this interval.- For (x) in ([-1, 3)), let's choose (x = 0): [ (0 + 1)(0 - 3) = (1)(-3) = -3 leq 0 ] The inequality holds in this interval.- At (x = 3): [ (3 + 1)(3 - 3) = (4)(0) = 0 leq 0 ] The inequality holds at this point.Therefore, the solution to the inequality is ([-1, 3]).Next, I'll calculate the probability by finding the ratio of the length of the interval where the inequality holds to the total length of the interval ([-2, 3]):- Length of the interval where the inequality holds: (3 - (-1) = 4)- Total length of the interval: (3 - (-2) = 5)Finally, the probability is:[frac{4}{5}]

💡First, I recognize that each die has 8 sides, so there are a total of 8 × 8 = 64 possible outcomes when rolling two dice.Next, I need to determine how many of these outcomes result in a sum that is 13 or more. I'll list the possible combinations for each sum:- **Sum of 13:** (5,8), (6,7), (7,6), (8,5) → 4 combinations- **Sum of 14:** (6,8), (7,7), (8,6) → 3 combinations- **Sum of 15:** (7,8), (8,7) → 2 combinations- **Sum of 16:** (8,8) → 1 combinationAdding these up, there are 4 + 3 + 2 + 1 = 10 combinations that result in a sum of 13 or more.The probability of rolling a sum of 13 or more is therefore 10/64, which simplifies to 5/32.Finally, to find the probability of rolling a sum less than 13, I subtract the probability of rolling 13 or more from 1:1 - 5/32 = 27/32