💡First, I need to determine the probability of having at least one boy and one girl in a family with four children. To do this, I'll consider the total number of possible gender combinations for the four children.Each child has two possible genders: boy (B) or girl (G). Since the births are independent and equally likely, there are (2^4 = 16) possible combinations of genders for the four children.Next, I'll identify the scenarios that do not meet the condition of having at least one boy and one girl. These scenarios are:1. All four children are boys: BBBB2. All four children are girls: GGGGThere are 2 such scenarios.To find the number of favorable outcomes (having at least one boy and one girl), I subtract the unfavorable outcomes from the total number of possible outcomes:[16 - 2 = 14]Finally, the probability of having at least one boy and one girl is the number of favorable outcomes divided by the total number of possible outcomes:[frac{14}{16} = frac{7}{8}]

💡Alright, so I need to find the range of the function ( y = frac{x^2 + 2x + 1}{x + 2} ). Hmm, okay. Let me think about how to approach this.First, I remember that the range of a function is the set of all possible output values (y-values) it can produce. To find the range, I might need to analyze the function and see what values of y are possible given the domain of the function.Looking at the function ( y = frac{x^2 + 2x + 1}{x + 2} ), I notice that the denominator is ( x + 2 ), which means ( x ) cannot be -2 because that would make the denominator zero, and division by zero is undefined. So, ( x neq -2 ).Next, I see that the numerator is a quadratic expression: ( x^2 + 2x + 1 ). I wonder if this can be factored. Let me try factoring it:( x^2 + 2x + 1 = (x + 1)^2 ).Oh, nice! So the function simplifies to:( y = frac{(x + 1)^2}{x + 2} ).Now, since ( (x + 1)^2 ) is always non-negative (because any real number squared is non-negative), and ( x + 2 ) can be positive or negative depending on the value of x, the overall sign of y will depend on the denominator.But before I get into the signs, maybe I should try to express y in terms of x and then see what constraints there are on y.Let me set up the equation:( y = frac{(x + 1)^2}{x + 2} ).I can cross-multiply to get rid of the denominator:( y(x + 2) = (x + 1)^2 ).Expanding both sides:Left side: ( yx + 2y ).Right side: ( x^2 + 2x + 1 ).So, putting it all together:( yx + 2y = x^2 + 2x + 1 ).Now, let's bring all terms to one side to form a quadratic equation in terms of x:( x^2 + 2x + 1 - yx - 2y = 0 ).Simplify the terms:( x^2 + (2 - y)x + (1 - 2y) = 0 ).Okay, so now I have a quadratic equation in x: ( x^2 + (2 - y)x + (1 - 2y) = 0 ).For this equation to have real solutions for x, the discriminant must be non-negative. The discriminant ( D ) of a quadratic equation ( ax^2 + bx + c = 0 ) is given by ( D = b^2 - 4ac ).So, let's calculate the discriminant for our quadratic equation:( D = (2 - y)^2 - 4(1)(1 - 2y) ).Expanding ( (2 - y)^2 ):( (2 - y)^2 = 4 - 4y + y^2 ).Now, expanding the second term:( 4(1 - 2y) = 4 - 8y ).So, putting it all together:( D = (4 - 4y + y^2) - (4 - 8y) ).Simplify the expression:( D = 4 - 4y + y^2 - 4 + 8y ).Combine like terms:( D = y^2 + 4y ).So, the discriminant is ( y^2 + 4y ).For the quadratic equation to have real solutions, the discriminant must be greater than or equal to zero:( y^2 + 4y geq 0 ).Let's solve this inequality:Factor the left side:( y(y + 4) geq 0 ).This inequality holds when both factors are positive or both are negative.So, the critical points are ( y = 0 ) and ( y = -4 ).Let's test the intervals:1. When ( y < -4 ): - ( y ) is negative, ( y + 4 ) is negative. - Negative times negative is positive. - So, ( y(y + 4) geq 0 ) holds.2. When ( -4 < y < 0 ): - ( y ) is negative, ( y + 4 ) is positive. - Negative times positive is negative. - So, ( y(y + 4) geq 0 ) does not hold.3. When ( y > 0 ): - ( y ) is positive, ( y + 4 ) is positive. - Positive times positive is positive. - So, ( y(y + 4) geq 0 ) holds.At the critical points:- When ( y = -4 ): - ( y(y + 4) = (-4)(0) = 0 ), which satisfies the inequality.- When ( y = 0 ): - ( y(y + 4) = 0(4) = 0 ), which satisfies the inequality.So, the solution to the inequality ( y(y + 4) geq 0 ) is ( y leq -4 ) or ( y geq 0 ).But wait, I need to make sure that these y-values are achievable by the original function.Let me check for ( y = 0 ):If ( y = 0 ), then from the equation ( y = frac{(x + 1)^2}{x + 2} ):( 0 = frac{(x + 1)^2}{x + 2} ).This implies that ( (x + 1)^2 = 0 ), so ( x = -1 ).But ( x = -1 ) is allowed because it doesn't make the denominator zero. So, ( y = 0 ) is attainable.Now, let's check ( y = -4 ):From ( y = frac{(x + 1)^2}{x + 2} ), set ( y = -4 ):( -4 = frac{(x + 1)^2}{x + 2} ).Multiply both sides by ( x + 2 ):( -4(x + 2) = (x + 1)^2 ).Expand both sides:Left side: ( -4x - 8 ).Right side: ( x^2 + 2x + 1 ).Bring all terms to one side:( x^2 + 2x + 1 + 4x + 8 = 0 ).Combine like terms:( x^2 + 6x + 9 = 0 ).Factor:( (x + 3)^2 = 0 ).So, ( x = -3 ).Check if ( x = -3 ) is allowed: ( x + 2 = -1 neq 0 ), so it's valid.Therefore, ( y = -4 ) is also attainable.So, from the discriminant analysis, the possible y-values are ( y leq -4 ) or ( y geq 0 ).But wait, I should also consider the behavior of the function as ( x ) approaches infinity or negative infinity.As ( x ) approaches infinity, the function behaves like ( frac{x^2}{x} = x ), which goes to infinity.As ( x ) approaches negative infinity, the function behaves like ( frac{x^2}{x} = x ), which goes to negative infinity.But since we have a quadratic over linear, the function will have an oblique asymptote, which is a line. Let me find that.Divide ( x^2 + 2x + 1 ) by ( x + 2 ):Using polynomial long division:( x + 2 ) goes into ( x^2 + 2x + 1 ) how many times?First term: ( x^2 / x = x ).Multiply ( x + 2 ) by x: ( x^2 + 2x ).Subtract from the original polynomial:( (x^2 + 2x + 1) - (x^2 + 2x) = 1 ).So, the division gives ( x ) with a remainder of 1.Therefore, ( frac{x^2 + 2x + 1}{x + 2} = x + frac{1}{x + 2} ).So, as ( x ) approaches infinity or negative infinity, ( frac{1}{x + 2} ) approaches 0, and the function approaches ( y = x ).This means that the function has an oblique asymptote at ( y = x ).But how does this affect the range? Well, since the function approaches ( y = x ) at the extremes, but it's not exactly equal to ( y = x ), there might be some y-values that are not attainable.Wait, but earlier we found that the range is ( y leq -4 ) or ( y geq 0 ). Is that correct?Let me think again. The discriminant analysis suggests that y can be any real number less than or equal to -4 or greater than or equal to 0.But I should also consider if there are any restrictions based on the function's behavior.For example, when ( y = 1 ), does the equation ( y = frac{(x + 1)^2}{x + 2} ) have a solution?Let me test ( y = 1 ):( 1 = frac{(x + 1)^2}{x + 2} ).Multiply both sides by ( x + 2 ):( x + 2 = (x + 1)^2 ).Expand the right side:( x + 2 = x^2 + 2x + 1 ).Bring all terms to one side:( x^2 + 2x + 1 - x - 2 = 0 ).Simplify:( x^2 + x - 1 = 0 ).Calculate the discriminant:( D = 1^2 - 4(1)(-1) = 1 + 4 = 5 ).Since the discriminant is positive, there are two real solutions:( x = frac{-1 pm sqrt{5}}{2} ).So, ( y = 1 ) is attainable.Wait, but according to our earlier discriminant analysis, y can be any real number less than or equal to -4 or greater than or equal to 0. So, y = 1 is within the range.But what about y = -2?Let me test ( y = -2 ):( -2 = frac{(x + 1)^2}{x + 2} ).Multiply both sides by ( x + 2 ):( -2(x + 2) = (x + 1)^2 ).Expand:Left side: ( -2x - 4 ).Right side: ( x^2 + 2x + 1 ).Bring all terms to one side:( x^2 + 2x + 1 + 2x + 4 = 0 ).Simplify:( x^2 + 4x + 5 = 0 ).Calculate the discriminant:( D = 16 - 20 = -4 ).Since the discriminant is negative, there are no real solutions. So, y = -2 is not attainable.This suggests that not all y-values between -4 and 0 are attainable. So, my earlier conclusion might be incomplete.Wait, but according to the discriminant, y can be less than or equal to -4 or greater than or equal to 0. But in reality, when I tested y = -2, which is between -4 and 0, there were no real solutions.So, perhaps the range is actually y ≤ -4 or y ≥ 0, but excluding some values in between.But wait, when y = -4, we saw that x = -3 is a solution. So, y = -4 is attainable.Similarly, y = 0 is attainable at x = -1.But for values between -4 and 0, like y = -2, there are no real solutions.So, the range is actually y ≤ -4 or y ≥ 0.But let me double-check.Let me consider the function ( y = frac{(x + 1)^2}{x + 2} ).Since ( (x + 1)^2 ) is always non-negative, and ( x + 2 ) can be positive or negative, y can be positive or negative.But when ( x + 2 > 0 ) (i.e., x > -2), the denominator is positive, so y is non-negative because the numerator is non-negative.When ( x + 2 < 0 ) (i.e., x < -2), the denominator is negative, so y is non-positive because the numerator is non-negative.So, for x > -2, y ≥ 0.For x < -2, y ≤ 0.But we also need to consider the behavior near x = -2.As x approaches -2 from the right (x → -2+), the denominator approaches 0 from the positive side, and the numerator approaches ( (-2 + 1)^2 = 1 ). So, y approaches positive infinity.As x approaches -2 from the left (x → -2-), the denominator approaches 0 from the negative side, and the numerator approaches 1. So, y approaches negative infinity.So, combining these observations:- For x > -2, y ≥ 0, and y can take any value from 0 to positive infinity.- For x < -2, y ≤ 0, but we need to see what values y can take.Wait, earlier when I set y = -2, there were no real solutions. So, perhaps y cannot take all values between negative infinity and 0, but only up to y = -4.Wait, let's think about this.When x < -2, y is negative. Let me see what is the minimum value of y in this region.Let me set x = -3, which is less than -2.Then y = ( frac{(-3 + 1)^2}{-3 + 2} = frac{4}{-1} = -4 ).So, y = -4 is attainable at x = -3.Now, what happens as x approaches -2 from the left?As x approaches -2-, the denominator approaches 0 from the negative side, and the numerator approaches 1. So, y approaches negative infinity.But wait, earlier when I set y = -2, there were no real solutions, meaning that y cannot be -2. So, perhaps the function only takes y-values less than or equal to -4 and greater than or equal to 0.Wait, but when x is less than -2, y is negative, but can y take any negative value?Wait, let's consider the function for x < -2.Let me express y in terms of x:( y = frac{(x + 1)^2}{x + 2} ).Let me make a substitution: let t = x + 2, so t < 0 because x < -2.Then x = t - 2.So, ( y = frac{(t - 2 + 1)^2}{t} = frac{(t - 1)^2}{t} ).Since t < 0, let me write t = -s, where s > 0.So, ( y = frac{(-s - 1)^2}{-s} = frac{(s + 1)^2}{-s} = -frac{(s + 1)^2}{s} ).Now, let's analyze this expression:( y = -frac{(s + 1)^2}{s} ).Since s > 0, the denominator is positive, and the numerator is positive, so y is negative.Let me see what values y can take.Let me set ( y = -k ), where k > 0.So, ( -k = -frac{(s + 1)^2}{s} ).Multiply both sides by -1:( k = frac{(s + 1)^2}{s} ).Simplify:( k = frac{s^2 + 2s + 1}{s} = s + 2 + frac{1}{s} ).So, ( k = s + 2 + frac{1}{s} ).Now, since s > 0, let's find the minimum value of k.Take the derivative of k with respect to s:( dk/ds = 1 - frac{1}{s^2} ).Set derivative equal to zero to find critical points:( 1 - frac{1}{s^2} = 0 ).So, ( 1 = frac{1}{s^2} ).Multiply both sides by ( s^2 ):( s^2 = 1 ).So, ( s = 1 ) (since s > 0).Now, check the second derivative to confirm if it's a minimum:( d^2k/ds^2 = frac{2}{s^3} ).At s = 1, ( d^2k/ds^2 = 2 > 0 ), so it's a minimum.So, the minimum value of k is:( k = 1 + 2 + 1 = 4 ).Therefore, the minimum value of y in this region is ( y = -4 ).So, for x < -2, y can take any value from negative infinity up to y = -4.But wait, earlier when I set y = -2, there were no real solutions, which suggests that y cannot be between -4 and 0.Wait, but according to this analysis, y can be less than or equal to -4.So, combining both regions:- For x > -2, y ≥ 0.- For x < -2, y ≤ -4.Therefore, the range of the function is ( y leq -4 ) or ( y geq 0 ).But earlier, when I set y = 1, which is greater than 0, there were real solutions, so y = 1 is attainable.Similarly, y = -4 is attainable at x = -3.But what about y = -5?Let me test y = -5:( -5 = frac{(x + 1)^2}{x + 2} ).Multiply both sides by ( x + 2 ):( -5(x + 2) = (x + 1)^2 ).Expand:Left side: ( -5x - 10 ).Right side: ( x^2 + 2x + 1 ).Bring all terms to one side:( x^2 + 2x + 1 + 5x + 10 = 0 ).Simplify:( x^2 + 7x + 11 = 0 ).Calculate the discriminant:( D = 49 - 44 = 5 ).Since the discriminant is positive, there are two real solutions:( x = frac{-7 pm sqrt{5}}{2} ).So, y = -5 is attainable.Therefore, the range is indeed all real numbers y such that y ≤ -4 or y ≥ 0.But wait, earlier I thought y = -2 was not attainable, which is correct because y cannot be between -4 and 0.So, the range is ( (-infty, -4] cup [0, infty) ).But let me double-check with another value, say y = -3.Set y = -3:( -3 = frac{(x + 1)^2}{x + 2} ).Multiply both sides by ( x + 2 ):( -3(x + 2) = (x + 1)^2 ).Expand:Left side: ( -3x - 6 ).Right side: ( x^2 + 2x + 1 ).Bring all terms to one side:( x^2 + 2x + 1 + 3x + 6 = 0 ).Simplify:( x^2 + 5x + 7 = 0 ).Calculate the discriminant:( D = 25 - 28 = -3 ).Negative discriminant, so no real solutions. Therefore, y = -3 is not attainable.This confirms that y cannot be between -4 and 0.So, putting it all together, the range of the function ( y = frac{x^2 + 2x + 1}{x + 2} ) is all real numbers y such that y ≤ -4 or y ≥ 0.Therefore, the range is ( (-infty, -4] cup [0, infty) ).

💡Okay, so I have this problem here: Between 7 and 8 o'clock, when is the angle between the hour and minute hands exactly 84 degrees? The options are given as (A) 7:23 and 7:53, (B) 7:20 and 7:50, (C) 7:22 and 7:53, (D) 7:23 and 7:52, and (E) 7:21 and 7:49. I need to figure out which one is correct.First, I remember that the angle between the hour and minute hands can be calculated using a formula. I think it's something like |30H - 5.5M|, where H is the hour and M is the minutes past the hour. Let me confirm that. Yeah, I think that's right because the hour hand moves at 30 degrees per hour and the minute hand moves at 6 degrees per minute, so the relative speed is 5.5 degrees per minute.So, in this case, the hour is 7, so H is 7. The angle is given as 84 degrees. So, I can set up the equation: |30*7 - 5.5M| = 84. That simplifies to |210 - 5.5M| = 84.This absolute value equation means that 210 - 5.5M can be either 84 or -84. So, I need to solve both equations:1. 210 - 5.5M = 842. 210 - 5.5M = -84Let me solve the first equation:210 - 5.5M = 84Subtract 210 from both sides:-5.5M = 84 - 210-5.5M = -126Now, divide both sides by -5.5:M = (-126)/(-5.5)M = 126 / 5.5Hmm, 126 divided by 5.5. Let me calculate that. 5.5 goes into 126 how many times? 5.5 times 20 is 110, so 126 - 110 is 16. 5.5 goes into 16 about 2.909 times. So, 20 + 2.909 is approximately 22.909 minutes. So, around 22.91 minutes past 7, which is approximately 7:23.Now, let's solve the second equation:210 - 5.5M = -84Subtract 210 from both sides:-5.5M = -84 - 210-5.5M = -294Divide both sides by -5.5:M = (-294)/(-5.5)M = 294 / 5.5Again, let me calculate that. 5.5 goes into 294 how many times? 5.5 times 50 is 275, so 294 - 275 is 19. 5.5 goes into 19 about 3.454 times. So, 50 + 3.454 is approximately 53.454 minutes. So, around 53.45 minutes past 7, which is approximately 7:53.So, the two times when the angle between the hour and minute hands is 84 degrees between 7 and 8 o'clock are approximately 7:23 and 7:53. Looking at the options, that's option (A).Wait, let me double-check my calculations to make sure I didn't make a mistake. For the first equation, 210 - 5.5M = 84, solving for M:210 - 84 = 5.5M126 = 5.5MM = 126 / 5.5Yes, that's 22.909, which is approximately 23 minutes.For the second equation, 210 - 5.5M = -84:210 + 84 = 5.5M294 = 5.5MM = 294 / 5.5Which is 53.454, approximately 53 minutes.So, yes, 7:23 and 7:53 are the correct times. Therefore, the answer should be option (A).**Final Answer**boxed{A}

💡First, I need to determine Amy's hourly wage based on her summer earnings. She earned 3200 by working 40 hours per week for 8 weeks. To find the hourly rate, I'll divide the total earnings by the total number of hours worked.Next, with the hourly rate established, I'll calculate the total number of hours Amy needs to work during the school year to earn 6400. This involves dividing the desired earnings by the hourly rate.Finally, to find out how many hours Amy must work each week during the school year, I'll divide the total required hours by the number of weeks she plans to work, which is 24 weeks.

💡Okay, so I have this equation here: x squared minus 6x minus 1 equals zero. I need to rewrite it in the form of (x plus m) squared equals n using the method of completing the square. Hmm, I remember that completing the square is a technique used to solve quadratic equations by turning them into a perfect square trinomial, which can then be easily solved. Let me try to recall the steps involved.First, I think I need to get the equation into a form where the coefficient of x squared is 1, which it already is here. So, that's good. The equation is x² - 6x - 1 = 0. I should probably move the constant term to the other side to make it easier to work with. So, I'll add 1 to both sides. That would give me x² - 6x = 1.Now, the next step is to complete the square. To do this, I need to take the coefficient of x, which is -6, divide it by 2, and then square it. Let me write that down. So, the coefficient of x is -6. Dividing that by 2 gives me -3, and squaring that gives me 9. So, I need to add 9 to both sides of the equation to maintain the balance.Adding 9 to both sides: On the left side, x² - 6x + 9, and on the right side, 1 + 9, which is 10. So now, the equation becomes x² - 6x + 9 = 10.Now, the left side should be a perfect square trinomial. Let me check if that's the case. The expression x² - 6x + 9 can be factored into (x - 3) squared because (x - 3)(x - 3) equals x² - 6x + 9. So, that's correct.Therefore, the equation can be rewritten as (x - 3)² = 10. So, in the form (x + m)² = n, m is -3 and n is 10. That makes sense because when you expand (x - 3)², you get x² - 6x + 9, which matches the left side of the equation after completing the square.Let me just double-check my steps to make sure I didn't make any mistakes. Starting with x² - 6x - 1 = 0, I moved the -1 to the other side by adding 1 to both sides, resulting in x² - 6x = 1. Then, I took half of -6, which is -3, squared it to get 9, and added 9 to both sides. That gave me x² - 6x + 9 = 10, which factors to (x - 3)² = 10. Yep, that all checks out.I think I've got it right. So, the equation in the form (x + m)² = n is (x - 3)² = 10. I don't see any errors in my calculations, so I feel confident about this answer.

💡Okay, so I have this problem where a body is moving along the Ox axis, and there's a graph showing how its speed changes over time. I need to figure out how far the body has traveled after 6 seconds. Hmm, let me think about how to approach this.First, I remember that distance traveled can be found by calculating the area under the velocity-time graph. That makes sense because velocity is the rate of change of position, so integrating velocity over time gives the total displacement. But wait, is it displacement or total distance? I think since the problem asks for the distance traveled, I need to consider the absolute value of the velocity, right? So even if the body moves backward, it still counts towards the total distance.Alright, so I need to look at the graph and figure out the areas of each section. Let me try to visualize the graph based on the description. It says the speed depends on time, so the graph has time on the x-axis and speed on the y-axis.From what I understand, the graph probably has different segments where the speed changes at certain times. Maybe it starts at a certain speed, then changes after a second, then another change, and so on. I need to identify these segments and calculate the area for each.Let me break it down second by second since the total time is 6 seconds. That might make it easier.1. **First second (t=0 to t=1):** I think the speed is constant here. Maybe it's 2 m/s? So, the area for this part would be a rectangle with height 2 and width 1, so 2*1=2 meters.2. **Second second (t=1 to t=2):** The speed might decrease here. Maybe it's 1 m/s? So, another rectangle with height 1 and width 1, giving 1*1=1 meter.3. **Third and fourth seconds (t=2 to t=4):** Hmm, perhaps the speed drops to zero here? So, the area would be zero for these two seconds.4. **Fifth second (t=4 to t=5):** Maybe the speed increases again to 2 m/s? So, another rectangle with height 2 and width 1, which is 2 meters.5. **Sixth second (t=5 to t=6):** It might drop back to zero. So, again, zero area.Wait, let me make sure I'm interpreting this correctly. If the graph has different segments, I need to ensure I'm accounting for each correctly. Maybe I should sketch a rough graph based on these assumptions.- From 0 to 1 second: 2 m/s- From 1 to 2 seconds: 1 m/s- From 2 to 4 seconds: 0 m/s- From 4 to 5 seconds: 2 m/s- From 5 to 6 seconds: 0 m/sIf I plot this, it would have a step-like pattern, changing speed at each second mark except between 2-4 and 5-6 seconds where it's zero.Now, calculating the distance for each interval:1. **0-1 seconds:** 2 m/s * 1 s = 2 meters2. **1-2 seconds:** 1 m/s * 1 s = 1 meter3. **2-4 seconds:** 0 m/s * 2 s = 0 meters4. **4-5 seconds:** 2 m/s * 1 s = 2 meters5. **5-6 seconds:** 0 m/s * 1 s = 0 metersAdding these up: 2 + 1 + 0 + 2 + 0 = 5 meters.Wait, that seems straightforward, but let me double-check. Did I miss any segments where the speed might be negative? The problem mentions the body moves along the Ox axis, but it doesn't specify direction. If speed is given, it's the magnitude, so even if it's moving backward, it still contributes positively to the total distance.But in this case, the speeds are all positive or zero, so I don't have to worry about subtracting any areas. That simplifies things.Is there any chance the graph has a different shape? Like maybe a triangular section or a curve? The problem doesn't specify, so I think assuming constant speeds at each interval is safe.Also, since the total time is 6 seconds, and I've accounted for each second, I think I've covered the entire period.So, summing up the distances:- 2 meters (first second)- 1 meter (second second)- 0 meters (third and fourth seconds)- 2 meters (fifth second)- 0 meters (sixth second)Total distance = 2 + 1 + 0 + 2 + 0 = 5 meters.I think that's it. The body travels a total distance of 5 meters in 6 seconds based on the given speed-time graph.