answer：First, I need to evaluate the inner function ( f(-4) ). Since (-4) is less than (0), I use the second part of the piecewise function, which is ( left( frac{1}{2} right)^x ). Plugging in (-4), I get ( left( frac{1}{2} right)^{-4} ).Next, I simplify ( left( frac{1}{2} right)^{-4} ). A negative exponent means taking the reciprocal, so this becomes ( 2^4 ), which equals (16).Now, I need to evaluate ( f(16) ). Since (16) is greater than or equal to (0), I use the first part of the piecewise function, which is ( sqrt{x} ). Plugging in (16), I get ( sqrt{16} ).Finally, I calculate ( sqrt{16} ), which equals (4). Therefore, the value of ( f[f(-4)] ) is (4).

answer：To determine the probability that point ( A(m, n) ) lies in the fourth quadrant, we first identify the conditions for a point to be in the fourth quadrant: the x-coordinate ( m ) must be positive, and the y-coordinate ( n ) must be negative.Given the numbers on the cards are ( 0 ), ( -1 ), ( 2 ), and ( -3 ), we draw one card for the x-coordinate and another for the y-coordinate without replacement. This results in a total of ( 4 times 3 = 12 ) possible outcomes.We then analyze the pairs where ( m ) is positive and ( n ) is negative:- When ( m = 2 ), the possible negative values for ( n ) are ( -1 ) and ( -3 ), resulting in the pairs ( (2, -1) ) and ( (2, -3) ).Thus, there are 2 favorable outcomes out of 12 possible outcomes. Therefore, the probability is:[frac{2}{12} = frac{1}{6}]

answer：First, I'll denote the height of the first person as ( h ).Since the difference between the first and second person is 2 inches, the height of the second person is ( h + 2 ) inches.Similarly, the height of the third person is ( h + 4 ) inches.The difference between the third and fourth person is 6 inches, so the height of the fourth person is ( h + 10 ) inches.The average height of the four people is 76 inches. Therefore, the total sum of their heights is ( 4 times 76 = 304 ) inches.Setting up the equation for the sum of the heights:[h + (h + 2) + (h + 4) + (h + 10) = 304]Simplifying the equation:[4h + 16 = 304]Solving for ( h ):[4h = 288 h = 72]Finally, calculating the height of the fourth person:[text{Height of fourth person} = 72 + 10 = 82 text{ inches}]

answer：Okay, so I have this problem here: If ( x = -1 ) is a point of local minimum for the function ( f(x) = x(x - a)^2 ), then what is the value of ( a )? The options are A: 0, B: -1, C: -2, D: -3.Hmm, let's see. I remember that to find local minima or maxima, I need to take the derivative of the function and set it equal to zero. Then, I can use the second derivative test to check if it's a minimum or maximum.First, let me write down the function again: ( f(x) = x(x - a)^2 ). I need to find its first derivative. I'll use the product rule for differentiation because it's a product of two functions: ( u(x) = x ) and ( v(x) = (x - a)^2 ).The product rule says that ( f'(x) = u'(x)v(x) + u(x)v'(x) ). Let's compute each part.First, ( u'(x) ) is the derivative of ( x ), which is 1. Then, ( v(x) = (x - a)^2 ), so ( v'(x) ) is 2(x - a) times the derivative of the inside, which is 1. So, ( v'(x) = 2(x - a) ).Putting it all together: ( f'(x) = 1 cdot (x - a)^2 + x cdot 2(x - a) ).Simplify that: ( f'(x) = (x - a)^2 + 2x(x - a) ). Maybe I can factor out ( (x - a) ) from both terms. Let's try that:( f'(x) = (x - a)[(x - a) + 2x] ).Simplify inside the brackets: ( (x - a) + 2x = 3x - a ).So, ( f'(x) = (x - a)(3x - a) ).Wait, is that right? Let me check my steps again. I had ( (x - a)^2 + 2x(x - a) ). Factoring out ( (x - a) ) gives ( (x - a)[(x - a) + 2x] ). Then, inside the brackets: ( x - a + 2x = 3x - a ). Yeah, that seems correct.So, ( f'(x) = (x - a)(3x - a) ).Now, since ( x = -1 ) is a point of local minimum, the first derivative at ( x = -1 ) should be zero. So, let's plug ( x = -1 ) into ( f'(x) ):( f'(-1) = (-1 - a)(3(-1) - a) = 0 ).So, either ( (-1 - a) = 0 ) or ( (-3 - a) = 0 ).Solving ( -1 - a = 0 ) gives ( a = -1 ).Solving ( -3 - a = 0 ) gives ( a = -3 ).So, possible values for ( a ) are -1 and -3.But we need to check which one actually gives a local minimum at ( x = -1 ). For that, we can use the second derivative test.First, let's find the second derivative ( f''(x) ).We have ( f'(x) = (x - a)(3x - a) ). Let's expand this first:( f'(x) = 3x^2 - a x - 3a x + a^2 = 3x^2 - 4a x + a^2 ).Wait, is that correct? Let me multiply it out:( (x - a)(3x - a) = x cdot 3x + x cdot (-a) - a cdot 3x + (-a) cdot (-a) = 3x^2 - a x - 3a x + a^2 = 3x^2 - 4a x + a^2 ). Yeah, that's correct.Now, take the derivative of ( f'(x) ) to get ( f''(x) ):( f''(x) = 6x - 4a ).Now, evaluate ( f''(x) ) at ( x = -1 ):( f''(-1) = 6(-1) - 4a = -6 - 4a ).For a local minimum, the second derivative should be positive. So, ( f''(-1) > 0 ).Let's plug in the possible values of ( a ):First, ( a = -1 ):( f''(-1) = -6 - 4(-1) = -6 + 4 = -2 ). That's negative, which would indicate a local maximum, not a minimum. So, ( a = -1 ) is not the answer.Next, ( a = -3 ):( f''(-1) = -6 - 4(-3) = -6 + 12 = 6 ). That's positive, so ( x = -1 ) is indeed a local minimum when ( a = -3 ).Therefore, the correct answer is D: -3.Wait, let me just double-check my steps to make sure I didn't make a mistake.1. Calculated the first derivative correctly using the product rule, got ( f'(x) = (x - a)(3x - a) ).2. Plugged in ( x = -1 ) and found ( a = -1 ) and ( a = -3 ).3. Calculated the second derivative ( f''(x) = 6x - 4a ).4. Evaluated at ( x = -1 ) for both ( a ) values: - For ( a = -1 ), got ( f''(-1) = -2 ) (local maximum). - For ( a = -3 ), got ( f''(-1) = 6 ) (local minimum).Everything seems to check out. So, yeah, the answer should be D: -3.

answer：First, I need to determine the range of the height data. The shortest height is 142 cm, and the tallest is 168 cm. This gives a total range of 26 cm.Next, I'll divide this range by the class interval of 4 cm to estimate the number of classes. 26 cm divided by 4 cm per class equals 6.5. Since we can't have half a class, I'll round up to 7 classes.To ensure accuracy, I'll list out the class intervals to cover the entire range:- 142–146 cm- 146–150 cm- 150–154 cm- 154–158 cm- 158–162 cm- 162–166 cm- 166–170 cmThis confirms that 7 classes are needed to cover all the data points without overlap.

answer：Alright, so I have this geometry problem about a square ABCD. E and F are midpoints of sides AB and AD respectively, and P is the intersection point of lines CF and DE. The problem has two parts: first, I need to show that DE is perpendicular to CF, and second, I need to determine the ratio of CF to PC to EP. Let me start with part (a). I remember that to show two lines are perpendicular, their slopes should multiply to -1. So maybe I can assign coordinates to the square and calculate the slopes of DE and CF. That sounds like a plan.Let's assume the square has side length 2 units for simplicity, so that the midpoints will have integer coordinates. Let me place the square on a coordinate system with point D at the origin (0,0). Then, the coordinates of the other points would be: A at (2,0), B at (2,2), and C at (0,2). Now, E is the midpoint of AB, so AB goes from (2,0) to (2,2). The midpoint E would be at (2,1). Similarly, F is the midpoint of AD, which goes from (2,0) to (0,0). Wait, no, AD goes from A(2,0) to D(0,0), so the midpoint F would be at (1,0). Okay, so E is (2,1) and F is (1,0). Now, I need to find the equations of lines DE and CF. First, line DE connects points D(0,0) and E(2,1). The slope of DE is (1 - 0)/(2 - 0) = 1/2. So the equation of DE is y = (1/2)x.Next, line CF connects points C(0,2) and F(1,0). The slope of CF is (0 - 2)/(1 - 0) = -2. So the equation of CF is y - 2 = -2(x - 0), which simplifies to y = -2x + 2.Now, to check if DE and CF are perpendicular, I can multiply their slopes. The slope of DE is 1/2, and the slope of CF is -2. Multiplying these gives (1/2)*(-2) = -1. Since the product of the slopes is -1, DE is indeed perpendicular to CF. That takes care of part (a).Moving on to part (b), I need to find the ratio CF : PC : EP. Hmm, okay. Let me visualize the square again. Points C, F, P, and E are all part of this. Since P is the intersection of DE and CF, I can find the coordinates of P by solving the equations of DE and CF.We already have the equations: DE is y = (1/2)x and CF is y = -2x + 2. Setting them equal to find the intersection:(1/2)x = -2x + 2Multiply both sides by 2 to eliminate the fraction:x = -4x + 4Bring -4x to the left:x + 4x = 45x = 4x = 4/5Now, substitute x back into DE's equation to find y:y = (1/2)(4/5) = 2/5So, point P has coordinates (4/5, 2/5).Now, I need to find the lengths of CF, PC, and EP. Let's start with CF. Points C(0,2) and F(1,0). The distance formula is sqrt[(x2 - x1)^2 + (y2 - y1)^2].So, CF = sqrt[(1 - 0)^2 + (0 - 2)^2] = sqrt[1 + 4] = sqrt[5].Next, PC. Points P(4/5, 2/5) and C(0,2). Using the distance formula:PC = sqrt[(0 - 4/5)^2 + (2 - 2/5)^2] = sqrt[(16/25) + ( (8/5)^2 )] = sqrt[16/25 + 64/25] = sqrt[80/25] = sqrt[16*5/25] = (4/5)sqrt(5).Wait, let me double-check that calculation. (2 - 2/5) is 8/5, so squared is 64/25. (0 - 4/5) is -4/5, squared is 16/25. So total is 16/25 + 64/25 = 80/25 = 16/5. So sqrt(16/5) is 4/sqrt(5) which rationalizes to (4 sqrt(5))/5. So PC is (4 sqrt(5))/5.Now, EP. Points E(2,1) and P(4/5, 2/5). Using the distance formula:EP = sqrt[(4/5 - 2)^2 + (2/5 - 1)^2] = sqrt[(4/5 - 10/5)^2 + (2/5 - 5/5)^2] = sqrt[(-6/5)^2 + (-3/5)^2] = sqrt[(36/25) + (9/25)] = sqrt[45/25] = sqrt[9/5] = 3/sqrt(5) = (3 sqrt(5))/5.So now, we have:CF = sqrt(5)PC = (4 sqrt(5))/5EP = (3 sqrt(5))/5To find the ratio CF : PC : EP, let's express them all in terms of sqrt(5):CF = sqrt(5) = (5 sqrt(5))/5PC = (4 sqrt(5))/5EP = (3 sqrt(5))/5So, the ratio is (5 sqrt(5))/5 : (4 sqrt(5))/5 : (3 sqrt(5))/5, which simplifies to 5 : 4 : 3.Wait, but hold on. The problem asks for CF : PC : EP. From our calculations, CF is sqrt(5), PC is (4 sqrt(5))/5, and EP is (3 sqrt(5))/5. So in terms of multiples of sqrt(5)/5, CF is 5*(sqrt(5)/5), PC is 4*(sqrt(5)/5), and EP is 3*(sqrt(5)/5). So the ratio CF : PC : EP is 5 : 4 : 3.But let me double-check because sometimes ratios can be tricky. If I think about the lengths:CF = sqrt(5) ≈ 2.236PC = (4 sqrt(5))/5 ≈ 1.789EP = (3 sqrt(5))/5 ≈ 1.342So, CF is the longest, followed by PC, then EP. So the ratio from largest to smallest is CF : PC : EP = 5 : 4 : 3.But wait, in the problem statement, it's just CF : PC : EP, not necessarily in any order. But since CF is the entire length, PC is a part of CF, and EP is another segment, the ratio should be CF : PC : EP as 5 : 4 : 3.Alternatively, sometimes ratios are expressed in the order given, so CF : PC : EP would be 5 : 4 : 3.But let me think again. If CF is 5 units, PC is 4 units, and EP is 3 units, then CF : PC : EP is 5:4:3.Yes, that seems correct.So, summarizing:a) DE is perpendicular to CF because the product of their slopes is -1.b) The ratio CF : PC : EP is 5 : 4 : 3.Wait, but in my earlier calculation, CF was sqrt(5), PC was (4 sqrt(5))/5, and EP was (3 sqrt(5))/5. So, if I factor out sqrt(5)/5, I get:CF = 5*(sqrt(5)/5)PC = 4*(sqrt(5)/5)EP = 3*(sqrt(5)/5)So, the ratio is 5:4:3.Yes, that makes sense.I think that's the answer.**Final Answer**a) boxed{DE perp CF}b) boxed{5 : 4 : 3}