answer:To solve for the predicted value of y when x=5 using the empirical regression equation hat{y}=-x+hat{a}, we first need to determine the value of hat{a}. This involves calculating the means of x and y from the given data. 1. Calculate the mean of x values: overline{x} = frac{-2-1+0+1+2}{5} = frac{0}{5} = 0 2. Calculate the mean of y values: overline{y} = frac{5+4+2+2+1}{5} = frac{14}{5} = 2.8 Given these means, the coordinates of the center of the sample points are (0, 2.8). This point must satisfy the regression equation hat{y}=-x+hat{a}. 3. Substitute overline{x}=0 and overline{y}=2.8 into the regression equation to find hat{a}: 2.8 = -0 + hat{a} Thus, hat{a} = 2.8. 4. With hat{a} determined, the regression equation becomes: hat{y} = -x + 2.8 5. To find the predicted value of y when x=5, substitute x=5 into the equation: hat{y} = -5 + 2.8 = -2.2 Therefore, the predicted value of y when x=5 is boxed{-2.2}, which corresponds to choice boxed{D}.

answer:To prove: [ frac{operatorname{tg} alpha}{alpha} < frac{operatorname{tg} beta}{beta} ] when alpha and beta are acute angles and alpha < beta, we proceed as follows: 1. **Geometric Setup**: - Draw a unit circle centered at point O with points K, A and B on the circumference such that angle AOK = alpha and angle BOK = beta. 2. **Construct Perpendiculars**: - Drop a perpendicular AH from point A to the line OK. - Let point C be the intersection of this perpendicular with the line OB. The configuration looks like this: ``` A | | | | O-----B ``` 3. **Compare Areas**: - Compare the areas of sector OAB and triangle OAC. Using the fact that the area of a sector is frac{1}{2} r^2 theta and the area of a triangle is frac{1}{2} times text{base} times text{height}, for our unit circle (r = 1): - The area of sector OAB is frac{1}{2} (beta - alpha). - The area of triangle OAH is frac{1}{2} times OH times tanalpha. - The area of triangle OAC is frac{1}{2} times OH times (tanbeta - tanalpha). Since the triangle area must be less than the corresponding circle sector, we get: [ frac{1}{2} OH cdot (tanbeta - tanalpha) < frac{1}{2}(beta - alpha) ] Simplifying: [ OH (tanbeta - tanalpha) < beta - alpha ] 4. **Area of Sector OAK and Triangle OAH**: - The area of the sector OAK is frac{1}{2}alpha. - The area of triangle OAH is frac{1}{2}OH cdot tanalpha. Since the area of the sector must be greater: [ frac{1}{2}alpha > frac{1}{2} OH cdot tanalpha ] Simplifying: [ alpha > OH cdot tanalpha ] 5. **Combine Inequalities**: From beta - alpha < OH cdot (tanbeta - tanalpha) and alpha > OH cdot tanalpha, we get: [ frac{beta - alpha}{alpha} < frac{tanbeta - tanalpha}{tanalpha} ] which rearranges to: [ frac{beta}{alpha} < frac{tanbeta}{tanalpha} ] 6. **Conclusion**: Therefore, we conclude: [ frac{tanalpha}{alpha} < frac{tanbeta}{beta} ] Thus, we have proved that if alpha and beta are acute angles and alpha < beta, then: [ boxed{frac{operatorname{tg} alpha}{alpha} < frac{operatorname{tg} beta}{beta}} ]

answer:To find the correct formula relating x and y, the values of x from the table will be substituted into each formula choice to verify the corresponding y values. Checking Choice (A) y = x^3 - 1 - For x = 1, y = 1^3 - 1 = 0 (not 5) - Fails for x = 1, hence not a valid choice. Checking Choice (B) y = x^4 + x^3 - x^2 + x + 1 - For x = 1, y = 1^4 + 1^3 - 1^2 + 1 + 1 = 3 (not 5) - Fails for x = 1, hence not a valid choice. Checking Choice (C) y = x^2 + 3x + 1 - For x = 1, y = 1^2 + 3*1 + 1 = 5 - For x = 2, y = 2^2 + 3*2 + 1 = 13 - For x = 3, y = 3^2 + 3*3 + 1 = 25 - For x = 4, y = 4^2 + 3*4 + 1 = 41 - Succeeds for all x, thus a valid choice. Checking Choice (D) y = x^3 + x^2 + x + 1 - For x = 1, y = 1^3 + 1^2 + 1 + 1 = 4 (not 5) - Fails for x = 1, hence not a valid choice. # Conclusion: After checking, the correct formula relating x and y is given by choice (C), which is y = x^2 + 3x + 1. Thus, the answer is text{(C)}. The correct answer is boxed{(C)}, boxed{y = x^2 + 3x + 1}.

answer:To find out how many seedlings should be in each packet, you divide the total number of seedlings by the number of packets needed. So, you would divide 420 seedlings by 60 packets: 420 seedlings ÷ 60 packets = 7 seedlings per packet Therefore, there should be boxed{7} seedlings in each packet.