answer:Let's denote the defective rate of the products checked by worker y as ( d_y ) (in percentage). We know that worker x checked ( 1 - 0.1666666666666668 = 0.8333333333333332 ) (or ( frac{5}{6} )) fraction of the products, and worker y checked ( 0.1666666666666668 ) (or ( frac{1}{6} )) fraction of the products. The total defective rate is a weighted average of the defective rates of worker x and worker y, based on the fraction of products they checked. The formula for the total defective rate is: [ 0.55% = 0.8333333333333332 times 0.5% + 0.1666666666666668 times d_y ] Now we can solve for ( d_y ): [ 0.55% = 0.8333333333333332 times 0.5% + 0.1666666666666668 times d_y ] [ 0.55% = 0.4166666666666666% + 0.1666666666666668 times d_y ] [ 0.55% - 0.4166666666666666% = 0.1666666666666668 times d_y ] [ 0.1333333333333334% = 0.1666666666666668 times d_y ] [ d_y = frac{0.1333333333333334%}{0.1666666666666668} ] [ d_y = 0.8% ] Therefore, the defective rate of the products checked by worker y is boxed{0.8%} .

answer:To find the coordinates of the vertex of the parabola given by the equation y = x^2 - 2, we can use the fact that the vertex form of a parabola is given by y = a(x-h)^2 + k, where (h, k) are the coordinates of the vertex. For the given parabola, y = x^2 - 2, it is already in a form where we can identify the vertex directly. The coefficient a is 1, and there is no (x-h)^2 term, which means h = 0. The constant term -2 tells us that k = -2. Therefore, the coordinates of the vertex are (h, k) = (0, -2). So, step by step, we have: 1. Identify the form of the parabola: y = x^2 - 2. 2. Recognize that there is no (x-h)^2 term, so h = 0. 3. Identify the constant term -2 as k, so k = -2. 4. Conclude that the vertex of the parabola is at (h, k) = (0, -2). Therefore, the correct answer is boxed{D}.

answer:1. Let ( pH_1 = 3.55 ) and ( pH_2 = 9.6 ). The charges at these pH values are ( Q_1 = -frac{1}{3} ) and ( Q_2 = frac{1}{2} ), respectively. 2. Since the charge increases linearly with pH, we can use the linear interpolation formula to find the isoelectric point (pH at which the charge is zero). 3. The linear relationship between charge ( Q ) and pH can be expressed as: [ Q = m cdot (text{pH}) + b ] where ( m ) is the slope and ( b ) is the y-intercept. 4. To find the slope ( m ): [ m = frac{Q_2 - Q_1}{pH_2 - pH_1} = frac{frac{1}{2} - left(-frac{1}{3}right)}{9.6 - 3.55} = frac{frac{1}{2} + frac{1}{3}}{6.05} = frac{frac{3}{6} + frac{2}{6}}{6.05} = frac{frac{5}{6}}{6.05} = frac{5}{6 cdot 6.05} = frac{5}{36.3} ] 5. Simplify the slope: [ m = frac{5}{36.3} approx 0.1377 ] 6. Using the point-slope form of the linear equation, we can find the y-intercept ( b ): [ Q_1 = m cdot pH_1 + b implies -frac{1}{3} = 0.1377 cdot 3.55 + b ] [ -frac{1}{3} = 0.4888 + b implies b = -frac{1}{3} - 0.4888 approx -0.8211 ] 7. The equation of the line is: [ Q = 0.1377 cdot (text{pH}) - 0.8211 ] 8. To find the isoelectric point (where ( Q = 0 )): [ 0 = 0.1377 cdot (text{pH}) - 0.8211 implies 0.1377 cdot (text{pH}) = 0.8211 implies text{pH} = frac{0.8211}{0.1377} approx 5.97 ] The final answer is ( boxed{5.97} ).

answer:1. **Assign Variables:** Let the shorter segment of the side of length 90 be denoted as x, and the altitude to this side be y. Hence, the longer segment will be 90 - x. 2. **Apply the Pythagorean Theorem:** - For the right triangle with sides 40, y, and x: [ 40^2 = x^2 + y^2 quad text{(Equation 1)} ] - For the right triangle with sides 50, y, and 90-x: [ 50^2 = (90-x)^2 + y^2 quad text{(Equation 2)} ] 3. **Eliminate y^2 and Solve for x:** Subtract Equation 1 from Equation 2: [ 50^2 - 40^2 = (90-x)^2 - x^2 ] Simplifying both sides: [ 2500 - 1600 = 900 = 8100 - 180x + x^2 - x^2 ] [ 180x = 3600 ] [ x = frac{3600}{180} = 20 ] 4. **Find the Larger Segment:** [ text{Larger segment} = 90 - 20 = 70 ] The larger segment cut off on the side of length 90 is 70. The final answer is boxed{textbf{(C)} 70}