The weight distribution of parcels sent in a certain manner is normal with mean of 12 lb and standard deviation of 3 lb. The parcel service wishes to establish a weight value c beyond which there will be a surcharge. What value of c is such that 99% of all parcels are under the surcharge weight?

Evaluate 8x2 + 9x − 1 2x3 + 3x2 − 2x dx. solution since the degree of the numerator is less than the degree of the denominator, we don't need to divide. we factor the denominator as 2x3 + 3x2 − 2x = x(2x2 + 3x − 2) = x(2x − 1)(x + 2). since the denominator has three distinct linear factors, the partial fraction decomposition of the integrand has the form† 8x2 + 9x − 1 x(2x − 1)(x + 2) = correct: your answer is correct. to determine the values of a, b, and c, we multiply both sides of this equation by the product of the denominators, x(2x − 1)(x + 2), obtaining 8x2 + 9x − 1 = a correct: your answer is correct. (x + 2) + bx(x + 2) + cx(2x − 1).