PLEASE HELP ME! Tom surveyed students in his English class and found that the mean time spent reading outside school is 35 minutes per day. Select all the inferences that are NOT valid. * A. All students in Tom’s school read for 35 minutes per day.
B. All students in Tom’s English class read for 35 minutes per day.
C. Most students in Tom’s English class read for 35 minutes per day.
D. Tom reads for 35 minutes a day.
E. Some students in Tom’s English class read for more than 35 minutes per day

Simplify.7x + 3y - 2 + 6x + 1 + y squared 2 you! ; )

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).

(co 3) the times that customers spend in a book store are normally distributed with a mean of 39.5 minutes and a standard deviation of 9.4 minutes. a random sample of 25 customers has a mean of 36.1 minutes or less. would this outcome be considered unusual, so that the store should reconsider its displays? no, the probability of this outcome at 0.035, would be considered usual, so there is no problem yes, the probability of this outcome at 0.035, would be considered unusual, so the display should be redone no the probability of this outcome at 0.359 would be considered usual, so there is no problem yes, the probability of this outcome at 0.965 would be considered unusual, so the display should be redone