Consider the graph with four lines below. On a coordinate plane, line a has a positive slope and goes through points (negative 1, 0) and (1, 2), line b has a negative slope and goes through (negative 2, 2) and (negative 1, negative 1), line c has a negative slope and goes through (0, 3) and (1, 0), and line d is horizontal at y = 1. By inspection, which system would have no solution? line a and line b line a and line c line b and line c line b and line d

Benito is selecting samples of five values from the table, which row will result in the greatest mean? population data row 1 4 2 2 3 3 row 2 3 3 4 3 2 row 3 2 4 3 4 3 row 4 3 4 4 7 3 row 1 row 2 r

Determine if the following statement is true or false. the normal curve is symmetric about its​ mean, mu. choose the best answer below. a. the statement is false. the normal curve is not symmetric about its​ mean, because the mean is the balancing point of the graph of the distribution. the median is the point where​ 50% of the area under the distribution is to the left and​ 50% to the right.​ therefore, the normal curve could only be symmetric about its​ median, not about its mean. b. the statement is true. the normal curve is a symmetric distribution with one​ peak, which means the​ mean, median, and mode are all equal.​ therefore, the normal curve is symmetric about the​ mean, mu. c. the statement is false. the mean is the balancing point for the graph of a​ distribution, and​ therefore, it is impossible for any distribution to be symmetric about the mean. d. the statement is true. the mean is the balancing point for the graph of a​ distribution, and​ therefore, all distributions are symmetric about the mean.

Asample of 200 rom computer chips was selected on each of 30 consecutive days, and the number of nonconforming chips on each day was as follows: the data has been given so that it can be copied into r as a vector. non.conforming = c(10, 15, 21, 19, 34, 16, 5, 24, 8, 21, 32, 14, 14, 19, 18, 20, 12, 23, 10, 19, 20, 18, 13, 26, 33, 14, 12, 21, 12, 27) #construct a p chart by using the following code. you will need to enter your values for pbar, lcl and ucl. pbar = lcl = ucl = plot(non.conforming/200, ylim = c(0,.5)) abline(h = pbar, lty = 2) abline(h = lcl, lty = 3) abline(h = ucl, lty = 3)