An experiment was conducted to investigate the the effectiveness of several insecticides. One-hundred insects of a particular species were put into a chamber and exposed to an insecticide for 15 seconds. The procedure was applied in random order six times for each of the four brands of insecticides (Brand A, Brand B, Brand C, Brand D). The number of dead insects out of 100 were observed for each chamber after exposure. To analyze the results, the investigators used ANOVA (Analysis of Variance). For this setting, what is the correct value for the degrees of freedom for the treatments

no, x cannot be the same number. in all of them, x is 1. for one input there is another output. g is not a function

give brainliest

irst, find the z score of a score of 80. this is given by the formula z = (x-u)/o, where x is the score, u is the mean, and o is the standard deviation. this gives

z = (80-78)/12 = 1/6 = .1667

now, find what percentile a z score of 1/6 correlates to in the normal distribution.

a z score of 1/6 = .1667 is equivalent to the 56.36th percentile of the distribution. so the probability a score is over this is 1-.5636 = .4364

the mean of a sample mean is equal to the original mean. however, the standard deviation of a sample mean is equal to the original standard deviation divided by the square root of n (sample size). so the new standard deviation is 12/sqrt(16) = 3.

the new z score is 80-78/3 = 2/3 = .667. this is equivalent to the 74.86th percentile. so the probability that the mean will be greater than this score is 1-.7486 = .2514.

the answer is a and d

if i did my math right then the answer should but 108