A 99% confidence interval for the mean μ of a population is computed from a random sample and found to be 6 ± 3. We may conclude that:

A. there is a 99% probability that μ is between 3 and 9.
B. there is a 99% probability that the true mean is 6, and there is a 99% chance that the true margin of error is 3.
C. All of the above.
D. if we took many, many additional random samples, and from each computed a 99% confidence interval for μ, approximately 99% of these intervals would contain μ

120 seems like the most sensible answer.

step-by-step explanation:

if there is only one topping on all caramel apples and there's 120 of them. then all apples should/will be covered in sprinkles. make sense?

in 1995, potato plants were taken into space aboard the space shuttle columbia.

step-by-step explanation:

no oblique asymptote

step-by-step explanation:

to find the oblique asymptote of f(x)

here numerator has degree 1 and denomintor degree 2

degree of numerator < degree of denominator

hence there is no oblique asymptote

when the degree of the denominator > that of the numerator, we saw that the value in the denominator got so much bigger, so quickly,   giving us a horizontal asymptote of the x-axis. or we can write

f(x) = a polynomial of x + a proper fraction.

in this case only we have oblique asymptotes but not when numerator has less degree.

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.