Two independent random samples have been selected. 100 observations from population 1 and 100 from population 2. sample means ¯x_1=70 and ¯x_2=50 were obtained. from previous experience with these populations, it is known that the variances are σ_1^2=100 and σ_2^2=64. determine the rejection region for the test of h_0: μ_1-μ_2=5 vs h_a: μ_1-μ_2≠5 using α=0.05. (a) z > 1.96 (b) z < -1.96 (c) z < -1.96 and z > 1.96 (d) z < -1.645 and z > 1.645

john earns $14.00 for two hours of work.

b. 6

step-by-step explanation:

we are given the table representing an exponential function.

we can see from the table that, as x is increasing, the value of y is increasing in the power of 6.

i.e. the following relation is obtained,

x = 1, y = 6

x = 2, y = 36 =

x = 3, y = 216 =

x = 4, y = 1296 =

x = 5, y = 7776 =

so, the geometric series obtained is .

thus, the common ratio is given by = = 6.

hence, the common ratio is 6.

thus, the first answer choice, y = sec x, is the correct one

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.