Monthly incomes of employees at a particular company have a mean of $5954. the distribution of sample means for samples of size 70 is normal with a mean of $5954 and a standard deviation of $259. suppose you take a sample of size 70 employees from the company and find that their mean monthly income is $5747. how many standard deviations is the sample mean from the mean of the sampling distribution?

b. 2 and 4

step-by-step explanation:

246

step-by-step explanation:

7.9m and 4.3°

step-by-step explanation:

let east be +x and north be +y, x = distance * cos θ, y = distance * sin θ  

for 3.5 south, y = -3.5  

for 8.2, x = 8.2 * cos 30 and y = 8.2 * sin 30  

for 15 west, x = -15  

total x = 8.2 * cos 30 – 15 ≈ -7.9  

total y = -3.5 + 8.2 * sin 30 = 0.6  

magnitude = [(8.2 * cos 30 – 15)^2 + 0.6^2)]^0.5  

the magnitude is approximately 7.9 m  

tan θ = y/x = 0.6/-7.9  

the angle is approximately 4.3˚ north of west

your answer is

a.

5

step-by-step explanation: