A researcher is interested in finding a 90% confidence interval for the mean number of times per day that college students text. The study included 147 students who averaged 44.7 texts per
day. The standard deviation was 17.9 texts. Round answers to 3 decimal places where possible.
a. To compute the confidence interval use a tv distribution.
b. With 90% confidence the population mean number of texts per day is between
and
texts.

a c

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