An economist wants to estimate the mean per capita income (in thousands of dollars) for a major city in california. suppose that the mean income is found to be $22.7$⁢22.7 for a random sample of 183183 people. assume the population standard deviation is known to be $6.3$⁢6.3. construct the 85%85% confidence interval for the mean per capita income in thousands of dollars. round your answers to one decimal place.

(22.0297, 23.3703)

Step-by-step explanation:

Given that an economist wants to estimate the mean per capita income (in thousands of dollars) for a major city in California.

Let X be per capita income (in thousands of dollars) for a major city in California.

Mean = 22.7

n = 183

Population std dev = 6.3

Since population std dev is known we can use Z critical value.

Std error =

Z critical =1.44

Marginof error = ±1.44*0.4657=0.6706

Confidence interval 85%

=

too many. (jk)

31 v bucks

step-by-step explanation:

since you bought 25 on friday and 6 on tuesday, you can just add them up.

25+6=31

check your work.

31-6=25

i hope this !

-yand3r3

answer: cos [   arcsin (3/5) - arccos (1/2) ] =

step-by-step explanation: × = 3/5 = 0.6--> x = 36.87 and x = 180 - 36.87 = 143.13 deg

cos y = 1/2 --> y = +- 60 deg a. cos (x - y) = cos (143.13 - 60) = cos 83,13  

=

0.12

b. cos (36.87 - 60) = cos (-23.23) = 0.92