Suppose that the numbers of residents in zip codes are normally distributed with an unknown mean and standard deviation. The numbers of residents of 50 randomly sampled zip codes are used to estimate the mean of the population. What t-score should be used to find the 99% confidence interval for the population mean?

Step-by-step explanation:

The required t score would be determined from the t distribution table. In order to use the t distribution, we would determine the degree of freedom, df for the sample.

df = n - 1 = 50 - 1 = 49

Since confidence level = 99% = 0.99, α = 1 - CL = 1 – 0.99 = 0.01

α/2 = 0.01/2 = 0.005

the area to the right of z0.005 is 0.005 and the area to the left of z0.005 is 1 - 0.005 = 0.995

Looking at the t distribution table,

t score = 2.678

r(0,90)(x,y) . t(-5,0)

last option

step-by-step explanation:

∆def to ∆d"e"f"

- rotate 90 counterclockwise degrees

- then translated 5 to the left (x --> x - 5)

answer is the last option