Apopulation has a standard deviation of 50. a random sample of 100 items from this population is selected. the sample mean is determined to be 600. calculate the margin of errors for 95%, 90% and 99% confidence interval.

Formula for Margin of error :

, where n= sample size

= sample standard deviation

= two-tailed z-value for confidence level of ().

Given : tex]\sigma=50[/tex]

n= 100

a) Confidence level = 95%

Critical z-value for 95% confidence = . (Using z-value table)

Then , Margin of error :

E=9.8

Hence, Margin of error for 95%= 9.8

b) Confidence level = 90%

Critical z-value for 90% confidence = .(Using z-value table)

Then , Margin of error :

E=8.225

Hence, Margin of error for 90%= 8.225

c) Confidence level = 99%

Critical z-value for 99% confidence = .(Using z-value table)

Then , Margin of error :

E=12.88

Hence, Margin of error for 99%= 12.88

answer: idk

step-by-step explanation:

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

the factor is 2

step-by-step explanation:

let f(x)=8(3^(x−2)+2)

to stretch a graph by 3, it would be 3f(x)

g(x) =3f(x)

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

x     −2     0     2     4     6

f(x) 1/64 1/16   1/4     1     4

1/64 * z * z = 1/16

i multiplies by z twice because it its 2 steps from -2 to 0

1/64 z^2 = 1/16

multiply by 64 to clear the fraction on the left side

1/4 * 64 z^2 =1/16 *64

z^2 = 4

take the square root of each side

sqrt(z^2) = ±sqrt(4)

z = ±2  

we are multiplying by 2 each time.