Suppose germination periods, in days, for grass seed are normally distributed and have a known population standard deviation of 6 days and an unknown population mean. a random sample of 21 types of grass seed is taken and gives a sample mean of 34 days. find the margin of error for the confidence interval for the population mean with a 99% confidence level.

3.725

Step-by-step explanation:

Formula of Margin of Error for (n<30):-

Given : Sample size : n= 21

Level of confidence = 0.99

Significance level :

By using the t-distribution table ,

Critical value :

Standard deviation:

Then, we have

Hence, the margin of error for the confidence interval for the population mean with a 99% confidence level=3.725

let's solve your equation step-by-step.

4x+2=2x+12

step 1: subtract 2x from both sides.

4x+2−2x=2x+12−2x

2x+2=12

step 2: subtract 2 from both sides.

2x+2−2=12−2

2x=10

step 3: divide both sides by 2.

2x

2

=

10

2

x=5

step-by-step explanation: