Confidence interval for the population mean z-test or normal distribution solution: given: n=20, x̄ =89, s=2.0, 99% formula: x̄ -z_alpha/2(s/square root of n)< µ< x̄ +z_alpha/2(s/square root of n) 89-(2.575)(2.0/square root of 20)< µ< 89-(2.575)(2.0/square root of 20) 87.848< µ< 90.152

margin of error: solution: the critical value is 2.575. the standard deviation is 2.0 (from the question), but as this is a sample, we need the standard error for the mean. the formula for the se of the mean is standard deviation / √(sample size), so: 2.0/ √(20)=0.447. me=2.575 * 0.447 = 1.151.

therefore, the margin of error is 1.151 with the confidence interval of 87.848< µ< 90.152.

z-test or normal distribution

solution:

given:

n=20, x̄ =89, s=2.0, 99%

formula:

x̄ -z_alpha/2(s/square root of n)< µ< x̄ +z_alpha/2(s/square root of n)

89-(2.575)(2.0/square root of 20)< µ< 89-(2.575)(2.0/square root of 20)

87.848< µ< 90.152

margin of error:

solution:

the critical value is 2.575.

the standard deviation is 2.0 (from the question), but as this is a sample, we need the standard error for the mean.

the formula for the se of the mean is standard deviation / √(sample size), so: 2.0/ √(20)=0.447.

me=2.575 * 0.447 = 1.151.

therefore, the margin of error is 1.151 with the confidence interval of 87.848< µ< 90.152.

1350

answer:

Please reframe from posting questions that you know the answer to...

Thank you!

The number is 45.

Step-by-step explanation:

A few ways to do this...but the simplest way I think is to list the two-digit numbers whose tens digit is even and digits add to nine...we have

27

45

63

81

The only one where the ones digit is one more than the tens digit is 45.

answer by texttutoring(324) About Me (Show Source):

You can put this solution on YOUR website!

Let x= tens digit

Let y = ones digit

Equation 1: x+y = 9

Equation 2: y=x+1

Substitute Equation 2 into Equation 1:

x+ (x+1) = 9

2x + 1 = 9

2x = 8

x = 4

So the tens digit is 4. This means that the ones digit is y =4+1 = 5