Arandom variable x has a normal distribution with an unknown mean and a standard deviation of 12. suppose that we take a random sample of size n = 36 and find a sample mean of ¯ x = 98 . what is a 95% confidence interval for the mean of x ?

The 95% confidence interval for the mean of x is (94.08, 101.92) .

Step-by-step explanation:

We are given that ,

A random variable x has a Normal distribution with an unknown mean and a standard deviation of 12.

i.e.

Also, it is given that , Sample mean having sample size : n= 36

For 95% confidence ,

Significance level :

By using the z-value table , the two-tailed critical value for 95% Confidence interval :

We know that the confidence interval for unknown population mean is given by :-

, where = Sample mean

= Population standard deviation

= Critical z-value.

Substitute all the given values, then the required confidence interval will be :

Therefore, the 95% confidence interval for the mean of x is (94.08, 101.92) .

a box 3 3/4 feet long or 4+ feet.

step-by-step explanation:

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.