Which one?????????????????
Answer:
Step-by-step explanation:
Given that for a sample size of 25, sample mean = 102 lbs. and sample std deviation = 10 lbs.

(Right tailed test)
Given that we assume population is normally distributed
Test statistic:
Mean difference = +2
Std error of sample = 
t = test statistic = Mean diff/Std error 
degree of freedom
Critical value of t = 1.711
Our test statistic >1.71
Accept alternate hypothesis
Because there is significant difference between the test statistic and test statistic lies above critical value.
Answer:
The 90% confidence level is 
Step-by-step explanation:
From the question we are told that
The sample size is 
The mean age is 
The standard deviation is 
Generally the degree of freedom for this data set is mathematically represented as
substituting values
Given that the level of confidence is 90% the significance level is mathematically evaluated as

10 %

Now 
Since we are considering a on tail experiment
The critical value for half of this significance level at the calculated degree of freedom is obtained from the critical value table as

The margin for error is mathematically represented as

substituting values


he 90% confidence interval for the true average age of all students in the university is evaluated as follows

substituting values


Answer:
x=7 and m<LMN = 120
Step-by-step explanation:
if MO bisects LMN then 13x - 31 must be equal to x + 53
13x - x = 53 + 31
12x = 84
x = 7
and
13x - 31 + x + 53 = m<LMN
14x + 22 = m<LMN
since x is 7
14×7 + 22 = 120
Step-by-step explanation:
green: 6/4
red: 7/4
this is all i know