T-6≤8 would be the answer
Answer:
The 95% confidence interval estimate of the population proportion of managers who have caught salespeople cheating on an expense report is (0.5116, 0.6484).
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence level of
, we have the following confidence interval of proportions.

In which
z is the zscore that has a pvalue of
.
In the survey on 200 managers, 58% of the managers have caught salespeople cheating on an expense report.
This means that 
95% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
The lower limit of this interval is:

The upper limit of this interval is:

The 95% confidence interval estimate of the population proportion of managers who have caught salespeople cheating on an expense report is (0.5116, 0.6484).
Answer:
D.SQUARE I searched it I'm 100 % sure!!!!
Step-by-step explanation:
Plzz mark brainlist
Answer:
yes
Step-by-step explanation:
a 90 percent would be 108
Answer:
$6.50
Step-by-step explanation:
Let the cost of a bowl and a jug be $b and $j respectively.
3b +j= 13.70 -----(1)
j= b +4.10 -----(2)
Substitute (2) into (1):
3b +b +4.10= 13.70
4b +4.10= 13.70 <em>(</em><em>simplify</em><em>)</em>
4b= 13.70 -4.10 <em> (-4.10 on both sides)</em>
4b= 9.60
b= 9.60 ÷4 <em>(÷4 on both sides)</em>
b= 2.40
Substitute b= 2.40 into (2):
j= 2.40 +4.10
j= 6.50
Thus, the jug costs $6.50.