Using the z-distribution, it is found that the smallest sample size required is given by:
B) 7 vehicles.
<h3>What is the margin of error for a z-distribution confidence interval?</h3>
It is given by:

In which:
is the population standard deviation.
In this problem, we have a 90% confidence level, hence
, z is the value of Z that has a p-value of
, so the critical value is z = 1.645.
The margin of error and the population standard deviation are given by, respectively:
.
Then, we solve for n to find the needed sample size.






Rounding up, 7 vehicles have to be sampled, hence option B is correct.
More can be learned about the z-distribution at brainly.com/question/25890103
Its either 2 or 12 im not too sure most likely 2
Answer: x= 1/3 - 2/3y
Step-by-step explanation:
Answer:
$1400
Step-by-step explanation:
x= profit of 6th month
(200+300+400+500+500+x)
-------------------------------------- = 550
6
(1900+x)=3300
x=3300-1900
x=1400