Answer:
We need a sample size of at least 719
Step-by-step explanation:
We have that to find our
level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of
.
So it is z with a pvalue of
, so 
Now, find the margin of error M as such

In which
is the standard deviation of the population and n is the size of the sample.
How large a sample size is required to vary population mean within 0.30 seat of the sample mean with 95% confidence interval?
This is at least n, in which n is found when
. So






Rouding up
We need a sample size of at least 719
Answer:
part A 17/23
part B 31/40
part C 23/40
part D Yes, sorry you gotta figure that out. Im to lazy to write a sentence
Step-by-step explanation:
The given points are the vertices of the quadrilateral

By Green's theorem, the line integral is


The normal distribution curve is shown in the diagram below
<span>The percentage of time that his commute time exceeds 61 minutes is equal to the area under the standard normal curve that lies to the RIGHT of X=61
Standardising X=61 to find z-score
</span>

<span>
from the z-table
</span>

<span>
</span>
Answer:
A = 250(1 + 0.016)^0.75.
Step-by-step explanation:
9 months = 0.75 years
So A = 250(1 + 0.016)^0.75.