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
A=Annual amount=2000
i=annual interest=0.0205
n=number of years=3
Present value
=A((1+i)^n-1)/(i(1+i)^n)
=2000(1.0205^3-1)/(.0205(1.0205^3))
=5762.15
It means 2.5m/s multiplied by 2500 to get the total distance in meters
Answer: answer is in the ss i put
Source:trust me bro
Answer:
0.216 = 0.6³
Step-by-step explanation:
using the rule of logarithms
•
x = n ⇔ x = 
0.216 = 3 ⇒ 0.216 = 