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
So we can turn the fraction into a decimal to make it easier
2.5 lbs (pounds)
since the bananas cost $1.89, we need to divide it with 2.5
1.89/2.5
=0.756
the answer would be $0.76
Answer:
Step-by-step explanation:
Answer:
A = (-7,6)
B = (2,4)
C = (-6,0)
Step-by-step explanation:
A = (-9 + 2, 2 + 4)
B = (0 + 4, 0 + 4)
C = (-8 + 2, -4 + 4)
Answer:
Yes.
Step-by-step explanation:
I have attached the work to your problem.
Please see the attachment below.
I hope this helps!
