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:
Step-by-step explanation:
you just need to add both equations
you will get 3y=12+6=18
3y=18
y=6
now replace y=6
3x+12=6
3x=6-12
3x=-6
x=-2
(-2,6)
Already factored:
Factors of 12X^3y^3 - 24x^2y^3
Answer:
The Factors of 12 are: 1, 2, 3, 4, 6, 12
The factor pairs of 12 are:
1(12) = 12
2(6) = 12
3(4) = 12
Hope this helps.!!!