Answer: 1
Step-by-step explanation:
Answer:ITS THE SECOND ONE
Step-by-step explanation:
Answer:
We need to sample at least 1069 parents.
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 many parents do you have to sample?
We need to sample at least n parents.
n is found when
. So






Rounding up
We need to sample at least 1069 parents.
Answer:
$1125000
Step-by-step explanation:
45,000 x 25 = 1125000
Since the question didn't specify what values to calculate, I will assume that it is how much money was issued in total.