Answer:
We need a sample size of 564.
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of 
, and a confidence level of 
, we have the following confidence interval of proportions.
In which
z is the zscore that has a pvalue of 
.
For this problem, we have that:
The margin of error is:
95% confidence level
So 
, z is the value of Z that has a pvalue of 
, so 
.
Based upon a 95% confidence interval with a desired margin of error of .04, determine a sample size for restaurants that earn less than $50,000 last year.
We need a sample size of n
n is found when 
So
Rounding up
We need a sample size of 564.
<span>Growth rate = (Present - Past)/Past
Plugging in what we know
Growth rate =(5600â’4420)/4420
Thus
Growth rate=.26697
Now we can plug the growth rate into our first formula which gives us
P=5600e^(.26697â‹…2)
Solve for P and we get
P
=
9551.58
however since you can not have .58 of a person we round down to 9551.
So Youngtown will have 9551 citizens in the year 2000</span>
Answer:
5.96 x 10 2
Step-by-step explanation:
Answer:
See below.
Step-by-step explanation:
1/2 = shrink by a factor of 1/2.
+7 = moved up 7 units.
Answer:
15
Step-by-step explanation:
5/6 * 18
Rearranging
5 * 18/6
5 *3
15
