B=number of ticets sold before
a=number of tickets sold after
cost of a ticket=number of tickets times cost per ticket
beforecost=39.95b
aftercost=54.95a
total cost=925000
39.95b+54.95a=925000
total number tickets=20000
b+a=20000
we have
39.95b+54.95a=925000
b+a=20000
multiply second equation by -39.95 and add to first equatin
39.95b+54.95a=925000
<u>-39.95b-39.95a=-799000 +</u>
0b+15a=126000
15a=126000
divide bot sides by 15
a=8400
sub back
b+a=20000
b+8400=20000
minus 8400 both sides
b=11600
11,600 tickets sold before
8400 tickets sold after
The answer is, there are 320 possible combinations of outfits
Answer:
n = 66.564
Step-by-step explanation:
- Because the population is unknown, we will apply the following formula to find the sample size:

Where:
z = confidence level score.
S = standard deviation.
E = error range.
2. We will find each of these three data and replace them in the formula.
"z" theoretically is a value that measures how many standard deviations an element has to the mean. For each confidence level there is an associated z value. In the question, this level is 99%, which is equivalent to a z value of 2.58. To find this figure it is not necessary to follow any mathematical procedure, it is enough to make use of a z-score table, which shows the values for any confidence interval.
The standard deviation is already provided by the question, it is S = 100.
Finally, "E" is the acceptable limit of sampling error. In the example, we can find this data. Let us note that in the end it says that the director wishes to estimate the mean number of admissions to within 1 admission, this means that she is willing to tolerate a miscalculation of just 1 admission.
Once this data is identified, we replace in the formula:

3. The corresponding mathematical operations are developed:


n= 66.564
%6 of the sixth graders are left handed