Answer:
The 85% onfidence interval for the population proportion of new car buyers who prefer foreign cars over domestic cars is (0.151, 0.205).
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:
Sample of 421 new car buyers, 75 preferred foreign cars. So 
85% confidence level
So 
, z is the value of Z that has a pvalue of 
, so 
.
The lower limit of this interval is:
The upper limit of this interval is:
The 85% onfidence interval for the population proportion of new car buyers who prefer foreign cars over domestic cars is (0.151, 0.205).
The question is asking to find the variance for the said samples in the problem ans use the sample data to determine each variance, and base on my further computation and further calculation, I would say that the answer would be the following:
#1. 3.3 -> 1 and 3 -> 2/9
#2. 1-> 0->3/9
#3. 6.3 - > 8 and 3-> 2/9
#4 49-> 1 and 8-> 2/9
Answer:
or 718.3775
Step-by-step explanation:
Formula:
we have
Answer:
0.125 as a fraction is 1/8 and as a percentage is 12 1/2%
18/19 as a decimal is 0.94 (rounded to the nearest hundredths) and as a percentage is 94.73% (rounded to nearest hundredths)
Hope I helped :)
Answer:
C. 25x+12y+4≤ 130
Step-by-step explanation:
Each ticket cost $25, so x would be the number of tickets she has to buy.
25x
Each T-shirt cost $12, so y would be the number of T-shirts she can buy.
12y
And since there is an service fee for the entree purchase, that's another $4.
+4
The cost of everything has to be less than $130, so the equation would be:
25x+12y+4≤ 130
