Answer:
h
Step-by-step explanation:
Answer:
The 95% confidence interval estimate for the true proportion of adults residents of this city who have cell phones is (0.81, 0.874).
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:
95% 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 95% confidence interval estimate for the true proportion of adults residents of this city who have cell phones is (0.81, 0.874).
Ex 1:
2/5 x 4/5
Multiply both the numerators together and then multiply both the denominators together. You would multiply 2x4 which is 8 and 5x5 which is 25. Which leaves you with 8/25 and it will stay that way because you cannot simplify it further.
Ex 2:
2/5 and 5/3
Multiply across 2x5/5x3 and you end up with 10/15 which can be simplified/reduced to 2/3 because they share a common factor of 5.
Since there are 300 students you have to use proportions to solve the percents. First for strawberry you need to set up a proportion such as x/300=30/100 to show you are trying to find 30%of 300. Then you must cross multiply so 300 times 30 and 100 times x. Then you have 9,000 and 100, so you divide 9,000 by 100 and then you get 90 which is the portion of kids who liked strawberry flavored ice cream. You should do the exact same thing for vanilla ice cream like this: x/300=29/100 but since you’re finding 29% you need to put a 29 over 100. Then you cross multiply and do 300 times 29 and do x times 100. Which makes you get 8,700 and 100, and then you divide 8,700 by 100 and get 87 for vanilla ice cream. Lastly you need to subtract 87 from 90 since it asked how many more students liked strawberry than vanilla: 90-87=
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