Using the z-distribution and the formula for the margin of error, it is found that:
a) A sample size of 54 is needed.
b) A sample size of 752 is needed.
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 z-score that has a p-value of
.
The margin of error is of:

90% confidence level, hence
, z is the value of Z that has a p-value of
, so
.
Item a:
The estimate is
.
The sample size is <u>n for which M = 0.03</u>, hence:






Rounding up, a sample size of 54 is needed.
Item b:
No prior estimate, hence 






Rounding up, a sample of 752 should be taken.
A similar problem is given at brainly.com/question/25694087
14: 14, 28, 42, 56, 70, 84, 98, 112, 126, 140, 154, 168, 182, 196, 210, 224,…
35: 35, 70, 105, 140, 175, 210, 245, 280, 315, 350, 385, 420, 455, 490, 525,…
LCM(14, 35) = 70
The answer is B.
8x+5y=$24
6x+2y=$16.60
16x+10y=$48
-30x-10y=-$83
-14x=-$35
x=$2.50
8($2.50)+5y=$24
$20.00+5y=$24
5y=$4
y=$0.80
Hamburgers costs $2.50 each
Fries cost $0.80 each
If you're multiplying the 2 exponents with the same base then you add the exponents.



Hope this helps :)