Tenths. Hope this helped.
Answer:
a) The margin of error for a 90% confidence interval when n = 14 is 18.93.
b) The margin of error for a 90% confidence interval when n=28 is 12.88.
c) The margin of error for a 90% confidence interval when n = 45 is 10.02.
Step-by-step explanation:
The t-distribution is used to solve this question:
a) n = 14
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 14 - 1 = 13
90% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 13 degrees of freedom(y-axis) and a confidence level of
. So we have T = 1.7709
The margin of error is:

In which s is the standard deviation of the sample and n is the size of the sample.
The margin of error for a 90% confidence interval when n = 14 is 18.93.
b) n = 28
27 df, T = 1.7033

The margin of error for a 90% confidence interval when n=28 is 12.88.
c) The margin of error for a 90% confidence interval when n = 45 is
44 df, T = 1.6802

The margin of error for a 90% confidence interval when n = 45 is 10.02.
Answer:
C : 89%
Step-by-step explanation:
If they sold 235 scoops of marshmallow ice cream and in total of all day they sold a total of 265 scoops of ice cream. 265 - 235 = 30
My answer is C : 89%
<span>(4 · 2^5) ÷ (2^3 · 1/16 )
</span>= (2^2 · 2^5) ÷ (2^3 · 2^-4 )
= (2^7) ÷ (2^-1)
= 2^8