2(n+3)
two times the sum of a number (n) and 3
Answer: 95% confidence interval would be (0.344,0.456).
Step-by-step explanation:
Since we have given that
n = 295
x = 118
so, 
At 95% confidence, z = 1.96
So, margin of error would be

so, 95% confidence interval would be

Hence, 95% confidence interval would be (0.344,0.456).
In
order to solve for a nth term in an arithmetic sequence, we use the formula
written as:<span>
an = a1 + (n-1)d
where an is the nth term, a1 is the first value
in the sequence, n is the term position and d is the common difference.
First, we need to calculate for d from the given
values above.
<span>a3 = 20.5 and a8 = 13
</span>
an = a1 + (n-1)d
20.5 = a1 + (3-1)d
</span>an = a1 + (n-1)d
13 = a1 + (8-1)d
<span>
a1 = 23.5
d = -1.5
The 11th term is calculated as follows:
a11 = a1 + (n-1)d
a11= 23.5 + (11-1)(-1.5)
a11 =
8.5</span>