Answer:
0.0208<p<0.0592
Step-by-step explanation:
-Given the sample size is 400 and the desired proportion is 16.
-The confidence interval can be determined as follows:
#We the use this proportion to find the CI at 95%:
![CI=0.04\pm 1.96\times \sqrt{\frac{0.04(1-0.04)}{400}}\\\\=0.04\pm 0.0192\\\\=[0.0208,0.0592]](https://tex.z-dn.net/?f=CI%3D0.04%5Cpm%201.96%5Ctimes%20%5Csqrt%7B%5Cfrac%7B0.04%281-0.04%29%7D%7B400%7D%7D%5C%5C%5C%5C%3D0.04%5Cpm%200.0192%5C%5C%5C%5C%3D%5B0.0208%2C0.0592%5D)
Hence, the 95% confidence interval is 0.0208<p<0.0592
Answer: fourth option
Explanation:1) the pair x = 3 f(x) = 0, leads you to probe this:
f(3) = 0 = A [4 ^ (3 - 1) ] + C = 0
=> A [4^2] = - C
A[16] = - C
if A = 1/4
16 / 4 = 4 => C = - 4
That leads you to the function f(x) = [1/4] 4 ^( x - 1) - 4
2) Now you verify the images for that function for all the x-values of the table:
x = 2 => f(2) + [1/4] 4 ^ (2 - 1) - 4 = [1/4] 4 - 4 = 4 / 4 - 4 = 1 - 4 = - 3 => check
x = 3 => f(3) = [1/4] 4^ (3 - 1) - 4 = [1/4] 4^2 - 4 = 16 / 4 - 4 = 4 - 4 = 0 => check
x = 4 -> f(4) = [1/4] 4^ (4-1) - 4 = [1/4] 4^(3) - 4 = (4^3) / 4 - 4 = 4^2 - 4 = 16 - 4 = 12 => check.
Therefore, you have proved that the answer is the fourth option.
Answer:
Horizontal Asymptote: y=2
Vertical Asymptote: x=3
Step-by-step explanation:
One way: Look at graph
Look at for a horizontal or vertical line that both lines don't touch.
Second Way:
When you look for vertical asymptotes, you set the denominator equal to zero and solve.
For example, the second question has a denominator of x-3.
x-3=0
x=3 (This is the vertical asymptote)
When you look for horizontal asymptote, you do long division.
For example, the first question 
The answer will be 2. 2 is the quotient. You only want quotient not remainder.
y=2.
Answer: D. This was a random sample. It may have included anyone in attendance.
Step-by-step explanation:
The options are:
A. This was a biased sample. Jim should interview all in attendance.
B. This was a census. Any guest may have participated.
C. This was a random sample. It may not have included anyone in attendance.
D. This was a random sample. It may have included anyone in attendance.
A random sampling is simply referred to as a subset of individuals that are picked from a larger set of individuals.
With regards to the question, Jim wanted to find out what the audience thought about the debate and after the event, he stood at the exit to survey every fifth guest.
This means that it was a random sampling and anyone could have been picked, the sampling wasn't bias.
