Given:
n = 150, sample size
Denote the sample proportion by q (normally written as

).
That is,
q = 60/150 = 0.4, sample proportion.
At the 96% confidence level, the z* multiplier is about 2.082, and the confidence interval for the population proportion is
![q \pm z^{*}[ \frac{q(1-q)}{ \sqrt{n} } ]](https://tex.z-dn.net/?f=q%20%5Cpm%20z%5E%7B%2A%7D%5B%20%5Cfrac%7Bq%281-q%29%7D%7B%20%5Csqrt%7Bn%7D%20%7D%20%5D)
That is,
0.4 +/- 2.082* √[(0.4*0.6)/150]
= 0.4 +/- 0.0833
= (0.3167, 0.4833)
= (31.7%, 48.3%)
Answer: The 96% confidence interval is about (32% to 48%)
Answer:
2a2b-6ab2+9ab+5 is right answer
Answer:
Dimension of the enclosure is 56.56 ft * 14.14 ft
Step-by-step explanation:
Given -
Let us suppose that x is the length of steel fence and y is the length of one side of pine boards
We know that - xy = 800.
y = 800/x
Let us say that C is the cost of fence
= 3x + 6(2y) = 3x + 12y = 3x + 12(800/x)
C = 3x + 9600x-1, x > 0
C' = 3 - 9600x-2 = (3x2-9600)/x2
C' = 0 when 3x2 = 9600
x2 = 3200
x = √3200 ft ≈ 56.56 ft
So, the cost is minimized when x ≈ 56.56 ft and y = 800/x ≈ 14.14 ft