Answer:
The 99% confidence interval for the proportion of readers who would like more coverage of local news is (0.3685, 0.4315).
Step-by-step explanation:
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 zscore that has a pvalue of
.
For this problem, we have that:

99% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
The lower limit of this interval is:

The upper limit of this interval is:

The 99% confidence interval for the proportion of readers who would like more coverage of local news is (0.3685, 0.4315).
Answer:
33.48
Step-by-step explanation:
30*0.07=2.10
30-2.10=27.90
27.90*0.20=5.58
27.90+5.58
33.48
I think the answer is <span>1100</span>
Answer:

Step-by-step explanation:
![(\sqrt{3} +4)(1+\sqrt{3})\\\\= \sqrt{3}(1+\sqrt{3} )+4(1+\sqrt{3})\\\\= \sqrt{3} + (\sqrt{3} )^2 + 4 + 4\sqrt{3} \\\\= \sqrt{3} + 3+4+4\sqrt{3} \\\\= 7 + \sqrt{3} + 4\sqrt{3} \\\\Take \ \sqrt{3} \ common\\\\= 7 + \sqrt{3} (1+4)\\\\= 7 + \sqrt{3}(5)\\\\= 7 + 5\sqrt{3} \\\\\rule[225]{225}{2}](https://tex.z-dn.net/?f=%28%5Csqrt%7B3%7D%20%2B4%29%281%2B%5Csqrt%7B3%7D%29%5C%5C%5C%5C%3D%20%5Csqrt%7B3%7D%281%2B%5Csqrt%7B3%7D%20%29%2B4%281%2B%5Csqrt%7B3%7D%29%5C%5C%5C%5C%3D%20%5Csqrt%7B3%7D%20%2B%20%28%5Csqrt%7B3%7D%20%29%5E2%20%2B%204%20%2B%204%5Csqrt%7B3%7D%20%5C%5C%5C%5C%3D%20%5Csqrt%7B3%7D%20%2B%203%2B4%2B4%5Csqrt%7B3%7D%20%20%5C%5C%5C%5C%3D%207%20%2B%20%5Csqrt%7B3%7D%20%20%2B%204%5Csqrt%7B3%7D%20%5C%5C%5C%5CTake%20%5C%20%5Csqrt%7B3%7D%20%5C%20common%5C%5C%5C%5C%3D%207%20%2B%20%5Csqrt%7B3%7D%20%281%2B4%29%5C%5C%5C%5C%3D%207%20%2B%20%5Csqrt%7B3%7D%285%29%5C%5C%5C%5C%3D%207%20%2B%205%5Csqrt%7B3%7D%20%5C%5C%5C%5C%5Crule%5B225%5D%7B225%7D%7B2%7D)
Hope this helped!
<h3>~AH1807</h3>