Answer:
vertex = 
Intercepts for x = 
intercepts for y = (0,-6)
The answer is 36.
You subtract or decrease the number by 8 every time.
Answer:
The claim that less than 50% of internet users get their health questions answered online is correct, given that according to the survey, 45.98% of people do.
Step-by-step explanation:
Since an online health and medicine company claims that about 50% of Internet users would go to an online sight first for information regarding health and medicine, and a random sample of 1318 Internet users was asked where they will go for information the next time they need information about health or medicine; 606 said that they would use the Internet, to test the claim that less than 50% of Internet users get their health questions answered online, the following calculation must be performed:
1318 = 100
606 = X
606 x 100/1318 = X
60600/1318 = X
45.978 = X
Therefore, the claim that less than 50% of internet users get their health questions answered online is correct, given that according to the survey, 45.98% of people do.
Answer:
(a) The sample sizes are 6787.
(b) The sample sizes are 6666.
Step-by-step explanation:
(a)
The information provided is:
Confidence level = 98%
MOE = 0.02
n₁ = n₂ = n

Compute the sample sizes as follows:



Thus, the sample sizes are 6787.
(b)
Now it is provided that:

Compute the sample size as follows:

![n=\frac{(z_{\alpha/2})^{2}\times [\hat p_{1}(1-\hat p_{1})+\hat p_{2}(1-\hat p_{2})]}{MOE^{2}}](https://tex.z-dn.net/?f=n%3D%5Cfrac%7B%28z_%7B%5Calpha%2F2%7D%29%5E%7B2%7D%5Ctimes%20%5B%5Chat%20p_%7B1%7D%281-%5Chat%20p_%7B1%7D%29%2B%5Chat%20p_%7B2%7D%281-%5Chat%20p_%7B2%7D%29%5D%7D%7BMOE%5E%7B2%7D%7D)
![=\frac{2.33^{2}\times [0.45(1-0.45)+0.58(1-0.58)]}{0.02^{2}}\\\\=6665.331975\\\\\approx 6666](https://tex.z-dn.net/?f=%3D%5Cfrac%7B2.33%5E%7B2%7D%5Ctimes%20%5B0.45%281-0.45%29%2B0.58%281-0.58%29%5D%7D%7B0.02%5E%7B2%7D%7D%5C%5C%5C%5C%3D6665.331975%5C%5C%5C%5C%5Capprox%206666)
Thus, the sample sizes are 6666.
222,702 rounded to the nearest ten thousand would be 220,000