The correct answer is 75,000. Hope this helps!!!! ;)
Answer:
11.6cm
Step-by-step explanation:
a^2+b^2=c^2
11^2+b^2=16^2
121+b^2=256
b^2=135
b=11.6
42:125, 42/125 or 42 to 125
Answer:
The smallest sample size n that will guarantee at least a 90% chance of the sample mean income being within $500 of the population mean income is 48.
Step-by-step explanation:
The complete question is:
The mean salary of people living in a certain city is $37,500 with a standard deviation of $2,103. A sample of n people will be selected at random from those living in the city. Find the smallest sample size n that will guarantee at least a 90% chance of the sample mean income being within $500 of the population mean income. Round your answer up to the next largest whole number.
Solution:
The (1 - <em>α</em>)% confidence interval for population mean is:

The margin of error for this interval is:

The critical value of <em>z</em> for 90% confidence level is:
<em>z</em> = 1.645
Compute the required sample size as follows:

![n=[\frac{z_{\alpha/2}\cdot\sigma}{MOE}]^{2}\\\\=[\frac{1.645\times 2103}{500}]^{2}\\\\=47.8707620769\\\\\approx 48](https://tex.z-dn.net/?f=n%3D%5B%5Cfrac%7Bz_%7B%5Calpha%2F2%7D%5Ccdot%5Csigma%7D%7BMOE%7D%5D%5E%7B2%7D%5C%5C%5C%5C%3D%5B%5Cfrac%7B1.645%5Ctimes%202103%7D%7B500%7D%5D%5E%7B2%7D%5C%5C%5C%5C%3D47.8707620769%5C%5C%5C%5C%5Capprox%2048)
Thus, the smallest sample size n that will guarantee at least a 90% chance of the sample mean income being within $500 of the population mean income is 48.