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.
Hey there! Just subtract 25 from 344. The answer is 319.
What do you need help with there is nothing showing.
Answer: Multiply (60 − 2x) and (2 + 0.25x) to create the equation y = -0.5x2 + 11x + 120
We know that David wants to increase the price per cup to increase his revenue. He found out that for every $0.25 grows( increase), x, in the price for each cup.
In the event of a price increase, 2 cups remain unsold, and doubling the cups is still not sold. Then the numbers are sold (60-2x). Depending on the choice:
Revenue= (60 -2x)(2 +0.25x)
60·2 +60·0.25x -2x·2 -2x·0.25x
= -0.5x² +11x +120
Answer:
(d) 78√3
hope it is right answer for it !!