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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.
Answer:
Solution: { -29/3 , -10/3 }
Step-by-step explanation:
2y = x + 3 ---------> equ 1
5y = x - 7 equ 2
equ 1 - equ s ====> -3y = 10
y = -10/3
Put y = -10/3 in equ 1
2 * (-10/3) = x + 3
-20/3 = x + 3
(-20/3) - 3 = x
(-20/3) - (9/3) = x
(-20 - 9) / 3 =x
x = -29/3
Solution: { -29/3 , -10/3 }
0.37 inches on each necklace or 37/100.
You just need to do 1.48 divided by 4.
Hope this helps !!!
You should go to sleep soon you have school tomorrow!!
Also make sure you ate today !
This is a kite due to the adjacent congruent sides (shown by the tickmarks). With any kite, the diagonals are perpendicular. So angle 3 is 90 degrees.
<h3>Answer: 90 degrees</h3>