Using the z-distribution, a sample of 609 nickels has to be weighed.
<h3>What is a z-distribution confidence interval?</h3>
The confidence interval is:
![\overline{x} \pm z\frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=%5Coverline%7Bx%7D%20%5Cpm%20z%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
The margin of error is:
![M = z\frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
In which:
is the sample mean.
is the standard deviation for the population.
For this problem, the parameters are:
![M = 20, \sigma = 300, z = 1.645](https://tex.z-dn.net/?f=M%20%3D%2020%2C%20%5Csigma%20%3D%20300%2C%20z%20%3D%201.645)
We solve for n to find the sample size, then:
![M = z\frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
![20 = 1.645\frac{300}{\sqrt{n}}](https://tex.z-dn.net/?f=20%20%3D%201.645%5Cfrac%7B300%7D%7B%5Csqrt%7Bn%7D%7D)
![20\sqrt{n} = 1.645 \times 300](https://tex.z-dn.net/?f=20%5Csqrt%7Bn%7D%20%3D%201.645%20%5Ctimes%20300)
![\sqrt{n} = 1.645 \times 15](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%20%3D%201.645%20%5Ctimes%2015)
![(\sqrt{n})^2 = (1.645 \times 15)^2](https://tex.z-dn.net/?f=%28%5Csqrt%7Bn%7D%29%5E2%20%3D%20%281.645%20%5Ctimes%2015%29%5E2)
n = 608.9
Rounding up, a sample of 609 nickels has to be weighed.
More can be learned about the z-distribution at brainly.com/question/25890103
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I would say b hope this help ☺
16, add 16 to 8 and then the rest is simple
Step-by-step explanation:
I hope this helps..........
Answer:
Step-by-step explanation:
13 - 2 * 5
13 - 10 = 3....ur answer
these types of problems are done using the order of operations....PEMDAS
(parenthesis, exponents, (multiplication/division), (addition/subtraction))