Answer:
The sample of students required to estimate the mean weekly earnings of students at one college is of size, 3458.
Step-by-step explanation:
The (1 - <em>α</em>)% confidence interval for population mean (<em>μ</em>) is:
The margin of error of a (1 - <em>α</em>)% confidence interval for population mean (<em>μ</em>) is:
The information provided is:
<em>σ</em> = $60
<em>MOE</em> = $2
The critical value of <em>z</em> for 95% confidence level is:
Compute the sample size as follows:
![n=[\frac{z_{\alpha/2}\times \sigma }{MOE}]^{2}](https://tex.z-dn.net/?f=n%3D%5B%5Cfrac%7Bz_%7B%5Calpha%2F2%7D%5Ctimes%20%5Csigma%20%7D%7BMOE%7D%5D%5E%7B2%7D)
![=[\frac{1.96\times 60}{2}]^{2}](https://tex.z-dn.net/?f=%3D%5B%5Cfrac%7B1.96%5Ctimes%2060%7D%7B2%7D%5D%5E%7B2%7D)
Thus, the sample of students required to estimate the mean weekly earnings of students at one college is of size, 3458.
Answer:
take 361times 3.14
Step-by-step explanation:
you will get the answer
I think it’s 32? i dont know good luck
To solve this problem, let us first assign some
variables. Let us say that:
x = pigs
y = chickens
z = ducks
From the problem statement, we can formulate the
following equations:
1. y + z = 30 --->
only chicken and ducks have feathers
2. 4 x + 2 y + 2 z = 120 --->
pig has 4 feet, while chicken and duck has 2 each
3. 2 x + 2 y + 2 z = 90 --->
each animal has 2 eyes only
Rewriting equation 1 in terms of y:
y = 30 – z
Plugging this in equation 2:
4 x + 2 (30 – z) + 2 z = 120
4 x + 60 – 2z + 2z = 120
4 x = 120 – 60
4 x = 60
x = 15
From the given choices, only one choice has 15 pigs. Therefore
the answers are:
She has 15 pigs, 12 chickens, and 18 ducks.
