Answer:
16
Step-by-step explanation:
+
+
+
16
Answer:
1.5s+2.5p<20
Step-by-step explanation:
Multiply by each kilo and then make sure it adds up to less than 20!
<em>Hope</em><em> </em><em>this</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>u</em><em>.</em><em>.</em><em>.</em><em>:</em><em>)</em>
Okay so the problem starts you off with the fraction (1/4). Sharing equally among 3 people would mean that you'd want to divide it by three so your equation would initially look like (1/4) / 3. Having two dividing signs is confusing so what you do is change the dividing into multiplying by the reciprocal of 3 = 1/3... This would make your expression (1/4) * (1/3) and when you simplify this you should get 1/12 as the answer.
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.