Answer:
The minimum sample size required is 207.
Step-by-step explanation:
The (1 - <em>α</em>) % confidence interval for population mean <em>μ</em> is:
The margin of error of this confidence interval is:
Given:
*Use a <em>z</em>-table for the critical value.
Compute the value of <em>n</em> as follows:
![MOE=z_{\alpha /2}\frac{\sigma}{\sqrt{n}}\\3=2.576\times \frac{29}{\sqrt{n}} \\n=[\frac{2.576\times29}{3} ]^{2}\\=206.69\\\approx207](https://tex.z-dn.net/?f=MOE%3Dz_%7B%5Calpha%20%2F2%7D%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%5C%5C3%3D2.576%5Ctimes%20%5Cfrac%7B29%7D%7B%5Csqrt%7Bn%7D%7D%20%5C%5Cn%3D%5B%5Cfrac%7B2.576%5Ctimes29%7D%7B3%7D%20%5D%5E%7B2%7D%5C%5C%3D206.69%5C%5C%5Capprox207)
Thus, the minimum sample size required is 207.
The answer is 3.75 hope this helped
Answer:
C
Step-by-step explanation:
3/4 + 6/10
Find a common denominator and find what you multiply to each denominator to get that number and multiply the numerator by the same amount
(3/4 * 5/5) + (6/10 * 2/2)=
15/20 + 12/20
Then add
15/20 + 12/20= 27/20
Then finally convert to mixed number
27/20 = 1 7/20
If George completed 3/5 of work in 9 days, it means that he needs 3 days to finish 1/5th of the work.
The remaining part is 2/5 because 1-3/5=2/5. 2/5 is also 6/15
From this, how much did George do? in the first 3 days he did 1/5th and then one more day was left, during which he did 1/5/3=1/15th.
So he did in total 1/15+1/5=1/15+3/15=4/15.
this means that Paul did the rest of 6/15, so Paul did 6/15-4/15=2/15.
So we know that he does 2/15 in 4 days, which means that every two days he can do 1/15 of the work
so he would need 15 times 2 days to finish the work - 30 days!
Answer:
1st one
Step-by-step explanation:
