What is it that we need to find?
Answer: 30 more servings
Explanation:
They had 90 tablespoons, but half are gone. 90/2 = 45
Each serving is 1.5 tablespoons.
45/1.5 = 30 more servings
Answer:
Part 1: There are 4.7*10^21 ways to select 40 volunteers in subgroups of 10
Part 2: The research board can be chosen in 32760 ways
Step-by-step explanation:
Part 1:
The number of ways in which we can organized n elements into k groups with size n1, n2,...nk is calculate as:
![\frac{ n!}{ n1!*n2!*...*nk! }](https://tex.z-dn.net/?f=%5Cfrac%7B%20n%21%7D%7B%20n1%21%2An2%21%2A...%2Ank%21%20%7D)
So, in this case we can form 4 subgroups with 10 participants each one, replacing the values of:
- n by 40 participants
- k by 4 groups
- n1, n2, n3 and n4 by 10 participants of every subgroups
We get:
![\frac{ 40!}{10!*10!*10!*10!} = 4.7*10^{21}](https://tex.z-dn.net/?f=%5Cfrac%7B%2040%21%7D%7B10%21%2A10%21%2A10%21%2A10%21%7D%20%3D%204.7%2A10%5E%7B21%7D)
Part 2:
The number of ways in which we can choose k element for a group of n elements and the order in which they are chose matters is calculate with permutation as:
![nPk = \frac{ n!}{(n-k)!}](https://tex.z-dn.net/?f=nPk%20%3D%20%5Cfrac%7B%20n%21%7D%7B%28n-k%29%21%7D)
So in this case there are 4 offices in the research board, those are director, assistant director, quality control analyst and correspondent. Additionally this 4 offices are going to choose from a group of 5 doctors.
Therefore, replacing values of:
- n by 15 doctors
- k by 4 offices
We get:
![\frac{ 15!}{ (15-4)! } = 32760](https://tex.z-dn.net/?f=%5Cfrac%7B%2015%21%7D%7B%20%2815-4%29%21%20%7D%20%3D%2032760)
Answer:
1 1/2, 3 1/3
<h3>
Step-by-step explanation:</h3>
For the denominators (3, 4) the least common multiple (LCM) is 12. Calculations to rewrite the original inputs as equivalent fractions with the LCD:
10/3 = 10/3 × 4/4 = 40/12
6/4 = 6/4 × 3/3 = 18/12
<h3 /><h3 />
Answer:
The correct answer to the following question will be "384".
Step-by-step explanation:
We are having,
![P(-0.05](https://tex.z-dn.net/?f=P%28-0.05%3Cp-p%3C0.05%29%3D0.95)
On solving, we get
⇒ ![P(\frac{-0.05}{\sqrt{\frac{pq}{n}}}](https://tex.z-dn.net/?f=P%28%5Cfrac%7B-0.05%7D%7B%5Csqrt%7B%5Cfrac%7Bpq%7D%7Bn%7D%7D%7D%3CZ%3C%5Cfrac%7B-0.05%7D%7B%5Csqrt%7B%5Cfrac%7Bpq%7D%7Bn%7D%7D%7D%3D0.95%29)
⇒ ![\frac{0.05}{\sqrt{\frac{p(1-p)}{n}}} =1.96](https://tex.z-dn.net/?f=%5Cfrac%7B0.05%7D%7B%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D%7D%20%3D1.96)
On applying cross-multiplication, we get
⇒ ![0.05^2=1.96^2\times \frac{P(1-p)}{n}](https://tex.z-dn.net/?f=0.05%5E2%3D1.96%5E2%5Ctimes%20%5Cfrac%7BP%281-p%29%7D%7Bn%7D)
⇒ ![0.05^2n=1.96^2-1.96^2p^2](https://tex.z-dn.net/?f=0.05%5E2n%3D1.96%5E2-1.96%5E2p%5E2)
⇒
...(equation 1)
When "p" seems to be the population proportion, (equation 1) approach, which is special
∴ ![b^2-4ac=0](https://tex.z-dn.net/?f=b%5E2-4ac%3D0)
On putting the values in the above expression, we get
⇒ ![(1.96)^4-4\times 0.05^2\times 1.96^2n=0](https://tex.z-dn.net/?f=%281.96%29%5E4-4%5Ctimes%200.05%5E2%5Ctimes%201.96%5E2n%3D0)
⇒ ![1.98^2-4\times 0.05^2n=0](https://tex.z-dn.net/?f=1.98%5E2-4%5Ctimes%200.05%5E2n%3D0)
⇒ ![n=\frac{1.96^2}{4\times 0.05^2}](https://tex.z-dn.net/?f=n%3D%5Cfrac%7B1.96%5E2%7D%7B4%5Ctimes%200.05%5E2%7D)
⇒ ![n=384.16=384](https://tex.z-dn.net/?f=n%3D384.16%3D384)