C
Because that is what you do first
Answer:
- Jim = 6 hours
- Regina = 12 hours
- Together at same rate = 3 hours
Step-by-step explanation:
- Let Jim's speed is J and Regina's speed is R and the work is W
<u>We have:</u>
<u>Substitute J with 2R:</u>
Regina needs 12 hours to complete work
and
<u>Substitute R with J/2:</u>
Jim needs 6 hours to complete same work
<u>If Regina works as quickly as Jim:</u>
Then
<u>Jim can complete 1/6 of work in one hour, they both can complete twice work:</u>
<u>Time to complete work:</u>
Hey bro attach a file with the pic then we can help you
You would take 3/4x=36
Then you would divide each side by 3/4 and 36÷0.75=48
There are 48 people in the club. Hope this helps!
Answer:
The number of rainfalls is ![n =96](https://tex.z-dn.net/?f=n%20%20%3D96)
The answer to the second question is no it will not be valid this because from the question we are told that the experiment require one pH reading per rainfall so getting multiply specimens(used for the pH reading) from one rainfall will make the experiment invalid.
Step-by-step explanation:
from the question we are told that
The standard deviation is ![\sigma = 0.5](https://tex.z-dn.net/?f=%5Csigma%20%20%3D%20%200.5)
The margin of error is ![E = 0.1](https://tex.z-dn.net/?f=E%20%20%3D%20%200.1)
Given that the confidence level is 95% then we can evaluate the level of significance as
![\alpha = 100 - 95](https://tex.z-dn.net/?f=%5Calpha%20%20%3D%20%20100%20-%20%2095)
![\alpha = 5 \%](https://tex.z-dn.net/?f=%5Calpha%20%20%3D%20%205%20%5C%25)
![\alpha =0.05](https://tex.z-dn.net/?f=%5Calpha%20%20%3D0.05)
Next we will obtain the critical value of
from the normal distribution table , the value is ![Z_{\frac{\alpha }{2} } = 1.96](https://tex.z-dn.net/?f=Z_%7B%5Cfrac%7B%5Calpha%20%7D%7B2%7D%20%7D%20%3D%20%201.96)
Generally the sample size is mathematically represented as
![n = [\frac{Z_{\frac{\alpha }{2} * \sigma }}{ E} ]^2](https://tex.z-dn.net/?f=n%20%20%3D%20%20%5B%5Cfrac%7BZ_%7B%5Cfrac%7B%5Calpha%20%7D%7B2%7D%20%2A%20%20%5Csigma%20%7D%7D%7B%20E%7D%20%5D%5E2)
substituting values
![n = [\frac{1.96 * 0.5 }{ 0.1} ]^2](https://tex.z-dn.net/?f=n%20%20%3D%20%20%5B%5Cfrac%7B1.96%20%2A%200.5%20%7D%7B%200.1%7D%20%5D%5E2)
![n =96](https://tex.z-dn.net/?f=n%20%20%3D96)
The answer to the second question is no the validity is null this because from the question we are told that the experiment require one pH reading per rainfall so getting multiply specimens(used for the pH reading) from one rainfall will make the experiment invalid