Find the lxwxh and then multiply all of them together and then u should get your answer
Answer: The goal was $500
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Work Shown:
x% = x/100
175% = 175/100 = 1.75
Let g be the goal, which is the amount of money the club wanted to raise
175% of goal = 175% of g = 1.75g
The expression 1.75g represents how much money was actually raised, which was $875. Set the two expressions equal to each other. Solve for g
1.75g = 875
g = 875/1.75 ....... divide both sides by 1.75
g = 500
The club's goal was to raise $500
Note how 75% of $500 is 0.75*500 = 375
When they raised 175% of the goal, this means they went 75% overboard and added on 375 additional dollars (on top of the 500 they wanted). So they got to 500+375 = 875 which lines up with the instructions. This helps verify the answer.
Or we can see that 1.75*g = 1.75*500 = 875 which helps confirm the answer as well.
Answer:
$200
Step-by-step explanation:
Half of $800 is $400, and a quarter of $800 is $200. $800-$400-$200=$200
Answer:
a) n= 1045 computers
b) n= 442 computers
c) A. Yes, using the additional survey information from part (b) dramatically reduces the sample size.
Step-by-step explanation:
Hello!
The variable of interest is
X: Number of computers that use the new operating system.
You need to find the best sample size to take so that the proportion of computers that use the new operating system can be estimated with a 99% CI and a margin of error no greater than 4%.
The confidence interval for the population proportion is:
p' ±
* 

a) In this item there is no known value for the sample proportion (p') when something like this happens, you have to assume the "worst-case scenario" that is, that the proportion of success and failure of the trial are the same, i.e. p'=q'=0.5
The margin of error of the interval is:
d=
* 



![n= [p'(1-p')]*(\frac{Z_{1-\alpha /2}}{d} )^2](https://tex.z-dn.net/?f=n%3D%20%5Bp%27%281-p%27%29%5D%2A%28%5Cfrac%7BZ_%7B1-%5Calpha%20%2F2%7D%7D%7Bd%7D%20%29%5E2)
![n=[0.5(1-0.5)]*(\frac{2.586}{0.04} )^2= 1044.9056](https://tex.z-dn.net/?f=n%3D%5B0.5%281-0.5%29%5D%2A%28%5Cfrac%7B2.586%7D%7B0.04%7D%20%29%5E2%3D%201044.9056)
n= 1045 computers
b) This time there is a known value for the sample proportion: p'= 0.88, using the same confidence level and required margin of error:
![n= [p'(1-p')]*(\frac{Z_{1-\alpha /2}}{d} )^2](https://tex.z-dn.net/?f=n%3D%20%5Bp%27%281-p%27%29%5D%2A%28%5Cfrac%7BZ_%7B1-%5Calpha%20%2F2%7D%7D%7Bd%7D%20%29%5E2)
![n= [0.88*0.12]*(\frac{{2.586}}{0.04})^2= 441.3681](https://tex.z-dn.net/?f=n%3D%20%5B0.88%2A0.12%5D%2A%28%5Cfrac%7B%7B2.586%7D%7D%7B0.04%7D%29%5E2%3D%20441.3681)
n= 442 computers
c) The additional information in part b affected the required sample size, it was drastically decreased in comparison with the sample size calculated in a).
I hope it helps!