I would go with football or hallway. If you choose footbal you would probably get a more random demographic but you might not get 200. If you choose hallways you could get 200 so hallway is probably more likely
You and your 4 other friends who rent a house owe $-50 for their water bill.They decide to split the cost of the water bill up amongst all of them.How much wil each of them owe?
Answer:
70.94 mm is the upper control level with a 99.7% level of confidence.
Step-by-step explanation:
We are given the following data:
69, 72, 71, 70, 68
Population mean = 70 mm
Population standard deviation = 1.25 mm
We have to find the upper control level with a 99.7% level of confidence.
![Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}](https://tex.z-dn.net/?f=Mean%20%3D%20%5Cdisplaystyle%5Cfrac%7B%5Ctext%7BSum%20of%20all%20observations%7D%7D%7B%5Ctext%7BTotal%20number%20of%20observation%7D%7D)
![Mean =\displaystyle\frac{350}{5} = 70](https://tex.z-dn.net/?f=Mean%20%3D%5Cdisplaystyle%5Cfrac%7B350%7D%7B5%7D%20%3D%2070)
99.7% Confidence interval:
![\mu \pm z_{critical}\frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=%5Cmu%20%5Cpm%20z_%7Bcritical%7D%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
Putting the values, we get,
![z_{critical}\text{ at}~\alpha_{0.003} = \pm 2.98](https://tex.z-dn.net/?f=z_%7Bcritical%7D%5Ctext%7B%20at%7D~%5Calpha_%7B0.003%7D%20%3D%20%5Cpm%202.98)
![70 \pm 2.98(\frac{1.25}{\sqrt{16}} ) = 70 \pm 0.93125 = (69.06875,70.93125) \approx (69.07,70.94)](https://tex.z-dn.net/?f=70%20%5Cpm%202.98%28%5Cfrac%7B1.25%7D%7B%5Csqrt%7B16%7D%7D%20%29%20%3D%2070%20%5Cpm%200.93125%20%3D%20%2869.06875%2C70.93125%29%20%5Capprox%20%2869.07%2C70.94%29)
Thus, 70.94 mm is the upper control level with a 99.7% level of confidence.
Answer:
$7499.82
Step-by-step explanation:
We have been given that a person places $6340 in an investment account earning an annual rate of 8.4%, compounded continuously. We are asked to find amount of money in the account after 2 years.
We will use continuous compounding formula to solve our given problem as:
, where
A = Final amount after t years,
P = Principal initially invested,
e = base of a natural logarithm,
r = Rate of interest in decimal form.
![8.4\%=\frac{8.4}{100}=\frac{8.4}{100}=0.084](https://tex.z-dn.net/?f=8.4%5C%25%3D%5Cfrac%7B8.4%7D%7B100%7D%3D%5Cfrac%7B8.4%7D%7B100%7D%3D0.084)
Upon substituting our given values in above formula, we will get:
![A=\$6340\cdot e^{0.084\cdot 2}](https://tex.z-dn.net/?f=A%3D%5C%246340%5Ccdot%20e%5E%7B0.084%5Ccdot%202%7D)
![A=\$6340\cdot e^{0.168}](https://tex.z-dn.net/?f=A%3D%5C%246340%5Ccdot%20e%5E%7B0.168%7D)
![A=\$6340\cdot 1.1829366106478107](https://tex.z-dn.net/?f=A%3D%5C%246340%5Ccdot%201.1829366106478107)
![A=\$7499.818111507119838](https://tex.z-dn.net/?f=A%3D%5C%247499.818111507119838)
Upon rounding to nearest cent, we will get:
![A\approx \$7499.82](https://tex.z-dn.net/?f=A%5Capprox%20%5C%247499.82)
Therefore, an amount of $7499.82 will be in account after 2 years.
Because the distance a car travels is directly related to the time it travels, the distance varies directly with the time: d=a*t. "a" would be the constant of variation. Just substituted the given values(300 and 5) into the equation, and you would get 300=5a. Divide both sides by 5 and you get a=60. So, the constant of variation is 60