5x = 1414
x = 1414 / 5
x = 282.80
Complete question:
He amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.3 minutes and standard deviation 1.4 minutes. Suppose that a random sample of n equals 47 customers is observed. Find the probability that the average time waiting in line for these customers is
a) less than 8 minutes
b) between 8 and 9 minutes
c) less than 7.5 minutes
Answer:
a) 0.0708
b) 0.9291
c) 0.0000
Step-by-step explanation:
Given:
n = 47
u = 8.3 mins
s.d = 1.4 mins
a) Less than 8 minutes:

P(X' < 8) = P(Z< - 1.47)
Using the normal distribution table:
NORMSDIST(-1.47)
= 0.0708
b) between 8 and 9 minutes:
P(8< X' <9) =![[\frac{8-8.3}{1.4/ \sqrt{47}}< \frac{X'-u}{s.d/ \sqrt{n}} < \frac{9-8.3}{1.4/ \sqrt{47}}]](https://tex.z-dn.net/?f=%20%5B%5Cfrac%7B8-8.3%7D%7B1.4%2F%20%5Csqrt%7B47%7D%7D%3C%20%5Cfrac%7BX%27-u%7D%7Bs.d%2F%20%5Csqrt%7Bn%7D%7D%20%3C%20%5Cfrac%7B9-8.3%7D%7B1.4%2F%20%5Csqrt%7B47%7D%7D%5D)
= P(-1.47 <Z< 6.366)
= P( Z< 6.366) - P(Z< -1.47)
Using normal distribution table,

0.9999 - 0.0708
= 0.9291
c) Less than 7.5 minutes:
P(X'<7.5) = ![P [Z< \frac{7.5-8.3}{1.4/ \sqrt{47}}]](https://tex.z-dn.net/?f=%20P%20%5BZ%3C%20%5Cfrac%7B7.5-8.3%7D%7B1.4%2F%20%5Csqrt%7B47%7D%7D%5D%20)
P(X' < 7.5) = P(Z< -3.92)
NORMSDIST (-3.92)
= 0.0000
Answer:
fiscal year
Step-by-step explanation:
The fiscal year is the accounting period for the federal government which begins on October 1 and ends on September 30. The fiscal year is designated by the calendar year in which it ends; for example, fiscal year 2013 begins on October 1, 2012 and ends on September 30, 2013.
Your original answer is correct.If you set up a ratio table, it works. For example:
So, they amount of money she's taxed on is $55. We get this by doing 175-20.
$55 is equal to 106% because 6% is tax
Now, you want to find just 100% which is the same as no tax. To get from 106 to 11, you divide by 1.06. So, you also divide 55 by 1.06 to get $51.89
She can't go over this because she won't have enough money.
So, the answer is x≤$51.89
Hope this helps!
By definition of covariance,
![\mathrm{Cov}(X,Y)=\mathbb E[(X-\mathbb E[X])(Y-\mathbb E[Y])]](https://tex.z-dn.net/?f=%5Cmathrm%7BCov%7D%28X%2CY%29%3D%5Cmathbb%20E%5B%28X-%5Cmathbb%20E%5BX%5D%29%28Y-%5Cmathbb%20E%5BY%5D%29%5D)
![\mathrm{Cov}(X,Y)=\mathbb E[XY-\mathbb E[X]Y-X\mathbb E[Y]+\mathbb E[X]\mathbb E[Y]]=\mathbb E[XY]-\mathbb E[X]\mathbb E[Y]](https://tex.z-dn.net/?f=%5Cmathrm%7BCov%7D%28X%2CY%29%3D%5Cmathbb%20E%5BXY-%5Cmathbb%20E%5BX%5DY-X%5Cmathbb%20E%5BY%5D%2B%5Cmathbb%20E%5BX%5D%5Cmathbb%20E%5BY%5D%5D%3D%5Cmathbb%20E%5BXY%5D-%5Cmathbb%20E%5BX%5D%5Cmathbb%20E%5BY%5D)
We have
![\mathbb E[(aX-b)(cY-d)]=\mathbb E[acXY-adX-bcY+bd]](https://tex.z-dn.net/?f=%5Cmathbb%20E%5B%28aX-b%29%28cY-d%29%5D%3D%5Cmathbb%20E%5BacXY-adX-bcY%2Bbd%5D)
![=ac\mathbb E[XY]-ad\mathbb E[X]-bc\mathbb E[Y]+bd](https://tex.z-dn.net/?f=%3Dac%5Cmathbb%20E%5BXY%5D-ad%5Cmathbb%20E%5BX%5D-bc%5Cmathbb%20E%5BY%5D%2Bbd)
![\mathbb E[aX-b]=a\mathbb E[X]-b](https://tex.z-dn.net/?f=%5Cmathbb%20E%5BaX-b%5D%3Da%5Cmathbb%20E%5BX%5D-b)
![\mathbb E[cY-d]=c\mathbb E[Y]-d](https://tex.z-dn.net/?f=%5Cmathbb%20E%5BcY-d%5D%3Dc%5Cmathbb%20E%5BY%5D-d)
![\mathbb E[aX-b]\mathbb E[cY-d]=ac\mathbb E[X]\mathbb E[Y]-ad\mathbb E[X]-bc\mathbb E[Y]+bd](https://tex.z-dn.net/?f=%5Cmathbb%20E%5BaX-b%5D%5Cmathbb%20E%5BcY-d%5D%3Dac%5Cmathbb%20E%5BX%5D%5Cmathbb%20E%5BY%5D-ad%5Cmathbb%20E%5BX%5D-bc%5Cmathbb%20E%5BY%5D%2Bbd)
Putting everything together, we find the covariance reduces to
![\mathrm{Cov}(aX-b,cY-d)=ac(\mathbb E[XY]-\mathbb E[X]\mathbb E[Y])=ac\mathrm{Cov}(X,Y)](https://tex.z-dn.net/?f=%5Cmathrm%7BCov%7D%28aX-b%2CcY-d%29%3Dac%28%5Cmathbb%20E%5BXY%5D-%5Cmathbb%20E%5BX%5D%5Cmathbb%20E%5BY%5D%29%3Dac%5Cmathrm%7BCov%7D%28X%2CY%29)
as desired.