Answer:
no
Step-by-step explanation:
threr is no 9
Answer:
Step-by-step explanation:
Assuming the number of tickets sales from Mondays is normally distributed. the formula for normal distribution would be applied. It is expressed as
z = (x - u)/s
Where
x = ticket sales from monday
u = mean amount of ticket
s = standard deviation
From the information given,
u = 500 tickets
s = 50 tickets
We want to find the probability that the mean will be greater than 510. It is expressed as
P(x greater than 510) = 1 - P(x lesser than or equal to 510)
For x = 510
z = (510 - 500)/50 = 0.2
Looking at the normal distribution table, the probability corresponding to the z score is 0.9773
P(x greater than 510) = 1 - 0.9773 = 0.0227
Answer:
D. ![xy\sqrt[3]{9y}](https://tex.z-dn.net/?f=xy%5Csqrt%5B3%5D%7B9y%7D)
Step-by-step explanation:
![\sqrt[3]{9x^3y^4}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B9x%5E3y%5E4%7D)
![\sqrt[3]{9}\sqrt[3]{x^3}\sqrt[3]{y^4}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B9%7D%5Csqrt%5B3%5D%7Bx%5E3%7D%5Csqrt%5B3%5D%7By%5E4%7D)
The
cancels out to become x:
![\sqrt[3]{9}x\sqrt[3]{y^4}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B9%7Dx%5Csqrt%5B3%5D%7By%5E4%7D)
Split the 
![\sqrt[3]{9}x\sqrt[3]{y^3}\sqrt[3]{y^1}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B9%7Dx%5Csqrt%5B3%5D%7By%5E3%7D%5Csqrt%5B3%5D%7By%5E1%7D)
![\sqrt[3]{y^3} =y](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7By%5E3%7D%20%3Dy)
![xy\sqrt[3]{9} \sqrt[3]{y}](https://tex.z-dn.net/?f=xy%5Csqrt%5B3%5D%7B9%7D%20%5Csqrt%5B3%5D%7By%7D)
Put the cube root of y and cube root of 9 together:
![xy\sqrt[3]{9y}](https://tex.z-dn.net/?f=xy%5Csqrt%5B3%5D%7B9y%7D)
11%=11/100
11/100=22/x
we have to solve for x now
11x=2200 divide by 11 both side
x=200
that the answer
Answer:
Its when you just swthc your thingy
Step-by-step explanation:
2+4=6
turns into
4+2=6