Step-by-step explanation:
You can't expect to get exactly 2500 out of 5000 tosses more than a few times . You will come pretty close, but that's only good in horseshoes.
Of course I'm answering this on the basis of a computer language and not actually performinig this a million tmes, each part of a million consisting of 5000 tosses.
Simulations and not completely unbiased, but based on experience, 5000 is a very small number and getting 2500 more than a couple of times is unlikely
Answer:
![\large\boxed{2.\ ab^{-3x}=a\left(\dfrac{1}{b}\right)^{3x}=a\left[\left(\dfrac{1}{b}\right)^3\right]^x}](https://tex.z-dn.net/?f=%5Clarge%5Cboxed%7B2.%5C%20ab%5E%7B-3x%7D%3Da%5Cleft%28%5Cdfrac%7B1%7D%7Bb%7D%5Cright%29%5E%7B3x%7D%3Da%5Cleft%5B%5Cleft%28%5Cdfrac%7B1%7D%7Bb%7D%5Cright%29%5E3%5Cright%5D%5Ex%7D)
Step-by-step explanation:
![Use:\ a^{-n}=\left(\dfrac{1}{a}\right)^n\\------------\\\\(4)^{-3x^2}=\left[(4)^{-1}\right]^{3x^2}=\left(\dfrac{1}{4}\right)^{3x^2}](https://tex.z-dn.net/?f=Use%3A%5C%20a%5E%7B-n%7D%3D%5Cleft%28%5Cdfrac%7B1%7D%7Ba%7D%5Cright%29%5En%5C%5C------------%5C%5C%5C%5C%284%29%5E%7B-3x%5E2%7D%3D%5Cleft%5B%284%29%5E%7B-1%7D%5Cright%5D%5E%7B3x%5E2%7D%3D%5Cleft%28%5Cdfrac%7B1%7D%7B4%7D%5Cright%29%5E%7B3x%5E2%7D)
![Use:\ a^{-n}=\left(\dfrac{1}{a}\right)^n\ and\ (a^n)^m=a^{nm}\\--------------------\\\\ab^{-3x}=a\cdot b^{-3x}=a\left[(b)^{-1}\right]^{3x}=a\left(\dfrac{1}{b}\right)^{3x}\\\\ab^{-3x}=a\left(\dfrac{1}{b}\right)^{3x}=a\left[\left(\dfrac{1}{b}\right)^3\right]^x](https://tex.z-dn.net/?f=Use%3A%5C%20a%5E%7B-n%7D%3D%5Cleft%28%5Cdfrac%7B1%7D%7Ba%7D%5Cright%29%5En%5C%20and%5C%20%28a%5En%29%5Em%3Da%5E%7Bnm%7D%5C%5C--------------------%5C%5C%5C%5Cab%5E%7B-3x%7D%3Da%5Ccdot%20b%5E%7B-3x%7D%3Da%5Cleft%5B%28b%29%5E%7B-1%7D%5Cright%5D%5E%7B3x%7D%3Da%5Cleft%28%5Cdfrac%7B1%7D%7Bb%7D%5Cright%29%5E%7B3x%7D%5C%5C%5C%5Cab%5E%7B-3x%7D%3Da%5Cleft%28%5Cdfrac%7B1%7D%7Bb%7D%5Cright%29%5E%7B3x%7D%3Da%5Cleft%5B%5Cleft%28%5Cdfrac%7B1%7D%7Bb%7D%5Cright%29%5E3%5Cright%5D%5Ex)
Answer:
A 21 /169
B independent
Step-by-step explanation:
The total number of candies is 7+3+3 = 13
P( 1st red) = red/total = 7/13
Since the candy is put back there are still 13 candies in the dish
P(2nd green) = green/total = 3/13
P( red, then green) = P( 1st red) * P(2nd green)
= 7/13* 3/13 = 21/169
The events are independent since the candies are put back
0.6/10^-2
move the decimal 2 places to the right
0.6 becomes 60
answer is 60
