I don’t know, what did he say?
Let X be a discrete random variable with geometric distribution.
Let x be the number of tests and p the probability of success in each trial, then the probability distribution is:
P (X = x) = p * (1-p) ^ (x-1). With x = (1, 2, 3 ... n).
This function measures the probability P of obtaining the first success at the x attempt.
We need to know the probability of obtaining the first success at the third trial.
Where a success is defined as a customer buying online.
The probability of success in each trial is p = 0.3.
So:
P (X = 3) = 0.3 * (1-0.3) ^ (3-1)
P (X = 3) = 0.147
The probability of obtaining the first success at the third trial is 14.7%
Answer:
Step-by-step explanation:
If you subtract the extra 12 quarters from the total, you have $4.55 that can be made up of groups of 1 quarter and 1 dime. Since those groups are worth $0.35, there are $4.55/$0.35 = 13 of them.
There are 13 dimes and 13+12 = 25 quarters in the purse.
Answer:
I think the better deal is $8.78.
Step-by-step explanation:
I think this because half of twenty is 10, and half of 12 is 6. The twelve pack price is closer to one dollar a can, and the 20 pack's is further. :) i hope this helps you!
(plz brainliest! :) )
Solution :
Given :

1. Cumulative distribution function




![$=\frac{1}{450}\left[\frac{x^3}{3}-\frac{x^2}{2}\right]_6^x$](https://tex.z-dn.net/?f=%24%3D%5Cfrac%7B1%7D%7B450%7D%5Cleft%5B%5Cfrac%7Bx%5E3%7D%7B3%7D-%5Cfrac%7Bx%5E2%7D%7B2%7D%5Cright%5D_6%5Ex%24)
![$=\frac{1}{450}\left[ \left( \frac{x^3}{3} - \frac{x^2}{2}\left) - \left( \frac{6^3}{3} - \frac{6^2}{2} \right) \right] $](https://tex.z-dn.net/?f=%24%3D%5Cfrac%7B1%7D%7B450%7D%5Cleft%5B%20%5Cleft%28%20%5Cfrac%7Bx%5E3%7D%7B3%7D%20-%20%5Cfrac%7Bx%5E2%7D%7B2%7D%5Cleft%29%20-%20%5Cleft%28%20%5Cfrac%7B6%5E3%7D%7B3%7D%20-%20%5Cfrac%7B6%5E2%7D%7B2%7D%20%5Cright%29%20%5Cright%5D%20%20%24)
![$=\frac{1}{450}\left[\frac{x^3}{3} - \frac{x^2}{2} - 54 \right]$](https://tex.z-dn.net/?f=%24%3D%5Cfrac%7B1%7D%7B450%7D%5Cleft%5B%5Cfrac%7Bx%5E3%7D%7B3%7D%20-%20%5Cfrac%7Bx%5E2%7D%7B2%7D%20-%2054%20%5Cright%5D%24)
2. Mean 


![$=\frac{1}{450} \left[\frac{x^4}{4} - \frac{x^3}{3} \right]_6^{12} \ dx$](https://tex.z-dn.net/?f=%24%3D%5Cfrac%7B1%7D%7B450%7D%20%5Cleft%5B%5Cfrac%7Bx%5E4%7D%7B4%7D%20-%20%5Cfrac%7Bx%5E3%7D%7B3%7D%20%5Cright%5D_6%5E%7B12%7D%20%5C%20dx%24)

![$=\frac{1}{450} [4608 - 252]$](https://tex.z-dn.net/?f=%24%3D%5Cfrac%7B1%7D%7B450%7D%20%5B4608%20-%20252%5D%24)
= 17.2857