X + 3 = 0
2x - 1 = 0
It would be D.
You can see how the difference between two consecutive terms is constantly increasing:
![a_1 = 4](https://tex.z-dn.net/?f=a_1%20%3D%204)
![a_2 = 7 = 4+3](https://tex.z-dn.net/?f=a_2%20%3D%207%20%3D%204%2B3)
![a_3 = 12 = 7+5](https://tex.z-dn.net/?f=a_3%20%3D%2012%20%3D%207%2B5)
![a_4 = 19 = 12+7](https://tex.z-dn.net/?f=a_4%20%3D%2019%20%3D%2012%2B7)
So, for the next terms we'll have to add +9, +11, +13 and so on.
Also, note that
is obtained by adding the 2nd odd number to
,
is obtained by adding the 3rd odd number to
, and so on.
So, the recursive formula is
![a_n = a_{n-1}+(2n-1),\quad a_1=1](https://tex.z-dn.net/?f=a_n%20%3D%20a_%7Bn-1%7D%2B%282n-1%29%2C%5Cquad%20a_1%3D1)
For the explicit formula, recall that the sum of the first n odd numbers is n squared. Taking into account the fact that we're not starting from 1, we have
![a_n = n^2+3](https://tex.z-dn.net/?f=a_n%20%3D%20n%5E2%2B3)
Least 10 digit whole number = 1,111,111,111
Answer:
#1 is 6 candy bars
#2 i think it's n≥15
#3 x= 19/7
Step-by-step explanation:
Answer:
a
![P(X = 4 ) = 0.1876](https://tex.z-dn.net/?f=P%28X%20%3D%204%20%29%20%3D%200.1876)
b
![P(X \le 4) = 0.8358](https://tex.z-dn.net/?f=P%28X%20%5Cle%204%29%20%20%3D%20%200.8358)
Step-by-step explanation:
From the question we are told that
The proportion that has outstanding balance is p = 0.20
The sample size is n = 15
Given that the properties of the binomial distribution apply, for a randomly selected number(X) of credit card
![X \ \ ~ Bin (n , p )](https://tex.z-dn.net/?f=X%20%5C%20%20%5C%20~%20Bin%20%28n%20%2C%20p%20%29)
Generally the probability of finding 4 customers in a sample of 15 who have "maxed out" their credit cards is mathematically represented as
![P(X = 4 ) = ^nC_4 * p^4 * (1 - p)^{n-4}](https://tex.z-dn.net/?f=P%28X%20%3D%204%20%29%20%3D%20%20%5EnC_4%20%2A%20p%5E4%20%2A%20%281%20-%20p%29%5E%7Bn-4%7D)
=> ![P(X = 4 ) = ^{15}C_4 * (0.20)^4 * (1 - 0.20)^{15-4}](https://tex.z-dn.net/?f=P%28X%20%3D%204%20%29%20%3D%20%20%5E%7B15%7DC_4%20%2A%20%280.20%29%5E4%20%2A%20%281%20-%200.20%29%5E%7B15-4%7D)
Here C stand for combination
=>
Generally the probability that 4 or fewer customers in the sample will have balances at the limit of the credit card is mathematically represented as
![P(X \le 4) = [ ^{15}C_0 * (0.20)^0 * (1 - 0.20)^{15-0}]+[ ^{15}C_1 * (0.20)^1 * (1 - 0.20)^{15-1}]+\cdots+[ ^{15}C_4 * (0.20)^4 * (1 - 0.20)^{15-4}]](https://tex.z-dn.net/?f=P%28X%20%5Cle%204%29%20%3D%20%20%5B%20%5E%7B15%7DC_0%20%2A%20%280.20%29%5E0%20%2A%20%281%20-%200.20%29%5E%7B15-0%7D%5D%2B%5B%20%5E%7B15%7DC_1%20%2A%20%280.20%29%5E1%20%2A%20%281%20-%200.20%29%5E%7B15-1%7D%5D%2B%5Ccdots%2B%5B%20%5E%7B15%7DC_4%20%2A%20%280.20%29%5E4%20%2A%20%281%20-%200.20%29%5E%7B15-4%7D%5D)
=> ![P(X \le 4) = 0.8358](https://tex.z-dn.net/?f=P%28X%20%5Cle%204%29%20%20%3D%20%200.8358)